ae eed”

Bem Gel Rim ~

 A SCHOOL GEOMETRY.
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A SCHOOL GEOMETRY
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 A
SCHOOL GEOMETRY
PARIS, ‘heVik

(Containing Plane and Solid Geometry, treated both

theoretically and graphically)

BY

PSs EAS MA
AND

F. H. STEVENS, M.A.

MWACMIELAN AND CO. LIMITED

ST. MARTINS’ STREET, LONDON
1920
 COPYRIGHT.

First Edition 1908.

Reprinted April and November 1904, 1905, 1906, 1907, 1908, 1909,
January and September 1910,
1911, 1912, 1913, 1914, 1916, 1917, 1918, 1919 (twice), 1920.

GLASGOW: PRINTED AT THK UNIVERSITY PRESS

BY ROBERT MACLEHOSE AND CO, LTD,
 PREFACE.
THE present work provides a course of Elementary Geometry
based on the recommendations of the Mathematical Association
and on the schedule recently proposed and adopted at Cambridge.

The principles which governed these proposals have been

confirmed by the issue of revised schedules for all the more
important Examinations, and they are now so generally accepted
by teachers that they need no discussion here. It is enough to
note the following points :
(i) We agree that a pupil should gain his first geometrical ideas
from a short preliminary course of a practical and experimental
character. A suitable introduction to the present book would
consist of Easy Exercises in Drawing to illustrate the subject
matter of the Definitions; Measurements of Lines and Angles ;
Use of Compasses and Protractor; Problems on Bisection, Per-
pendiculars, and Parallels; Use of Set Squares; The Construction
of Triangles and Quadrilaterals. These problems should be
accompanied by informal explanation, and the results verified
by measurement. Concurrently, there should be a series of ex-
ercises in Drawing and Measurement designed to lead inductively
to the more important Theorems of Part I. [Euc. I. 1-34].* While
strongly advocating some such introductory lessons, we may point
out that our book, as far as it goes, is complete in itself, and from
the first is illustrated by numerical and graphical examples of the
easiest types. Thus, throughout the whole work, a graphical
and experimental course is provided side by side with the usual
deductive exercises.
(ii) Theorems and Problems are arranged in separate but parallel
courses, intended to be studied pari passu. This arrangement is
made possible by the use, now generally sanctioned, of Hypothetical
Constructions. ‘These, before being employed in the text, are care-
fully specified, and referred to the Axioms on which they depend.
* Such an introductory course is now furnished by our Lessons in Experi:
mental and Practical Geometry.
vi PREFACK.

(iii) The subject is placed on the basis of Commensurable Mag-

nitudes. By this means, certain difficulties which are wholly
beyond the grasp of a young learner are postponed, and a wide
field of graphical and numerical illustration is opened. Moreover
the fundamental Theorems on Areas (hardly less than those on
Proportion) may thus be reduced in number, greatly simplified,
and brought into line with practical applications.
(iv) An attempt has been made to curtail the excessive body of
text which the demands of Examinations have hitherto forced as
“bookwork” on a beginner’s memory. Even of the Theorems
here given a certain number (which we have distinguished with
an asterisk) might be omitted or postponed at the discretion of the
teacher. And the formal propositions for which—as such—teacher
and pupil are held responsible, might perhaps be still further
limited to those which make the landmarks of Elementary Geo-
metry. Time so gained should be used in getting the pupil to
apply his knowledge; and the working of examples should be
made as important a part of a lesson in Geometry as it is so
considered in Arithmetic and Algebra,
Though we have not always followed Euclid’s order of Proposi-
tions, we think it desirable for the present, in regard to the
subject-matter of Euclid Book I. to preserve the essentials of his
logical sequence. Our departure from Euclid’s treatment of Areas
has already been mentioned ; the only other important divergence
in this section of the work is the position of I. 26 (Theorem 17),
which we place after I. 32 (Theorem 16), thus getting rid of the
tedious and uninstructive Second Case. In subsequent Parts a freer
treatment in respect of logical order has been followed,
As regards the presentment of the propositions, we have con-
stantly kept in mind the needs of that large class of students, who,
without special aptitude for mathematical study, and under no
necessity for acquiring technical knowledge, may and do derive
real intellectual advantage from lessons in pure deductive reasoning.
Nothing has as yet been devised as effective for this purpose as the
Euclidean form of proof; and in our opinion no excuse is needed
for treating the earlier proofs with that fulness which we have
always found necessary in our experience as teachers,
 PREFACE. Vii

The examples are numerous and for the most part easy. They
have been very carefully arranged, and are distributed throughout
the text in immediate connection with the propositions on which
they depend. A special feature is the large number of examples
involving graphical or numerical work. The answers to these
have been printed on perforated pages, so that they may easily be
removed if it is found that access to numerical results is a source
of temptation in examples involving measurement.
We are indebted to several friends for advice and suggestions.
In particular we wish to express our thanks to Mr. H. C. Playne
and Mr. H. C. Beaven of Clifton College for the valuable assist-
ance they have rendered in reading the proof sheets and checking.
the answers to some of the numerical exercises.
H. 8S. HALL
F. H. STEVENS.
November, 1903.

PREFATORY NOTE TO THE SECOND EDITION.

In the present edition some further steps have been taken towards |
the curtailment of bookwork by reducing certain less important
propositions (e.g. Euclid T. 22, 43, 44) to the rank of exercises.
Room has thus been. found for more numerical and graphical
exercises, and experimental work such as that leading to the
Theorem of Pythagoras.
Theorem 22 (page 62), in the shape recommended in the Cam-
bridge Schedule, replaces the equivalent proposition given as
Additional Theorem A (page 60) in previous editions.
In the case of a few problems (e.g. Problems 23, 28, 29) it has
been thought more instructive to justify the construction by a pre-
liminary analysis than by the usual formal proof.’
H. 8. HALL.
FE. H. STEVENS.
March, 1904.
 CONTENTS.

PAR ie
PAaGh

Axioms. Definitions. Postulates. - . - ° -

HyYporueticaL CONSTRUCTIONS - - , - :

INTRODUCTORY - = : c = 3 : : é ‘
SyMpoLs AND ABBREVIATIONS - . : £ 3 s

Lines and Angles,

THEOREM 1. [Euic. 1. 13.] The adjacent angles which one
straight line makes with another straight line on one side
of it are together equal to two right angles.
Cor. 1. If two straight lines cut one another, the four
angles so formed are together equal to foyr right angles.
Cor. 2. When any number of straight lines meet at a
point, the sum of the consecutive angles so formed is equal
to four right angles.
Cor. 3. (i) Supplements of the same angle are equal.
(ii) Complements of the same angle are equal.
Theorem 2, [Eue. I. 14.] If, at a point in a straight line,
two other straight lines, on opposite sides of it, make the
adjacent angles together equal to two right angles, then
these two straight lines are in one and the same straight line. - 12
TreorREM 3. [Iuc. I. 15.] If two straight lines cut one
another, the vertically opposite angles are equal. 14

Triangles.
DEFINITIONS Sede - - - - : - - - 16
Ti£ CoMPARISON OF Two TRIANGLES ~~ - . - - - 17
TuroremM 4. [Euc. I. 4.] If two triangles have two sides
of the one equal to two sides of the other, cach to each, and
the angles included by those sides equal, then the triangles
are equal in all respects. 18
H.8.G..L.-V1. h
x CONTENTS.

THrorEM 5. [Euc. I. 5.] The angles at the base of an isosceles

triangle are equal.
Cor. 1. If the equal sides of an isosceles triangle are pro-
duced, the exterior angles at the base are equal.
Cor. 2. Ifa triangle is equilateral, it is also equiangular.
THEorEM 6. [Euc. I. 6.] If two angles of a triangle are equal
to one another, then the sides which are opposite to the equal
angles are equal to one another.
TuEoRrEM 7. [Euc. I. 8.] If two triangles have the three sides
of the one equal to the three sides of the other, each to each,
they are equal in all respects.
TnrorREM 8. [Euc. I. 16.] If one side of a triangle is pro-
duced, then the exterior angle is greater than either of the
interior opposite angles.
Cor. 1. Any two angles of a triangle are together less
than two right angles.
Cor. 2. Every triangle must have at least two acute
angles.
Cor. 3. Only one perpendicular can be drawn to a
straight line from a given point outside it.
TueoreM 9, [Euc. I. 18.] If one side of a triangle is greater
than another, then the angle opposite to the greater side is
greater than the angle opposite to the less.
Turorem 10. [Euc. I. 19.] If one angle of a triangle is
greater than another, then the side opposite to the greater
angle is greater than the side opposite to the less.
THeroreM 11, [Euce. I. 20.] Any two sides of a triangle are
together greater than the third side.
TaHeoreM 12. Of all straight lines from a given point to a
given straight line the perpendicular is the least.
Cor. 1. If OC is the shortest straight line from O to the
straight line AB, then OC is perpendicular to AB.
Cor. 2. Two obliques OP, OQ, which cut AB at equal
distances from C the foot of the perpendicular, are equal.
Cor. 3. Of two obliques OQ, OR, if OR cuts AB at the
sreater distance from C the foot of the perpendicular, then
R is greater than OQ,

Parallels.
PLAYFATR’S AXIOM” -
THEOREM 13. [KEue. I. 27 and 28.] If a straight line cuts two
other straight lines so as te make (i) the alternate angles
 CONTENTS.

equal, or (ii) am exterior angle equal to the interior opposite

angle on the same side of the cutting line, or (iii) the interior
angles on the same side equal to two right angles; then in
_ each case the two straight lines are parallel. 36
THEOREM 14. [Euce. I. 29.] Ifa straight line cuts two parallel
lines, it makes (i) the alternate angles equal to one another ;
(ii) the exterior angle equal to the interior opposite angle on
the same side of the cutting line ; (iii) the two interior angles
on the same side together equal to two right angles. 38
PARALLELS ILLUSTRATED BY RoTATIOoN. HyporuEtTicaL Con-
STRUCTION = = > A = é e Z ‘ : 39
THEorEM 15. [Eue. I. 30.] Straight lines which are parallel
to the same straight line are parallel to one another. 40

Triangles continued.
THEOREM 16. (uc. I. 32.] The three angles of a triangle
are together equal to two right angles. 42
Cor. 1. All the interior angles of any rectilineal figure,
together with four right angles, are equal to twice as many
right angles as the figure has sides. 44
Cor. 2. If the sides of a rectilineal figure, which has no
re-entrant angle, are produced in ordex, then all the exterior
angles so formed are together equal to four right angles, 46
THEOREM 17. [Euc. I. 26.] If two triangles have two angles
of one equal to two angles of the other, each to each, and any
side of the first equal to the corresponding side of the other,
the triangles are equal in all respects.
On THE IDENTICAL EQUALITY oF TRIANGLES - - 2
THEOREM 18. Two right-angled triangles which have their
hypotenuses equal, and one side of one equal to one side of
the other, are equal in all respects.
THEOREM 19. [Euc. I. 24.] If two triangles have two sides of
the one equal to two sides of the other, each to each, but the
angle included by the two sides of one greater than the angle
included by the corresponding sides of the other; then the
base of that which has the greater angle is greater than the
base of the other. 52
CoNVERSE oF THEOREM 19 - = - : = ¥ 3 53

Parallelograms.
DEFINITIONS - -~ - oe ae : : : eee 56
THEOREM 20. [Euc. I. 33.] The straight lines which join the
extremities of two equal and parallel straight. lines towards
the same parts are themselves equal and parallel. 57
xi CONTENTS.

PaGz

THEOREM 2], [Euc. I. 34.] The opposite sides and angles of a

_ parallelogram are equal to one another, and each diagonal
isects the parallelogram. 58
Cor. 1. If one angle of a parallelogram is a right angle, ll
its angles are right angles. 59
Cor. 2. All the sides of a square are equal; and cll its
angles are right angles. 59
Cor. 3. The diagonals of a parallelogram bisect one
another. 59
THEOREM 22, If there are three or more parallel straight lines,
and the intercepts made by them on any transversal are equal,
then the corresponding intercepts on any other transversal
are also equal. 62
Cor. Ina triangle ABC, it a set of lines Pp, Qq, Rr,...,
drawn parallel to the base, divide one side AB into equal parts,
they also divide the other side AC into equal parts. 63
DIAGONAL SCALES” - - - - : E ; “ - 66

Practical Geometry. Problems.

{xtRopucTION. Necessary INSTRUMENTS - . - - 69
PROBLEMS ON LINES AND ANGLES.
ProsieM 1. ‘To bisect a given angle. 70
ProsieM 2. ‘To bisect a given straight line. 71
ProsieM 3. To draw a straight line perpendicular to a given
straight line at a given point in it. 72
Prosiem 4. ‘To draw a straight line perpendicular to a given
straight line from a given external point. 74
Prosiem 5, At a given point in a given straight line to make
an angle equal to a given angle. 76
Prostem 6. Through a given point to draw a straight line
parallel to a given straight line. 77
Prostem 7. To divide a given straight line into any number
of equal parts. 78
Tue Construction oF TRIANGLES.
Prosiem 8. To draw a triangle, having given the lengths of
the three sides, 80
ProsteM 9. ‘To construct a triangle having given two sides
and an angle opposite to one of them. 82
ProsteM 10. To construct a right-angled triangle having given
the hypotenuse and one side. 83
Tie CONSTRUCTION OF QUADRILATERALS.
Propiem 11. To construct a quadrilateral, given the lengths
of the four sides, and one Pry
 CONTENTS. xii

PAGE

PRosBLEM 12, To construct a parallelogram haying given two

adjacent sides and the included angle.
ProsuLeM 13. To construct a square on a given side.

Loci.
ProsLeM 14. To find the locus of a point P which moves so
that its distances from two fixed points A and B are always
equal to one another.
ProsiEM 15. To find the locus of a point P which moves so
that its perpendicular distances from two given straight lines
AB, CD are equal to one another.
INTERSECTION OF LOCI : : < : s E Z -
THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.
I. The perpendiculars drawn to the sides of a triangle from
their middle points are concurrent.
II. The bisectors of the angles of a triangle are concurrent.
III. The medians of a triangle are’ concurrent.
Cor. The three medians of a triangle cut one another at a
point of trisection, the greater segment in each being towards
the angular point.

EA gil:
Areas.
DEFINITIONS : - - - - - - : : : 99
THEOREM 23. AREA OF A RECTANGLE. 100
TuEoREM 24. [Euc. I. 35.] Parallelograms on the same base
and between the same parallels are equal in area. 104
AREA OF A PARALLELOGRAM : - - - - - - 105
THEOREM 25. AREA OF A TRIANGLE. 106
THEOREM 26. [Iuc. I. 37.]- Triangles on the same base and
between the same parallels (hence, of the same altitude) are
equal in area. 108
THEOREM 27. [Euc. I. 39.] If two triangles are equal in area,
and stand on the same base and on the same side of it, they
are between the same parallels. 108
THEOREM 28. AR#HA OF (i) A TRAPEZIUM. 112
(ii) ANY QUADRILATERAL. 112
AREA OF ANY RECTILINEAL FIGURE - - - - - - 114
TurorREM 29. [Euc. I. 47. PyruHacoras’s TuEorEM.] In a
right-angled triangle the square described on the hypotenuse
is equal to the sum of the squares described on the other two
sides. 118
xiv CONTENTS.
PAGE
EXPERIMENTAL PRoors oF PyTHAGORAS’s THEOREM - - 120
THEOREM 30. [Euce. I. 48.] If the square described on one side
of a triangle is equal to the sum of the squares described on
the other two sides, then the angle contained by these two
sides is a right angle. 122
ProsLEM 16. To draw squares whose areas shall be respectively
twice, three-times, four-times, ..., that of a given square. 124

Problems on Areas.
ProsteM 17. To describe a parallelogram equal to a given
triangle, and having one of its angles equal to a given angle.
ProsLtEM 18. To draw a triangle equal in area to a given
quadrilateral.
ProspieM 19. To draw a parallelogram equal in area to a given
rectilineal figure, and having an angle equal to a given angle.

Axes of Reference. Coordinates.

EXERCISES FOR SQUARED PAPER : . : - - .

PAR? TIL

The Circle. Definitions and First Principles. - - -

Symmetry. SYMMETRICAL PROPERTIES oF CIRCLES - -

Chords.
TuroreM 31. [Euc. III. 3.] If a straight line drawn from
the centre of a circle bisects a chord which does not pass
through the centre, it cuts the chord at right angles.
Conversely, if it cuts the chord at right angles, it bisects it.
Cor. 1. The straight line which bisects a chord at right
angles passes through the centre.
Cor. 2. A straight line cannot meet a circle at more than
two points.
Cor. 3. A chord of a circle lies wholly within it.
TuroREM 32. One circle, and only one, can pass through any
three points not in the same straight line.
Cor. 1. The size and position of a circle are fully deter-
mined if it is known to pass through three given points.
Cor. 2. Two circles cannot cut one another in more than
two points without coinciding entirely.
HyYpoTuHetTicaAL CoNSTRUCTION. - « “ ‘ m ‘ 3
 CONTENTS, XV

PAGE

THEOREM 33. [Euc. III. 9.] If from a point within a circle

more than two equal straight lines can be drawn to the
circumference, that point is the centre of the circle. 148
THEOREM 34. [Euc. III. 14.] Equal chords of a circle are equi-
distant from the centre.
Conversely, chords which are equidistant from the centre
are equal. 150
THEoREM 35. [Euce. III. 15.] Of any two chords of a circle,
that which is nearer to the centre is greater than one more
remote.
Conversely, the greater of two chords is nearer to the
centre than the less. 152
Cor. The greatest chord in a circle is a diameter. 153
THEOREM 36. [Eue. III. 7.] If from any internal point, not
the centre, straight lines are drawn to the circumference of a
circle, then the greatest is that which passes through the
centre, and the least is the remaining part of that diameter.
And of any other two such lines the greater is that which
subtends the greater angle at the centre. 154
THEOREM 37. [Euc. III. 8.] If from any external point
straight lines are drawn to the circumference of a circle, the
greatest is that which passes through the centre, and the
least is that which when produced passes through the centre.
And of any other two such lines, the greater is that which
subtends the greater angle at the centre. 156
4

Angles in a Circle.
THEOREM 38. [Euc. III. 20.] The angle at the centre of a
circle is double of an angle at the circumference standing on
the same are. 158
THEOREM 39. [Euc. III. 21.] Angles in the same segment of a
circle are equal. 160
ConVERSE OF THEOREM 39. Equal angles standing on the
same base, and on the same side of it, have their vertices on
an arc of a circle, of which the given base is the chord. 161
TurorEM 40. [Euc. III. 22.] The opposite angles of any
quadrilateral inscribed in a circle are together equal to two
right angles. 162
CoNVERSE oF THEOREM 40. If a pair of opposite angles
of a quadrilateral are supplementary, its vertices are con-
cyclic. 163
TurorEM 41. [Euc. III. 3l.] The angle in a semi-circle is a
right angle. 164
Cor. The angle in a segment greater than a semi circle is
acute; and the angle in a segment less than a semi-circle is
obtuse. : 165
Xvi CONTENTS.

PAGE
THeoREM 42. [Eue. III. 26.] In equal circles, ares which sub-
tend equal angles, either at the centres or at the cireum-
ferences, are equal. 166
Cor. In equal circles sectors which have equal angles are
equal. 166
THEOREM 43. [Euc. IIT. 27.] In equal circles angles, either at
the centres or at the circumferences, which stand on equal
ares are equal. 167
TuroreM 44. [KEuc. III. 28.] In equal circles, ares which are
cut off by equal chords are equal, the major are equal to the
major are, and the minor to the minor. 168
TuHeorEeM 45, [Euc. III. 29.] In equal circles chords which
cut off equal ares are equal.

Tangency.
DEFINITIONS AND Frrst PRINCIPLES - - - - - -
TrrorEM 46, The tangent at any point of a circle is perpendi-
cular to the radius drawn to the point of contact.
Cor. 1. One and only one tangent can be drawn to a
circle at a given point on the circumference.
Cor. 2. The perpendicular to a tangent at its point of
contact passes through the centre.
Cor. 3. The radius drawn perpendicular to the tangent
passes through the point of contact.
THEoREM 47. Two tangents can be‘drawn to a circle from an
external point.
Cor. The two tangents to a circle from an external point
are equal, and subtend equal angles at the centre.
TrroreM 48. If two circles touch one another, the centres and
the point of contact are in one straight line.
Cor. 1. If two circles touch externally the distance be-
tween their centres is equal to the sum of their radii.
Cor. 2. If two circles touch internally, the distance be-
tween their centres is equal to the difference of their radii.
TuroreM 49. [Euc. ITI. 32.] The angles made by a tangent
to a circle with a chord drawn from the point of contact are
respectively equal to the angles in the alternate segments of
the circle.

Problems.
GEOMETRICAL ANALYSIS - : ‘ - s 5 .
Propiem 20. Given a circle, or an are of a circle, to find its
centre.
Prosiem 21. To bisect a given are,
 CONTENTS. XVli

PAGE

ProsLeM 22. To draw a tangent toa circle from a given ex-

ternal point. 184
PROBLEM 23. To draw a common tangent to two circles. 185
THE CONSTRUCTION OF CIRCLES- : - - - - - 188
PROBLEM 24. Ona given straight line to describe a segment of
a circle which shall contain an angle equal to a given angle. 190
Cor. To cut off from a given circle a segment containing
a given angle, it is enough to draw a tangent to the circle,
and from the point of contact to draw a chord making with
the tangent an angle equal to the given angle. 191

Circles in Relation to Rectilineal Figures.

DEFINITIONS - - - : - - - - - - 192
PrRoBLEM 25. To circumscribe a circle about a given triangle. 193
ProsieM 26. To inscribe a circle in a given triangle. 194
PrRoBLEM 27. To draw an escribed circle of a given triangle. 195
PrRosieM 28. In a given circle to inscribe a triangle equi-
angular to a given triangle. 196
PresieM 29. About a given circle to cireumscribe a triangle
ey liangular to a given triangle. 197
ProsiEeM 30. To draw a regular polygon (i) in (ii) about a
given circle. 200
ProsiEeM 31, To draw a circle (i) in (ii) about a regular polygon. 201

Circumference and Area of a Circle - - - : = 202

Theorems and Examples on Circles and Eaneles,

THE ORTIHOCENTRE OF A TRIANGLE - - - : 2 207
Locr - = “ = 210
Srmson’s LINE 5 : 212
THE TRIANGLE AND ITS CIRCLES : 3 x < : : 213
Tue Ntine-Pornts CIRcLe - “ : 4 z ‘ Z 216

PAde LY,

Geometrical Equivalents of some Algebraical Formule.

DEFINITIONS : 2 P = : : : : P 3
THeorEM 50. [Euc. II. 1.] If of two straight lines, one is
divided into any number of parts, the rectangle contained by
XVill CONTENTS.

PAGE
the two lines is equal to the sum of the rectangles con-
tained by the undivided line and the several parts of the
divided line.
Corotuaries. [Eue. II. 2and3.]_ - - ss : 3
TuHeEorEM 51. [Euc. TH. 4.] If a straight line is divided in-
ternally at any point, the square on the given line is equal to
the sum of the squares on the two segments together with
twice the rectangle contained by the segments.
THEOREM 52. [Euce. II. 7.] If a straight line is divided
externally at any point, the square on the given line is equal
to the sum of the squares on the two segments diminished by
twice the rectangle contained by the segments.
TueoreM 53. [Euce. II. 5 and6.] The difference of the squares
on the two straight lines is equal to the rectangle contained
by their sum and difference.
Cor. If a straight line is bisected, and also divided (inter-
nally or externally) into two unequal segments, the rectangle
contained by these segments is equal to the difference of the
squares on half the line and on the line between the points of
section.
TueoreM 54, [Eue. II. 12.] In an obtuse-angled triangle, the
square on the side subtending the obtuse angle is equal to the
sum of the squares on the sides containing the obtuse angle
together with twice the rectangle contained by one of those
sides and the projection of the other side upon it.
TurorEM 55. [Euc. II. 13.] In every triangle the square on
the side subtending an acute angle is equal to the sum of the
squares on the ite containing that angle diminished by
twice the rectangle contained by one of those sides and the
projection of the other side upon it.
Tueorem 56. In any triangle the sum of the squares on two
sides is equal to twice the square on half the third side
together with twice the square on the median which bisects
the third side.

Rectangles in connection with Circles.

THroremM 57. [Euc. III. 35.] If two chords of a circle cut at a
point within it, the rectangles contained by their segments
are equal, 232
TuHEoREM 58. [Euc. III. 36.] If two chords of a circle, when
roduced, cut at a point outside it, the rectangles contained
te their segments are equal. And each rectangle is equal to
the square on the tangent from the point of intersection.
 CONTENTS.

TuroreM 59. [Eue. III. 37.] If from a point outside a circle

two straight lines are drawn, one of which cuts the circle, and
the other meets it; and if the rectangle contained by the
whole line-which cuts the circle and the part of it outside the
circle is equal to the square on the line which meets the circle,
then the hne which meets the circle is a tangent to it. 234

Problems.
ProspieM -32. To draw a square equal in area to a given
rectangle. 238
ProsieM 33. To divide a given straight line so that the
rectangle contained by the whole and one part may be equal
to the square on the other part. 240
Progsiem 34. To draw an isosceles triangle having each of the
angles at the base double of the vertical angle. 242
THE GRAPHICAL SOLUTION OF QUADRATIC EQUATIONS” - 244

PART V.
Proportion.
DEFINITIONS AND FIRST PRINCIPLES - - - - 247
IntRoructory Turorems I.-VI. - - - - - : 249

Proportional Division of Straight Lines.

TrEorEeM 60. [Euc. VI. 2.] A straight line drawn parallel to
one side of a triangle cuts the other two sides, or those sides
produced proportionally. 254
TuEoREM 61. [Euc. VI. 3 and A.] If the vertical angle of a
triangle is bisected internally or externally, the bisector
divides the base internally or externally into segments which
have the same ratio as the other sides of the triangle.
Conversely, if the base is divided internally or externally
into segments proportional to the other sides of the triangle,
the line joining the point of section to the vertex bisects
the vertical angle internally or externally. 256

Equiangular Triangles.
TurorEeM 62. [Euc. VI. 4.] If two triangles are equiangular
to one another, their corresponding sides are proportional 260
xx CONTENTS.

PAGE
THEOREM 63. [Euc. VI. 5.] If two triangles have their sides
proportional when taken in order, the triangles are equi-
angular to one another, and those angles are equal which are
opposite to corresponding sides. 261
THEOREM 64. [Euc. VI. 6.] If two triangles have one angle
of the one equal to one angle of the other, and the sides about
the equal angles proportionals, the triangles are similar. 265
THEOREM 65. [Euc. VI. 7.] If two triangles have one angle
of the one equal to one angle of the other, and the sides about
another angle in one proportional to the corresponding sides of
the other, then the third angles are either equal or supple-
mentary ; and in the former case the triangles are similar. 266
THEOREM 66. [Euc. VI. 8.] In a right-angled triangle, if a
perpendicular is drawn from the right angle to the hypo-
tenuse, the triangles on each side of it are similar to the
whole triangle and to one another.
Tuer TRIGONOMETRICAL RATIOS - - - - - -
GEOMETRICAL RESULTS IN TRIGONOMETRICAL FoRM . -

Problems.
Prosiem 35. To find the fourth proportional to three given
straight lines.
PrositEemM 36. To find the third proportional to two given
straight lines.
Prosiem 37. To divide a given straight line internally and
externally in a given ratio.
ProsiEeM 38. To find the mean proportional between two given
straight lines.

Similar Figures. ;
THEOREM 67. Similar polygons can be divided into the same
number of similar triangles ; and the lines joining correspond-
ing vertices in each figure are proportional. 280
PROBLEM 39. On a side of given length to draw a figure similar
to a given rectilineal figure. 281
THEOREM 68. Any two similar rectilineal figures may be so
placed that the lines joining corresponding vertices are con-
current.
TuroreM 69. [Euc. VI. 33.] In equal circles, angles, whether
at the centres or circumferences, have the same ratio as the
ares on which they stand.
 CONTENTS. xxi

PAGE
Proportion Applied to Areas.
THEOREM 70. [Euc. VI. 1.] The areas of triangles of equal
altitude are to one another as their bases. 286
JHEOREM 71. If two triangles have one angle of the one equal
to one angle of the other, their areas are proportional to the
rectangles contained by the sides about the equal angles. 288
THEOREM 72 [Euc. VI. 19.] The areas of similar triangles are
proportional to the squares on corresponding sides. 290
THEOREM 73. [Euc. VI. 20.] The areas of similar polygons
are proportional to the squares on corresponding sides, 292
THEoREM 74. [KEuc. VI. 31.] In a right-angled triangle, any
rectilineal figure described on the hypotenuse is equal to the
sum of the two similar and similarly described figures on the
sides containing the right angle. 296
PrositeM 40. To draw a figure similar toa given rectilineal
figure, and equal to a given fraction of it in area. 298

Rectangles in Connection with Circles,

TreEoREM 75. [Euc. III. 35 and 36.] If any two chords of a
circle cut one another internally or externally, the rectangle
contained by the segments of one is equal to the rectangle
contained by the segments of the other. 300
Cor. If from an external point a secant and a tangent are
drawn to a circle, the rectangle contained by the whole
secant and the part of it outside the circle is equal to the
square on the tangent. 301
THEOREM 76. If the vertical angle of a triangle is bisected by
a straight line which cuts the base, the rectangle contained
by the sides of the triangle is equal to the rectangle contained
by the segments of the base, together with the square on the
straight line which bisects the angle. 302
THEOREM 77. If from the vertical angle of a triangle a straight
line is drawn perpendicular to the base, the rectangle con-
tained by the sides of the triangle is equal to the rectangle
contained by the perpendicular and the diameter of the
circum-circle. 303
THrEoREM 78. [Ptolemy’s Theorem.] The rectangle contained
by the diagonals of a quadrilateral inscribed in a circle is
equal to the sum of the two rectangles contained by its
opposite sides. 304

Miscellaneous Theorems and Examples.

Somz CoNSTRUCTIONS OF CIRCLES” - : = é : 3)1
Maxima AND MINIMA - - - s : “ 314
xxil CONTENTS.

PAGE
GRAPHS. APPLICATION TO MAXIMA AND MINIMA - - - 319
Harmonic SECTION - 3 = F + 4 . E 7 323
CENTRES OF SIMILITUDE 2 : : 3 a 328
PoLE AND FoLaR - - : : = : ‘ b 331
THe RapicaL Axis - - : : = F : : 336
INVERSION . : : : : : E = ; : 340
Crva’s THEOREM - = : : : : _ 344
MENELAUS’ THEOREM - : : - é : 3 345

PART VF.
Lines and Planes.
DEFINITIONS AND First PRINCIPLES - = ‘ = : :
Turorem 79. [Euc. XI. 2.] One, and only one, plane may be
made to pass through any two intersecting straight lines.
Cor. Any three straight lines, of which each pair cut one
another, must be co-planar,
TuroreM 80. [Euc. XI. 3.] Two intersecting planes cut one
another in a straight line, and in no point outside it.
TuroreM 81. [Euc. XI. 4.] Ifa straight line is perpendicular
to each of two intersecting straight lines at their point of
intersection, it is also perpendicular to the plane in which
they lie.
TaroremM 82. [Euc. XI. 5.} All straight lines drawn perpen-
dicular to a given straight line at a given point are co-
planar.
Cor. If a right angle revolves about one of its arms, the
other arm generates a plane.
HyporHeticAL CoNSTRUCTION - - - - . -

THEOREM 83, [Euc. XI. 8.] If two straight lines are parallel,
and if one of them is perpendicular to a plane, then the other
is also perpendicular to the same plane. 356
Cor. If AB is perpendicular to a plane XY, and if from B,
the foot of the perpendicular, a line BE is drawn perpen-
dicular to any line CD in the plane, then the join AE is also
perpendicular to CD. 357
TnroreM 84. Through any point in or outside a plane there
can always be one, and only one, straight line perpendicular
to the plane.
 CONTENTS. Xxiit
PAGE

THEOREM 85. (i) Of all straight lines drawn from an external

point to a plane, the perpendicular is the shortest.
(ii) Of obliques drawn from the given point, those which
cut the plane at equal distances from the foot of the perpen-
dicular are equal. 360
TmEoREM 86. [Euc. XI. 9.] Straight lines which are parallel
to a given straight line are parallel to one another. 362
THEOREM 87. [Euc. XI. 10.] If two intersecting straight lines
are respectively parallel to two other intersecting straight
lines not in the same plane with them, then the first pair and
the second pair contain equal angles. 363
THEOREM 88. [Euc. XI. 14.] Planes to which the same
straight line is perpendicular are parallel to one another. _ 364
THEOREM 89. [Euc. XI. 16.] If two parallel planes are cut -by
a third plane, their lines of section with it are parallel. 365
THeoreM 90. [Euc. XI. 15.] If two intersecting straight lines
are parallel respectively to two other intersecting straight
lines which are not in the same plane with them, then the
plane containing the first pair is parallel to the plane con-
taining the second pair. 366
THEOREM 91. [Euc. XI. 17.] Straight lines which are cut by
parallel planes are cut proportionally. 367
THEOREM 92. The projection of a straight line on a plane is
itself a straight line. 368
THEOREM 93. If a straight line outside a given plane is parallel
to any straight line drawn on the plane, it is also parallel to
the plane itself. 370
Cor. Through either of two skew straight lines a plane
may be made to pass to which the other line is parallel. 371
TuEoREM 94. If two straight lines neither intersect nor are
parallel, then
(i) there is one straight line perpendicular to both of them ;
(ii) this common perpendicular is the shortest distance
between the given lines. 375bo

Dihedral Angles.
DEFINITIONS - - - - - 2 = - e = 2S
TuerorEM 95. [Euc. XI. 18.] If a straight line is perpendicular
to a plane, then any plane passing through the perpendicular
is also perpendicular to the given plane. ‘374
XXiV CONTENTS.

PAGE

Tu£oREM 96. [Euc. XI. 19.] If two intersecting planes are

each perpendicular to a third plane, their line of section is
also perpendicular to that plane.

Solid Angles.
DEFINITIONS - - - ~ . . : :
TuHEoREM 97. [Euc. XI. 20.] In a trihedral angle the sum of
any two of the face-angles is greater than the third.
TuEoREM 98. [Euc. XI. 21.] In a convex solid angle the
sum of the face-angles is less than four right angles.

Axes of Reference. Position of a Point in Space - :

Solid Figures.
DEFINITIONS AND PRELIMINARY THEOREMS - : .

Surfaces and Volumes of Solids.

RECTANGULAR SOLIDS” - - - - : . :
PRISMS. - . . - - : : . : :
PYRAMIDS - - - - - : -
Tus Five ReGuLtarR PoLYHEDRA- - : : :

Solids of Revolution.
The Cylinder - - - . 2 . ; ‘. :
SURFACE AND VOLUME of A CYLINDER- : . »

The Cone - : - : : ‘ ° " ¢ ‘ P

SURFACE AND VOLUME OF A CONE ° - 3 p
FrustuM OF A PYRAMID AND CoNnE-~ - . * : .

The Sphere,
FUNDAMENTAL PROPERTIES - - . - ° - .
SorFACE OF A SPHERE - é . é ‘ J e
FRustuM, SEGMENT, ZONE OF A SPHERE : -
VoLUME OF A SPHERE - - ‘ > é ‘. . :
LINES OF REFERENCE ON A SpueRR. Latrrvupk AND Lon@i-
TUDE ‘ 5 ‘ ; : z a ’ ,
LUNES OF A SPHERICAL SURFACE - : : -
SPHERICAL TRIANGLES - ‘ e ‘ . . s *

Answers to Numerical Exercises.

 GEOMETRY.

PsAelee als

AXIOMS.

ALL mathematical reasoning is founded on certain simple

principles, the truth of which is so evident that they are
accepted without proof. These self-evident truths are called
Axioms.
For instance :
_ Lhings which are equal to the same thing are equal to one
another.
The following axioms, corresponding to the first four Rules
of Arithmetic, are among those most commonly used in
geometrical reasoning.
Addition. Jf equals are added to equals, the sums are equal.
Subtraction. Jf equals are taken from equals, the remainders
are equal.
Multiplication. Things which are the same multiples of equals
are equal to one another.
For instance: Doubles of equals are equal to one another.
Division. Things which are the same parts of equals are equal
to one another.
For instance: Halves of equals are equal to one another.
The above Axioms are given as instances, and not as a
complete list, of those which will be used. They are said to
be general, because they apply equally to magnitudes of all
kinds. Certain specia] axioms relating to geometrical magni-
_tudes only will be stated from time to time as they are
required. ~
H.S.G, A &
2 GLOMETRY.

DEFINITIONS AND FIRST PRINCIPLES.

Every beginner knows in a general way what is meant by 4

point, a line, and a surface. but in geometry these terms are
used in a strict sense which needs some explanation.

1. A point has position, but is said to have 1o magnitude.

This means that we are to attach to a point no idea of size either as
to length or breadth, but to think only where it is situated. A dot
made with a sharp pencil may be taken as roughly representing a
joint ; but small as such a dot may be, it still has some length and
readth, and is therefore not actually a geometrical point. The smaller
the dot however, the more nearly it represents a point,

2. A line has length, but is said to have no breadth.

A line is traced out by a moving point. If the point of a pencil is
moved over a sheet of paper, the trace left represents a line. But such
a trace, however fincly drawn, has some degree of breadth, and is
therefore not itself a true geometrical line. The finer the trace left by
the moving pencil-point, the more nearly will it represent a line.

3. Proceeding in a similar manner from the idea of a line

to the idea of a surface, we say that
A surface has length and breadth, but no thickness.
And finally,
A solid has length, breadth, and thickness.
Solids, surfaces, lines and points are thus related to one another :
(i) A solid is bounded hy surfaces.
(ii) A surface is bounded by lines ; and surfaces meet in lines.
(iii) A line is bounded (or terminated) by points ; and lines meet in
points.

4. A line may be straight or curved.

A straight line has the same direction from point to point
throughout its whole length.
A curved line changes its direction continually from point
to point.
 DEFINITIONS. 3

Axiom. There can be only one straight line joining two given
points: that is,
Two straight lines cannot enclose a space.

5s OA plane is a flat surface, the test of flatness being that

if any twopoints are taken in the surface, the straight line
between them lies wholly in that surface.

6. When two straight lines meet at a

point, they are said to form an angle. B
The straight lines are called the arms of the
angle ; the point at which they meet is its vertex. in
The magnitude of the angle may be thus 9 A
explained :
Suppose that the arm OA is fixed, and that OB turns about
the point O (as shewn by the arrow). Suppose also that OB
began its turning from the position OA. ‘Then the size of the
angle AOB is measured by the amount of turning required to
bring the revolving arm from its first position OA into its
subsequent position OB.
Observe that the size of an angle docs not in any way depend on the
length of its arms.
Angles which lie on either side of C B
a common arm are said to be ad-
jacent.
For example, the angles AOB, BOC,
which have the common arm OB, are O A
adjacent.

When two straight lines such as AB, CD C

eross one another at O, the angles COA, BOD
are said to be vertically opposite. The 5 6 i
angles AOD, COB are also vertically opposite
to one another.
4 GEOMETRY.

7. When one straight line stands on an- cs

other so as to make the adjacent angles equal
to one another, each of the angles is called a
right angle; and each line is said to be per-
pendicular to the other. 5 6] A

Axioms. (i) If O is a point in a straight line AB, then a-line

OC, which twrns about O from the position OA to the position OB,
must pass through one position, and only one, in which i is
perpendicular to AB.
(ii) All right angles are equal.
A right angle is divided into 90 equal parts called degrees (°); each
degree into 60 equal parts called minutes (‘) ; each minute into 60 equal
parts called seconds (”).

In the above figure, if OC revolves about O from the

position OA into the position OB, it turns through two right
angles, or 180°.
If OC makes a complete revolution about O, starting from OA
and returning to its original position, it turns through four
right angles, or 360°.

8. An angle which is less than one right

angle is said to be acute. 4

That is, an acute angle is less than 90°. \

9. An angle which is greater g

than one right angle, but less than
two right angles, is said to be obtuse.
That is, an obtuse angle lies between
90° and 180°. 0 A

10. If one arm OB of an angle turns

until it makes a straight line with the os,
other arm OA, the angle so formed is I «ee
called a straight angle.
A straight angle=2 right angles= 180°,
 DEFINITIONS. 5

EF. An Sele which is greater

than two right angles, but less than te
four right angles, is said to be 2 7
reflex. ge Hi)
That is, a reflex angle lies between B
180° and 360°.
Norse. When two straight lines meet, two angles are formed, one
greater, and one less than two right angles. The first arises by -
supposing OB to have revolved from the position OA the longer way
round, marked (i); the other by supposing OB to have revolved the
shorter way round, marked (ii). Unless the contrary is stated, the
angle between two ‘straight lines will be considered to be that which is
less than two right angles.

12. Any portion of a plane surface bounded by one or

more lines is called a plane figure.

13. A circle is a plane figure contained

by a line traced out by a point which moves
so that its distance from a certain fixed
point is always the same. O
Here the point P moves so that its distance P
from the fixed point O is always the same.

The fixed point is called the centre, and the bounding lins
is called the circumference.

14. A radius of a circle is a straight line drawn from the

centre to the circumference. It follows that all radii of a
circle are equal.

15. A diameter of a circle is a straight line drawn through

the centre, and terminated both ways by the circumference.

16. An arc of a circle is any part of the circumference.

6 GEOMETRY.

17. A semi-circle is the figure bounded

by a diameter of a circle and the part of the as
circumference cut off by the diameter.

18. To bisect means to divide into two equal parts.

Axioms. (i) If a point O moves

from A to B along the straight line 4 O B
AB, it must pass through one post-
tion in which it divides AB into two equal parts.
That is to say:
Every finite straight line has a point of bisection.
(ii) Jf a line OP, revolving about O, turns
from OA to OB, it must pass through one
position in which it divides the angle AOB
into two equal parts.
That is to say:
Every angle may be supposed to have a line of bisection.

HyYpoTHETICAL CONSTRUCTIONS.

From the Axioms attached to Definitions 7 and 18, it

follows that we may suppose
(i) A straight line to be drawn perpendicular to a given
straight line from any point in it.
(ii) A finite straight line to be bisected at a point.
(iii) An angle to be bisected by a line.

SUPERPOSITION AND EQUALITY.

Axiom. Magnitudes which can be made to coincide with one

another are equal.
This axiom implies that any line, angle, or figure, may be taken up
from its position, and without change in size or form, laid down upon a
second line, angle, or figure, for the purpose of comparison, and it
states that two such magnitudes are scat when one can be exactly
placed over the other without overlapping.
This process is called superposition, and the first magnitude is said to
be applied to the other,
 POSTULATES. 7

POSTULATES.

In order to draw geometrical figures certain instruments

are required. These are, for the purposes of this book, (i) a
straight ruler, (11) a pair of compasses. The following Postulates
(or requests) claim the use of these instruments, and assume
that with their help the processes mentioned below may be
duly performed.
Let it be granted :
1. That a straight line may be drawn from any one point to
any other point.
2. That a FINITE (or terminated) straight line may be
PRODUCED (that is, prolonged) to any length in that straight
line.
3. That a circle may be drawn with any point as centre and
with a radius of any length.

Nores. (i) Postulate 3, as stated above, im-

plies that we may adjust the compasses to the
length of any straight line PQ, and with a radius re)
of this length draw a circle with any point O as
centre. That is to say, the compasses may be
used to transfer distances from one part of a
diagram to another.

(ii) Hence from AB, the greater of two

straight lines, we may cut off a part equal. |
to PQ the less. Ns. Xx B
For if with centre A, and radius equal
to PQ, we draw an arc of a circle cutting
AB at X, it is obvious that AX is equal pp Q
to PQ.
8 GEOMETRY.

INTRODUCTORY.

1. Plane geometry deals with the properties of such lines

and figures as may be drawn on a plane surface.

2. The subject is divided into a number of separate dis-

cussions, called propositions.

Propositions are of two kinds, Theorems and Problems.

A Theorem proposes to prove the truth of some geometrical
statement.

A Problem proposes to perform some geometrical construc-

tion, such as to draw some particular lne, or to construct
some required figure.

3. A Proposition consists of the following parts:

The General Enunciation, the Particular Enunciation, the
Construction, and the Proof.
(i) The General Enunciation is a preliminary statement,
describing in general terms the purpose of the proposition.
(ii) The Particular Enunciation repeats in special terms
the statement already made, and refers it to a diagram, which
enables the reader to follow the reasoning more easily.
(iii) The Construction then directs the drawing of such
straight lines and circles as may be required to effect the
purpose of a problem, or to prove the truth of a theorem.
(iv) The Proof shews that the object proposed in a problem
has been accomplished, or that the property stated in a theorem
1s true.

4, The letters Q.£.D. are appended to a theorem, and stand

for Quod erat Demonstrandum, which was to be proved.
 INTRODUCTORY. 9

5. <A Corollary is a statement the truth of which follows

readily from an established proposition; it is therefore
appended to the proposition as an inference or deduction,
which usually requires no further proof.

6. The following symbols and abbreviations are used in

the text of this book.
in Parts
.. for therefore, £ for angle,
= ,, is, or are, equal to, A ..,, triangle.
After Part I.
pt. for point, perp. for perpendicular,
st. line 5, siraight line, par” ,, parallelogram,
Bb, £ », right angle, rectil. ,, rectilineal,
par’ (or ||) ,, parallel, © » circle,
Sq. ,» square, Or ,, circumference ;
and all obvious contractions of commonly occurring words,
such as opp., adj., diag., etc., for opposite, adjacent, diagonal,
etc.
[For convenience of oral work, and to prevent the rather common
abuse of contractions by beginners, the above code of signs has been
introduced gradually, and at first somewhat sparingly.]

In numerical examples the following abbreviations will

be used.’
m. for metre, cm. for centimetre,
mm. ,, millimetre. km. ,, kilometre.
Also inches are denoted by the symbol (’).
Thus 5” means 5 inches.
10 GEOMETRY.

ON LINES AND ANGLES.

TuroreM 1. [Euclid I. 13.]
The adjacent angles which one straight line makes with another
straight line on one side of it, are together equal to two right angles.

i)

B oO
ecoucce—
een
—o

Let the straight line CO make with the straight line AB the
adjacent 2*AOC, COB.
It is required to prove that the ‘AOC, COB are together equal to
two right angles.
Suppose OD is at right angles to BA.
Proof. Then the 2*AOC, COB together
=the three 2"AOC, COD, DOB.
Also the 2*AOD, DOB together
=the three 2"AOC, COD, DOB.
_. the 2*AOC, COB together =the 2*AOD, DOB
= two right angles.
Q.E.D.

PROOF BY ROTATION. od

Suppose a straight line revolving about O turns from the position OA

into the position OC, and thence into the position OB; that is, let the
revolving line turn in succession through the 4s AOC, COB.
Now in passing from its first position OA to its final position OB,
the revolving line turns through two right angles, for AOB is a straight
line.
Hence the 2*AOC, COB together = two right angles.
 LINES AND ANGLES, ll

CoroLuARY 1. Jf two straight lines

cut one another, the four angles so formed
are together equal to four right angles. A
For example, C
4BOD+ LDOA+ZA0C+ LCOB=4 right angles.

CoroLuany 2. /Vhen any number of

straight lines meet at a point, the sum of
the consecutive angles so formed is equal
to four right angles.
: E
For a straight line revolving about O, and turning in succession
through the £?AOB, BOC, COD, DOE, EOA, will have made one
complete revolution, and therefore turned through four right angles.

DEFINITIONS.

(i) Two angles whose sum is two right angles, are said to
be supplementary ; and each is called the supplement of the
other.
Thus in the Fig. of Theor. 1 the angles AOC, COB are supplementary.
Again the angle 123° is the supplement of the angle 57°.

(ii) Two angles whose sum is one right angle are said to
be complementary ; and each is called the complement of the
other.
Thus in the Fig. of Theor. 1 the angle DOC is the complement of
the angle AOC. Again angles of 34° and 56° are complementary.

COROLLARY 3. (i) Supplements of the same angle ave equal.

(ii) Complements of the same angle are equal.
12 GEOMETRY.

THeoreM 2. [Euclid I. 14.]

Tf, at a point in a straight line, two other straight lines, on
opposite sides of it, make the adjacent angles together equal to two
right angles, then these two straight lines are in one and the same
straight line.
Cc

x
B O A
At O in the straight line CO let the two straight lines OA,
OB, on opposite sides of CO, make the adjacent 2" AOC, COB
together equal to two right angles: (that is, let the adjacent
4" AOC, COB be supplementary).
It is required to prove that OB and OA are in the same straight
line.
Produce AO beyond O to any point X: it will be shewn that
OX and OB are the same line.

Proof. Since by construction AOX is a straight line,

.. the COX is the supplement of the 2COA. Theor. 1.
But, by hypothesis,
the 2 COB is the supplement of the 2 COA.
*, the ~COX=the 2 COB;
-. OX and OB are the same line.
But, by construction, OX is in the same straight line
with OA;
hence OB is also in the same straight line with OA.
Q.E.D,
 LINES AND ANGLES, : 13

EXERCISES.

1. Write down the supplements of one-half of a right angle, four-

thirds of a right angle; also of 46°, 149°, 83°, 101° 15’.
2. Write down the complement of two-fifths of a right angle;
also of 27°, 38°16’, and 41° 29’ 30”.
3. If two straight lines intersect forming four angles of which one
is known to be a right angle, prove that the other three are also right
angles.

4, In the triangle ABC the angles ABC, ACB are given equal. If
the side BC is produced both ways, shew that the exterior angles so
formed are equal.
5. In the triangle ABC the angles ABC, ACB are given equal. If
AB and AC are produced beyond the base, shew that the exterior angles
so formed are equal.

DEFINITION. The lines which bisect an angle and the

adjacent angle made by producing one of its arms are called
the internal and external bisectors of the given angle.

Thus in the diagram, OX and OY are the

internal and external bisectors of the angle
AOB.

6. Prove that the bisectors of the adjacent angles which one

straight line makes with another contain a right angle. That is to
say, the internal and external bisectors of an angle are at right angles
to one another.

7. Shew that the angles AOX and COY in the above diagram are
complementary.
8. Shew that the angles BOX and COX are supplementary ; and
also that the angles AOY and BOY are supplementary.
9. If the angle AOB is 35°, find the angle COY.
14 GEOMETRY.

Tueorem 3. [Euclid 1. 15.]

If two straight lines cut one another, the vertically opposite angles
are equal,

A D

Let the straight lines AB, CD cut one another at the point O.
It is required to prove that
(i) the «AOC =the . DOB;
(ii) the © COB =the 2 AOD.

Proof. Because AO meets the straight line CD,

*, the adjacent 2* AOC, AOD together = two right angles;
that is, the 2 AOC is the supplement of the 2 AOD.
Again, because DO meets the straight line AB,
”, the adjacent 2* DOB, AOD together = two right angles;
that is, the 2 DOB is the supplement of the 2 AOD,
Thus cach of the 2*AOC, DOB is the supplement of the 2 AOD,
“. the 2AOC=the 2 DOB.
Similarly, the 2 COB =the z AOD.
Q.E.D,

PROOF BY ROTATION,

Suppose the line COD to revolve about O until OC turns into the
osition OA. Then at the same moment OD must reach the position
B (for AOB and COD are straight).
Thus the same amount of turning is required to close the AOC as to
close the 4 DOB.
* the LAOC=the LDOB.
 LINES AND ANGLES, 15

EXERCISES ON ANGLES.

(Numerical. )
1. Through what angles does the minute-hand of a clock turn in
(i) 5 minutes, (ii) 21 minutes, (ili) 484 minutes, (iv) 14 min. 10 sec.?
And how long will it take to turn through (v) 66°, (vi) 222°?
' 2. A clock is started at noon: through what angles will the hour-
hand have turned by (i) 3.45, (ii) 10 minutes past 5? And what will
be the time when it has turned through 1723°?
3. The earth makes a complete revolution about its axis in 24 hours.
Through what angle will it turn in 3 hrs. 20 min., and how long will it
tuke to turn through 130°?
4. In the diagram of Theorem 3 :
(i) If the LAOC=35°, write down (without measurement) the value
of each of the £48COB, BOD, DOA.
(ii) If the LSCOB, AOD together make up 250°, find each of the
ZsCOA, BOD.
(iii) If the 2SAOC, COB, BOD togethcr make up 274°, find each of
the four angles at O.

(Theoretical.)
« 5. If from O a point in AB two straight lines OC, OD are drawn on
opposite sides of AB so as to make the angle COB equal to the angle
AOD ; shew that OC and OD are in the same straight line.
6. Two straight lines AB, CD cross at O. If OX is the bisector of
the angle BOD, prove that XO produced bisects the angle AOC.
7. Two straight lines AB, CD cross at O. If the angle BOD is
bisected by OX, and AOC by OY, prove that OX, OY are in the same
straight line.
8. If OX bisects an angle AOB, shew that, by folding the diagram
about the bisector, OA may be made to coincide with OB.
How would OA fall with regard to OB, if
(i) the L AOX were greater than the L XOB;
(ii) the 2 AOX were less than the 2XOB?
9, AB and CD are straight lines intersecting at right angles at O ;
shew by folding the figure about AB, that OC may be made to fall
along OD.
10. A straight line AOB is drawn on paper, which is then folded
about O, so as to make OA fall along OB; shew that the crease left in
the paper is perpendicular to AB.
16 GEOMETRY.

ON TRIANGLES.

1. Any portion of a plane surface bounded by one or more

lines is called a plane figure.
The sum of the bounding lines is called the perimeter of the figure.
The amount of surface enclosed by the perimeter is called the area.

2. Rectilineal figures are those which are bounded by

straight lines.
3. A triangle is a plane figure bounded by three straight
lines.
4, A quadrilateral is a plane figure bounded by four straight
lines. +

5. <A polygon is a plane figure bounded by

more than four straight lines.

6. A rectilineal figure is said to be

equilateral, when all its sides are equal ;
equiangular, when all its angles are equal ;
regular, when it is both equilateral and equiangular.

7. Triangles are thus classified with regard to their sides:

A triangle is said to be
equilateral, when all its sides are equal ;
isosceles, when two of its sides are equal ;
scalene, — when its sides are all unequal.

Equilateral Triangle. Isosceles Triangle. Sculene Triangle.

In a triangle ABC, the letters A, B, C often denote

the magnitude of the several angles (as measured in ¢ 3
degrees); and the letters a, b, ¢ the lengths of the
opposite sides (as measured in inches, centimetres, or
8 a 6
some other unit of length),
 TRIANGLES. 17

Any one of the angular points of a triangle may be regarded as its

vertex ; and the opposite side is then called the base.
In an zsosceles triangle the term vertex is usually applied to the point
at which the equal sides intersect ;’and the vertical angle is the angle
imcluded by them.
8. ‘Triangles are thus classified with regard to their angles:
A triangle is said to be
right-angled, when one of its angles is a right angle;
obtuse-angled, when one of its angles is obtuse ;
acute-angled, when all thice of its angles are acute.
[It will be seen hereafter (Theorem 8. Cor. 1) that every triangle must
have at least two acute angles. |

| ee eee
Right-angled Triangle. Obtuse-angled Triangle. Acute-angled Triangle.

In a right-angled triangle the side opposite to the right angle is

called the ‘hypotenuse.

9. In any triangle the straight line joining a vertex to the

middle point of the opposite sideis called a median.

THE COMPARISON OF TWO TRIANGLES.

(i) The three sides and three angles of a triangle are called
its six parts. A triangle may also be considered with regard
to its area.
(ii) Two triangles are said to be equal in all respects,
when one may be so placed upon the other as to exactly
coincide with it; in which case each part of the first triangle
is equal to the corresponding part (namely that with which
it coincides) of the other; and the triangles are equal in area.
In two such triangles corresponding sides are opposite to equal
angles, and corresponding angles are opposite to equal sides.
Triangles which may thus be made to coincide by super-
position are said to be identically equal or congruent.
H.S.G. R
18 GEOMETRY.

THEOREM 4. [Euclid I. 4.]

If two tricngles have two sides of the one equal to two sides of the
other, each to each, and the angles included by those sides equal,
then the triangles are equal in all respects.
A D

Let ABC, DEF be two triangles in which

AB = DE,
AC = DF,
and the included angle BAC = the included angle EDF.
It 1s required to prove that the AABC=the ADEF in all
respects,

Proof. Apply the A ABC to the A DEF,

so that the point A falls on the point D,
and the side AB along the side DE.
Then because AB= DE,
.. the point B must coincide with the point E.
And because AB falls along DE,
and the 2 BAC =the 2 EDF,
.. AC must fall along DF.
And because AC = DF,
.“. the point C must coincide with the point F.
Then since B coincides with E, and C with F,
.*. the side BC must coincide with the side EF.
Hence the AABC coincides with the A DEF,
and is therefore equal to it in all respects.
Q.E.D.
 CONGRUENT TRIANGLES. 19

Obs. In this Theorem we must carefully observe what is

given and what is proved.
AB=DE,
Given that AC = DF,
and the 2 BAC=the 2 EDF.
From these data we prove that the triangles coincide on
superposition.
BC=EF,
Hence we conclude that the 2 ABC = the z DEF,
and the 2 ACB=the z DFE;
also that the triangles are equal in area.
Notice that the angles which are proved equal in the two triangles
are opposite to sides which were given equal.
D

Norz. The adjoining diagram shews

that in order to make two congruent
triangles coincide, it may be necessary -
to reverse, that is, turn over one of them GF
before superposition.

B =

EXERCISES.

1. Shew that the bisector of the vertical angle of an isosceles triangle

(i) bisects the base, (ii) is perpendicular to the base.
2. Let O be the middle point of a straight line AB, and let OC be
perpendicular to it. Then if P is any point in OC, prove that PA=PB.
;
3. Assuming that the four sides of a square are equal, and that its
angles are all right angles, prove that in the square ABCD, the
diagonals AC, BD are equal.
4. ABCD isa square, and L, M, and N are the middle points of AB,
BC, and CD: prove that
(i) LM=MN. (ii) AM=DM.
(iii) AN=AM. (iv) BN=DM.
[Draw a separate figure in each case.]
5. ABC is an isosceles triangle: from the equal sides AB, AC two
equal parts. AX, AY are cut off, and BY and CX are joined. Prove that
BY —CxX:
20 GEOMETRY.

THEOREM 5. [Euclid I. 5.]

The angles at the base of an isosceles triangie are equal,

Be
D -G

Let ABC be an isosceles triangle, in which the side AB= the

side AC.
It is required to prove that the 2 ABC = the 2 ACB.
Suppose that AD is the line which bisects the 2 BAC, and
Igt it meet BC in D.

Ist Proof: Then in the A* BAD, CAD,

BA=CA,
because AD is common to both triangles,
and the included 2 BAD=the included 2 CAD;
‘. the triangles are equal in all respects; Theor. 4.
so that the ~ABD=the 2 ACD.
Q.E.D.

2nd Proof. Suppose the A ABC to be folded about AD.

Then since the 2 BAD=the 2 CAD,
. AB must fall along AC.
And since AB=AC,
‘. B must fall on C, and consequently DB on DC.
.’. the ABD will coincide with the ACD, and is therefore
equal to it.
Q.E.D.
 ISOSCELES TRIANGLES. aT

Corotiary 1. Jf the equal sides AB, AC of

‘an isosceles triangle are produced, the exterior
angles EBC, FCB are equal; for they are the
supplements of the equal angles at the base. B C

= F
CoronLaRy 2. If a triangle is equilateral, it is also equi-
angular.

DEFINITION. A figure is said to be symmetrical about a

line when, on being ‘folded about that line, the parts of the
figure on each side of it can be brought into coincidence.
“The straight line is called an axis of symmetry*
That this may be possible, it is clear that the two parts of the figure
must have the same size and shape, and must be similarly placed with
regard to the axis.
Theorem 5 proves that an isosceles triangle is symmetrical about
the bisector of its VERTICAL angle.
An equilateral triangle is symmetrical about the bisector of ANY
ONE of us angles.

EXERCISES.

1. ABCD is a four-sided figure whose sides are all equal, and the
diagonal BD is drawn: shew that
(i) the angle ABD=the angle ADB;
(ii) the angle CBD =the angle CDB;
(iii) the angle ABC =the angle ADC.
2. ABC, DBC are two isosccles triangles drawn on the same base
BC, but on opposite sides of it: prove (by means of Theorem 5) that
the angle ABD =the angle ACD.
3. ABC, DBC are two isosceles triangles drawn on the same base
BC and on the same side of it: employ Theorem 5 to prove that
the angle ABD=the angle ACD.
4. AB, AC are the equal sides of an isosceles triangle ABC; and
L, M, N are the middle points of AB, BC, and CA respectively: prove
that
- (i) LM=NM. (ii) BN=CL.
(iii) the angle ALM=the angle ANM.
22 GEOMETRY.

THEOREM 6. [Euclid I. 6.]

If two angles of a triangle are equal to one another, then the .
sides which are opposite to the equal angles are equal to one another.

A
D

B Cc

Let ABC be a triangle in which

the 4ABC =the 2 ACB.
It is required to prove that the side AC =the side AB.
If AC and AB are hot-equal, suppose that AB is the greater
From BA cut off BD equal to AC.
Join DC.

Proof. Then in the A* DBC, ACB,

- DB=AC,
because BC is common to both,
lana the included 2 DBC =the included 2 ACB;
. the A\DBC=the AACB in area, Theor. 4.
the part ha to the whole ;which is absurd.
_ AB is not unequal to AC ;
that is, AB = AC.
Q.E.D.

CornoLiary. Hence if a triangle is equiangular it is also

equilateral,
 A THEOREM AND ITS CONVERSE, ~ 23

NOTE ON THEOREMS 5 AND 6.

Theorems 5 and 6 may be verified ex- sd

perimentally by cutting out the given in
AABC, and, after turning it over, fitting ‘
it thus reversed into the vacant space left 7 ‘
in the paper. (eee 5
B Gc B
Suppose A’B’C’ to be the original position of the AABC, and let
ACB represent the triangle when reversed.
In Theorem 5, it will be found on applying A to A’ that C may be
made to fall on B’, and B on C’.
In Theorem 6, on applying C to B’ and B to C’ we find that A will
fallon A’. .
In either case the given triangle reversed will coincide with its own
‘* trace,” so that the side and angle on the eft are respectively equal to
the side and angle on the right.

NOTE ON A THEOREM AND ITS CONVERSE.

The enunciation of a theorem consists of two clauses. The first

clause tells us what we are to asswme, and is called the hypothesis; the
second tells us what 2¢ is required to prove, and is called the conclusion.
For example, the enunciation of Theorem 5 assumes that in a certain
triangle ABC the side AB=the side AC: this is the hypothesis. From
this it is required to prove that the angle ABC=¢he angle ACB: this is
the conclusion.
If we interchange the hypothesis and conclusion of a theorem, we
enunciate a new theorem which is called the converse of the first.
For example, in Theorem 5 .
it is assumed that AB=AC;
it is required to prove that the angle ABC=the angle ACB.
Now in Theorem 6
it is asswmed that the angle ABC=the angle ACB ;
it is réquired to prove that AB=AC.
Thus we see that Theorem 6 is the converse of Theorem 5; for the
hypothesis of each is the conclusion of the other.
In Theorem 6 we employ an indirect method of proof frequently
used in geometry. It consists in shewing that the theorem cannot be
untrue; since, if it were, we should be led to some impossible conclusion.
This form of proof is known as Reductio ad Absurdum, and is most
commonly used in demonstrating the converse of some foregoing theorem.
It must not however be supposed that if a theorem is true, its con-
verse is necessarily true. [See p. 25.]
24 ; GEOMETRY.

THEOREM 7. [Euclid I. 8.]

If two triangles have the three sides of the one equal to the three
sides of the other, each to each, they are equal in all respects.
A D

B ewe F

G
Let ABC, DEF be two triangles in which
AB = DE,
AC = DF,
BC=EF.
It is required to prove that the triangles are equal in all respects.
Proof. Apply the A ABC to the A DEF,
so that B falls on E, and BC along EF, and
so that A is on the side of EF opposite to D.
Then because BC = EF, C must fall on F.
Let GEF be the new position of the A ABC.
Join DG.
Because ED = EG,
*, the .EDG=the 2 EGD. Theor. 5.
Again, because FD = FG,
*. the 2 FDG=the 2 FGD.
Hence the whole 2 EDF =the whole 2 EGF,
that is, the 2 EDF =the 2 BAC.
Then in the A* BAC, EDF;
BA = ED;
because AC = DF,
and the included 2 BAC = the included 2 EDF ;
.. the triangles are equal in all respects. Theor. 4,
Q.E.D.
 CONGRUENT TRIANGLES. 25

Obs. In this Theorem

it is given thab AB=DE, BC=EF, CA=FD;
and we prove that LC=LF, LA=LD, LB=ZE.
Also the triangles are equal in area.
Notice that the angles which are proved equal in the two triangles
are opposite to sides which were given equal.

Nore 1. We have taken the case in which DG falls within the

LEDE, EGF.
Two other cases might arise :
(i) DG might fall outside the 2s EDF, EGF [as in Fig. 1].
(ii) DG might coincide with DF, FG [as in Fig. 2].

A D

Fig.1. ee Fiig.2

These cases wiil arise only when the given triangles are obtuse-
angled or right-angled ; and (as will be seen hereafter) not even then,
if we begin by choosing for superposition the greatest side of the A ABC,
as in the diagram of page 24.

Notr}2. Two triangles'are said to be equiangular to one another

when the angles of one are respectively equal to the angles of the other.
Hence if two triangles have the three sides of one severally equal to the
three sides of the other, the triangles are equiangular to one another.
The student should state the converse theorem, and shew by a
diagram that the converse is not necessarily true.

i At this stage Problems 1-5 and 8 [see page 70} may

conveniently be taken, the proofs affording good illustrations of the
Identical Equality ofTwo Triangles. < «
26 : GEOMETRY.

EXERCISES.

On Tue IDENTICAL EquaLity oF Two TRIANGLES.

THEOREMS 4 AND 7.
(Theoretical. )
1. Shew that the straight line which joins the vertex of an isosceles
triangle to the middle point of the base,
(i) bisects the vertical angle :
(ii) is perpendicular to the base.
2. -If ABCD is a rhombus, that is, an equilateral foursided figure ;
shew, by drawing the diagonal AC, that
(i) the angle ABC=the angle ADC;
(ii) AC bisects each of the angles BAD, BCD.
3. If in a quadrilateral ABCD the opposite sides are equal, namely
AB=CD and AD=CB;; prove that the angle ADC=the angle ABC.
4. If ABC and DBC are two isosceles triangles drawn on the same
base BC, prove (by means of Theorem 7) that the angle ABD=the
angle ACD, taking (i) the case where the triangles are on the same side
of BC, (ii) the case where they are on opposite sides of BC.
5. If ABC, DBC are two isosceles triangles drawn on opposite
sides of the same base BC, and if AD be joined, prove that cach of the
angles BAC, BDC will be divided into two equal parts.
6. Shew that the straight lines which join the extremities of the
base of an isosceles triangle to the middle points of the opposite sides,
are equal to one another.
7. Two given points in the base of an isosceles triangle are equi-
distant from the extremities of the base: shew that they are also
equidistant from the vertex.
8. Shew that the triangle formed by joining the middle points
of the sides of an equilateral triangle is also equilateral.
9. ABC is an isosceles triangle having AB equal to AC; and the
angles at B and C are bisected by BO and CO: shew that
(i) BO=CO;
(ii) AO bisects the angle BAC.
10. Shew that the diagonals of a rhombus [see Ex. 2] bisect one
another at right angles,
11. The equal sides BA, CA of an isosceles triangle BAC are pro-
‘duced beyond the vertex A to the points E and F, so that AE is equal
to AF; and FB, EC are joined: shew that FB is equal to EC.
 TRIANGLES. PH

EXERCISES ON TRIANGLES.

(Numerical and Graphical.)

1. Draw a triangle ABC, having given a=2'0”, b=2°1", c=1°3".

Measure the angles, and find their sum.
2. In the triangle ABC, a=7°5 em., b=7:0 cm., and c=6'5 cm.
Draw and measure the perpendicular from B on CA.
3. Draw a triangle ABC, in which a=7 em., b=6 cm., C=65°.
How would you prove theoretically that any two triangles having
these parts are alike in size and shape? Invent some experimental
illustration.
4. Draw a triangle from the following data: b=2’, c=2°5", A=57° ;
and measure a, B, and C.
Draw a second triangle, using as data the values just found for a,
_ B, and C; and measure 0, c, and A. What conclusion do you draw?

5. A ladder, whose foot is placcd 12 feet from the base of a house,

reaches to 1 window 35 feet above the ground. Draw a plan in which
1” represents 10 ft.; and find by measurement the length of the ladder.
6. I go due North 99 metres, then duc East 20 metres. Plot my
course (scale 1 cm. to 10 metres), and find by measurement as nearly as
you can how far I am from my starting point.

7. When the sun is 42° above the horizon, a vertical pole casts a
shadow 30 ft. long. Represent this on a diagram (scale 1” to 10 ft.):
and find by measurement the approximate height of the pole.
8. From a point Aa surveyor goes 150 yards due East to B; then
300 yards due North to C; finally 450 yards due West to D. Plot his
course (scale 1” to 100 yards); and find roughly how far D is from A.
Measure the angle DAB, and say in what direction D bears from A.
9. B and C are two points, known to be 260 yards apart, on a
straight shore. A is a vessel at anchor. The angles CBA, BCA are
observed to be 33° and 81° respectively. Find graphically the approxi-
mate distance of the vessel from the points B and C, and from the
nearest point on shore.
10. In surveying a park it is required to find the distance between
two points A and B; but as a lake intervenes, a direct measurement
eannot be made. The surveyor therefore takes a third point C, from
which both A and B are accessible, and he finds CA=245 yards,
CB=820 yards, and the angle ACB=42°. Ascertain from a plan the
approximate distance between A and B.
28 GEOMETRY.

THEOREM 8. [Euclid I. 16.]

If one side of a triangle is produced, then the exterior angle is
greater than either of thegnierior opposite angles.
’

Let ABC be @ triangle, and let BC be produced to D.

It is required to prove that the exterior L ACD ts greater than
either of the interior opposite .* ABC, BAC.
Suppose E to be the middle point of AC.
Join BE ; and produce it to F, making-EF equal to BE.
Join FC.
Proof. Then in the A*AEB, CEF,
AE =CE,
because EB=EF,
and the 2 AEB=the vertically opposite 2 CEF;
.”. the triangles are equal in all respects; Theor. 4.
so that the 2 BAE =the 2 ECF.
But the 2 ECD is greater than the 2 ECF;
.. the ECD is greater than the 2 BAE;
that is, the 2 ACD is greater than the 2 BAC.
In the same way, if AC is produced to G, by supposing A to
be joined to the middle point of BC, it may be proved that
the 2 BCG is greater than the 2 ABC.
But the 2 BCG=the vertically opposite 2 ACD.
.. the ACD is greater than the 2 ABC.
QED.
 EXTERIOR AND INTERIOR TRIANGLES, 29

CoROLLARY 1. Any two angles of a triangle are together less

than two right angles. :
A
For the LABC is less than the LACD: Proved.
to each add the L ACB.
Then the 2s ABC, ACB are less than the Ls ACD, ACB,
therefore, less than two right angles.
+
B Cc D

COROLLARY 2. Every triangle must havg/at least two acute

angles.
For if one angle is obtuse or a right angle, then by Cor. 1 each of the
other angles must be less than a right angle.

COROLLARY 3. Only one perpendicular can be drawn to a

straight line from a given point outside tt.

If two perpendiculars could be drawn to AB from

P, we should have a triangle PQR in which each of
the 4sPOR, PRQ would be a right angle, which is
impossible.

EXERCISES. °

1. Prove Corollary 1 by joining the vertex A to any point in the

base BC.
2. ABC isa triangle and D any point within it. If BD and CD are
joined, the anyle BDC is greater than the angle BAC. Prove this
(i) by producing BD to meet AC,
(ii) by joining AD, and producing it towards the base.
3. If any side of a triangle is produced both ways, the exterior
angles so formed are together greater than two right angles.

4, Toa given straight line there cannot be drawn from a point out-
side it more than two straight lines of the same given length.
5. If the equal sides of an isosceles triangle are produced, the
exterior angles must be obtuse.
30 GEOMETRY.

THEOREM 9. [Euclid I. 18.]

If one side of a triangle is greater than another, then the angle
opposite to the greater side is greater than the angle opposite to the
less. .
A

Bo c

Let ABC be a triangle, in which the side AC is greater than

the side AB.
It is required to prove that the L ABC is greater than the L ACB.
From AC cut off AD equal to AB.
Join BD.

Proof. Because AB=AD,

.. the 2ABD =the ZADB. Theor. 5.
But the exterior ADB of the ABDC is greater than the
interior opposite 2 DCB, that is, greater than the 2 ACB.
.". the ZABD is greater than the 2 ACB.
Still more then is the ZABC greater than the 2 ACB.
Q.E.D.

Obs. The mode of demonstration used in the following Theorem

is known as the Proof by Exhaustion. It is applicable to cases in which
one of certain suppositions must necessarily be true ;and it consists in
shewing that each of these suppositions is false with one exception:
hence the truth of the remaining supposition is inferred.
 INEQUALITIES. 3]

THEoREM 10. [Euclid I. 19.]

Tf one angle of a triangle 1s greater than another, then the side
opposite to the greater angle is greater than the side opposite to the
less.
A

B Cc

Let ABC be a triangle, in which the LABC is greater than

the 2 ACB.
It is required to prove that the side AC is greater than the
side AB.
Proof. If AC is not greater than AB,
it must be either equal to, or less than AB.
Now if AC were equal to AB,
then the 2 ABC would be equal to the LACB; Theor. 5.
' but, by hypothesis, it is not.
Again, if AC were less than AB,
then the L ABC would be less than the 2ACB; Theor. 9.
but, by hypothesis, it is not.
That is, AC is neither equal to, nor less than AB.
‘. AC is greater than AB. Q.E.D,

{For Exercises on Theorems 9 and 10 see page 34.]

a2 GEOME'RY.

THEOREM 11. [Euclid I. 20.]

Any two sides of a triangle are together greater than the third
side.
D

B CG

Let ABC be a triangle.

ft is required, to prove that any two of its sides are together
areater than the third side.
Jt is enough to shew that if BC is the greatest side, then
BA, AC are together greater than BC.
Produce BA to D, making AD equal to AC.
Join DC.
Proof. Because AD = AC,
. .. the .ACD=the 2 ADC. Theor. 5.
But the 2 BCD is greater than the 2 ACD;
*. the BCD is greater than the 2 ADC,
that is, than the 2 BDC.
Hence from the A BDC,
BD is greater than BC. Theor. 10.
But BD=BA and AC together;
.*. BA and AC are together greater than BC.
Q.E.D.

Note. This proof may serve as an exercise, but the truth of the
Theorem is really self-evident. For to go from B to C along the
straight line BC is clearly shorter than to go from B to A and then
from Ato C. In other words
The shortest distance between two points is the straight line which joins
them.
 INEQUALITIES. oo

THEOREM 12.
Of all straight lines drawn fiom a given point to a given straight
line the perpendicular is the least.

a: eq “ey, UR B
Let OC be the perpendicular, and OP any oblique, drawn from
the given point O to the given straight line AB.
It is required to prove that OC 1s less than OP.
Proof. In the AOCP, since the 2 OCP is a right angle,
.. the 2 OFC is less than aright angle; Theor. 8. Cor.
that is, the L OPC is less than the 2 OCP.
.. OC is less than OP. Theor. 10.
@D:

CoroLLany 1. Hence conversely, since there can be only

one perpendicular and one shortest line from O to AB,
If OC is the shortest straight line from O to AB, then OC is
perpendicular to AB.
CorOLLARY 2. Two obliques OP, OQ, which cut AB at equal
distances from © the foot of the perpendicular, are equal.
The At OCP, OCQ may be shewn to be congruent by Theorem 4;
hence OP=OQ.

CoroLuaARY 3. Of two obliques OQ, OR, if OR cuts AB at the

greater distance from C the foot of the perpendicular, then OR ts
greater than OQ.
The LOQC is acute, .». the 2 OQR is obtuse;
. the LOQR is greater than the LORQ;
.. OR is greater than OQ.
H.S.G. o
34 GEOMETRY.

EXERCISES ON INEQUALITIES IN A TRIANGLE.

1. The hypotenuse is the greatest side of a right-angled triangle.

2. The greatest side of any triangle makes acute angles with each of
the other sides.
3. If from the ends of a side of a triangle, two straight lines are
drawn to a point within the triangle, then these straight lines are together
less than the other two sides of the triangle.
4. BC, the base of an isosceles triangle ABC, is produced to any
point D; shew that AD is greater tlian either of the equal sides.
5. If in a quadrilateral the greatest and least sides are opposite
to one another, then each of the angles, adjacent to the least side is
greater than its opposite angle.
6. In a triangle ABC, if AC is not greater than AB, shew that
any straight line drawn through the vertex A and terminated by the
base BC, is less than AB. °
7. ABC is a triangle, in which OB, OC bisect the angles ABC,
ACB respectively: shew that, if AB is greater than AC, then OB is
greater than OC.
8. The difference of any two sides of a triangle is less than the
third side.
9. The sum ot the distances of any point from the three angular
points of a triangle is greater than half its perimeter.
10. The perimeter of a quadrilateral is greater than the sum of
its diagonals.
ll. ABC is a triangle, and the vertical angle BAC is bisected by
a line which meets BC in X; shew that BA is greater than BX, and
CA greater than CX. Hence obtain a proof of Theorem 11.
_ 12. The sum of the distances of any point within a triangle from
its angular points is less than the perimeter of the triangle.
13, The sum of the diagonals of a quadrilateral is less than the
sum of the four straight lines drawn from the angular points to any
given point. Prove this, and point out the exceptional case.
14. In a@ triangle any two sides are together greater than twice the
median which bisects the remaining side.
[Produce the median, and complete the construction after the
manner of Theorem 8.]
15. In any triangle the sum of the medians is less than the perimeter,
 PARALLELS. ; 35

PARALLELS.

DEFINITION. Parallel straight lines are such as, being in

the same plane, do not meet however far they are produced
beyond both ends.
Notre. Parallel lines must be in the same plane. For instance, two
straight lines, one of which is drawn on a table and the other on the
floor, would never meet if produced ; but they are not for that reason
necessarily parallel.

AxIoM. Two intersecting straight lines cannot both be parallel

to a third straight line.
In other words:
Through a given point there can be only one straight line parallel
to a given straight line.
This assumption is known as Playfair’s Axiom.

Derinition. When two straight lines AB, CD are met by

a third straight line EF, ezght angles are formed, to which for
-the sake of distinction particular names are given.
Thus in the adjoining figure, A
1, 2, 7, 8 are called exterior angles,
3, 4, 5, 6 are called interior angles,
4 and 6 are said to be alternate angles ;
so also the angles 3 and 5 are alternate
to one another.
Of the angles 2 and 6, 2 is referred
to as the exterior angle, and 6 as the
interior opposite angle on the same side
of EF. Such angles are also known as corresponding angles.
Similarly 7 and 3, 8 and 4, 1 and 5 are pairs of corresponding
angles.
36 GEOMETRY.

THEOREM 13. [Euclid I. 27 and 28.]

If a straight line cuts two other straight lines so as to make
(i) the alternate angles equal,
or (ii) an exterior angle equal to the interior opposite angle on the
same side of the cutting line,
or (ili) the interior angles on the same side equal to two right
angles ;
then in each case the two straight lines are parallel.

(i) Let the straight line EGHF cut the two straight lines
AB, CD at G and H so as to make the alternate 2*>AGH, GHD
equal to one another.
It is required to prove that AB and CD are parallel.

Proof. If AB and CD are not parallel, they will meet, if

produced, either towards B and D, or towards A and C.
If possible, let AB and CD, when produced, meet towards B
and D, at the point K.

Then KGH is a triangle, of which one side KG is produced to A;

‘. the exterior 2 AGH is greater than the interior opposite
A4GHK; but, by hypothesis, it is not greater.
. AB and CD cannot meet when produced towards B and D.
Similarly it may be shewn that they cannot meet towards
Aand C:
.'. AB and CD are parallel.
 PARALLELS. 37

(ii) Let the exterior EGB =the interior opposite 2 GHD.

Lt is required to prove that AB and CD are parallel.
Proof. Because the 2 EGB=the 2 GHD,
and the 2 EGB=the vertically opposite 2 AGH ;
.-. the LAGH=the'2 GHD:
and these are alternate angles ;
... AB and CD are parallel.

(iii) Let the two interior 2*BGH, GHD be together equai to

two right angles. .
It is required to prove that AB and CD are parallel.

Proof. Because the 2*BGH, GHD together=two right

angles; —
and because the adjacent 2*BGH, AGH together=two right
angles ;
.. the Z°BGH, AGH together =the 2* BGH, GHD.
From these equals take the 2 BGH ;
then the remaining AGH = the remaining 2 GHD:
and these are alternate angles;
.. AB and CD are parallel.
Q.E.D.

DEFINITION. A straight line drawn across a set of given

lines is called a transversal.
For instance, in the above diagram the line EGHF, which crosses
the given lines AB, CD is a transversal.
38 GROMETRY.

THEOREM 14. [Euclid I. 29.]

Tf a straight line cuts two parallel lines, it makes
(i) the alternate angles equal to one another ;
(ii) the exterior angle equal to the interior opposite angle on the
same side of the cutting line ;
(iii) the two interior angles on the same side together equal to two
right angles,

Let the straight lines AB, CD be parallel, and let the

straight line EGHF cut them.
Ti is required to prove that
(i) the LAGH =the alternate L GHD ;
(ii) the exterior L EGB = the interior opposite L GHD ;
(iii) the two interior 2° BGH, GHD together= two right angles.

Proof. (i) If the 2 AGH is not equal to the 2 GHD,

suppose the 2 PGH equal to the 2 GHD, and alternate to it;
then PG and CD are parallel. Theor, 13.
But, by hypothesis, AB and CD are parallel ;
*. the two intersecting straight lines AG, PG are both parallel
to CD: which is impossible. Playfair’s Axiom.
.". the ZAGH is not unequal to the 2GHD ;
that is, the alternate 2"AGH, GHD are equal.
(ii) Again, because the LEGB=the vertically opposite
i. AGH ;
and the 2 AGH = the alternate < GHD ; Proved,
”, the exterior . EGB= the interior opposite 2 GHD.
 PARALLELS. 39

(iii) Lastly, the 2 EGB=the 2 GHD ; Proved.

add to each the 2 BGH ;
then the 2*EGB, BGH together =the angles BGH, GHD.
But the adjacent 2*EGB, BGH together= two right angles ;
.*. the two interior 2*BGH, GHD together =two right angles.
Q.E.D.

PARALLELS ILLUSTRATED BY ROTATION.

The direction of a straight line is determined by the angle which

it makes with some given line of reference.
Thus the direction of AB, relatively to the given line YX, is given by:
the angle APX.
Now suppose that AB and CD in
the adjoining diagram are parallel;
then we have learned that
theext. LAPX=theint. opp. -CQX;
that is, AB and CD make equal angles
with the line of reference YX.
This brings us to the leading idea
connected with parallels :
Parallel straight lines have the same
DIRECTION, bat differ in POSITION.
The same idea may be illustrated
thus:
Suppose AB to rotate about P through the LAPX, so as to take the
position XY. Thence Ict it rotate about Q the opposite way through
the equal LXQC: it will now take the position CD. Thus AB may be
brought into the position of CD by two rotations which, being equal
and opposite, involve no final change of direction.

HYPOTHETICAL CONSTRUCTION. In the above diagram let

AB be a fixed straight line, Q a fixed point, CD a straight
line turning about Q, and YQPX any transversal through Q.
Then as CD rotates, there must be one position in which the
4 CQX = the fixed 2 APX.
Hence through any given point we may assume a line to pass
parallel to any given straight line.
Obs. Jf AB is a straight line, movements from A towards
B, and from B towards A are said to be in opposite senses
of the line AB.
40 GEOMETRY.

THEoreM 15. [Euclid I. 30.]

Straight lines which are parallel to the same straight line are
parallel to one another.

Let the straight lines AB, CD be each parallel to the straight

line PQ.
It is required to prove that AB and CD are parallel to one
another.
Draw a straight line EF cutting AB, CD, and PQ in the
points G, H, and K.

Proof. Then because AB and PQ are parallel, and EF meets

them,
*, the . AGK =the alternate 2 GKQ.
And because CD and PQ are parallel, and EF meets them,
. the exterior 2 GHD =the interior opposite 2 GKQ.
.. the .AGH=the GHD;
and these are alternate angles;
.. AB and CD are parallel.
Q.E.D.

Norge. If PQ lies between AB and CD, tle Proposition needs no

proof; for it is inconceivable that two straight lines, which do not
meet an intermediate straight line, should meet one another.
The truth of this Proposition may be readily deduced from Playfair’s
Axiom, of which it is the converse.
For if AB and CD were not parallel, they would meet when produced.
Then there would be two intersecting straight lines both parallel to «
third straight line: which is impossible.
Therefore AB and CD never meet; that is, they are parallel,
 PARALLELS. 41

EXERCISES ON PARALLELS.

1. In the diagram of the previous page, if the angle EGB is 55°,

express in degrees each of the angles GHC, HKQ, QKF.
2. Straight lines which are perpendicular to the same straight line are
parallel to one another.

3. Ifa straight line meet two or more parallel straight lines, and is
perpendicular to one of them, wt is also perpendicular to all the others.

4. Angles of which the arms are parallel, each to each, are either
equal or supplementary.

5. Two straight lines AB, CD bisect one another at O. Shew that

the straight lines joining AC and BD are parallel.
6. Any straight line drawn parallel to the base of an isosceles tri-
ar.gle makes equal angles with the sides.

7. If from any point in the bisector of an angle a straight line is

drawn parallel to either arm of the angle, the triangle thus formed is
isosceles.

8. From X, a point in the base BC of an isosceles triangle ABC, a

straight line is drawn at right angles to the base, cutting AB in Y, and
CA produced in Z: shew the triangle AYZ is isosceles.

9. If the straight line which bisects an exterior angle of a triangle

is parallel to the opposite side, shew that the triangle is isosceles.

10. The straight lines drawn from any point in the bisector of an
angle parallel to the arms of the angle, and terminated by them, are
equal ; and the resulting figure is a rhombus.

11. AB and CD are two straight lines intersecting at D, and the

adjacent angles so formed are bisected: if through any point Xin DCa
straight line YXZ is drawn parallel to AB and meeting the biscctors in
Y and Z, shew that XY is equal to XZ.

12. Two straight rods PA, QB revolve about pivots at Pand Q, PA

making 12 complete revolutions a minute, and QB making 10. If they
start parallel and pointing the same way, how long will it be before they
are again parallel, (i) pointing opposite ways, (11) pointing the same way?
49 GEOMETRY.

THEOREM 16. [Euclid I. 32.]

The three angles of a triangle are together equal to two right
angles. '

Let ABC be a triangle.

It is required to prove that the three L*ABC, BCA, CAB together
=two right angles.
Produce BC to any point D; and suppose CE to be the line
through C parallel to BA.

Proof. Because BA and CE are parallel and AC mecis them,

. the LACE = the alternate 2 CAB.
Again, because BA and CE are parallel, and BD meets them,
‘, the exterior L ECD = the interior opposite Z ABC.
. the whole exterior LACD=the sum of the two interior opposite
L° CAB, ABC.
To each of these equals add the 2 BCA;
then the 2* BCA, ACD together =the three 2° BCA, CAB, ABC.
But the adjacent 2* BCA, ACD together= two right angles,
.". the 2* BCA, CAB, ABC together=two right angles.
Q.E.D.

Obs. In the course of this proof the following most im-

portant property has been established.
If a side of a triangle is produced the exterior angle is equal to
the sum of the two interior opposite angles.
Namely, the ext. ACD =the LCAB+the 2 ABC,
 THE ANGLES OF A TRIANGLE. 43

INFERENCES FROM THEOREM 16.

1. Jf A, B, and C denote the nwinber of degrees in the angles of

a triangle,
then A+B+C=180".
2. If two triangles have two angles of the one respectively equal
to two angles of the other, then the third angle of the one is equal to
the third angle of the other.
3. In any right-angled triangle the two acute angles are conrple-
_ mentary.
4. If one angle of a triangle is equal to the sum of the other
_ two, the triangle is vight-angled.
5. The sum of the angles of any quadrilateral figure is equal to
four right angles.

EXERCISES ON THEOREM 16.

1, Each angle of an equilateral triangle is two-thirds of a right
angle, or 60°,
2. In a right-angled isosceles triangle each of the equal angles
is 45°,
3. Two angles of a triangle are 36° and 123° respectively : deduce
the third angle ; and verify your result by measurement.
4. Ina triangle ABC, the 2B= 111°, the C=42°; deduce the LA,
and verify by measurement.
5. One side BC of a triangle ABC is produced to D. If the extericr
angle ACD is 134°; and the axgle BAC is 42° ; find each of the remaining
interior angles.
6. In the figure of Theorem 16, if the LACD=118°, and the
LB=5lI*, find the 48A and C; and check your results by measurement.
7. Prove that the three angles of a triangle are together equal to two
right angles by supposing a line drawn through the vertex parallel to the
base.
8. If two straight lines are perpendicular to two other neraight lines,
each to each, the acute angle CoS the first pair is equal to the acute
angle hetween the second pair.
44 GEOMETRY.

CoroLLARY 1. All the interior angles of any rectilineal figure,

together with four right angles, are equal to twice as many right

»
angles as the figure has sides.

<
Vase
Let ABCDE be a rectilineal figure of n sides.
It is required to prove that all the interior angles+4 rt. 2°
=2n rt. 2’.
Take any point O within the figure, and join O to each of.
its vertices.
Then the figure is divided into n triangles.
And the three 2* of each A together= 2 rt. 2".
Hence all the 2" of all the A* together= 2n rt. L*.
But all the z* of all the A* make up all the interior angles
of the figure together with the angles at O, which =4 rt. 2.
*. all the int. 2* of the figure +4 rt. 2°=2n rt. 2.
Q.E.D.
Derinition. A regular polygon is one which has all its
sides equal and all its angles equal.
Thus if D denotes the number of degrees in each angle of
a regular polygon of n sides, the above result may be stated
thus :
nD + 360° =n.180°.

EXAMPLE,

Vind the number of degrees in each angle of

(i) a regular hexagon (6 sides) ;
(ii) a regular octagon (8 sides) ;
(iii) a regular decagon (10 sides).
 THE ANGLES OF RECTILINEAL FIGURES. 45

EXERCISES ON THEOREM 16.

(Numerical and Graphical.)

1. ABC is a triangle in which the angles at B and C are re-

spectively double and treble of the angle at A: find the number of
degrees in each of these angles.
2. Express in degrees the angles of an isosceles triangle.in which
(i) Each base angle is double of the vertical angle ;
(ii) Each base angle is four times the vertical angle.
3. The base of a triangle is produced both ways, and the exterior
angles are found to be 94° and 126°; deduce the vertical angle. Con-
struct such a triangle, and check your result by measurement.
4. The sum of the angles at the base of a triangle is 162°, and their
difference is 60°: find all the angles.
5. The angles at the base of a triangle are 84° and 62°; deduce
(i) the vertical angle, (ii) the angle between the bisectors of the base
angles. Check your results by construction and measurement.
6. Ina triangle ABC, the angles at B and C are 74° and 62°; if AB
and AC are produced, deduce the angle between the bisectors of the
exterior angles. Check your result graphically.
7. Three angles of a quadrilateral are respectively 114°, 50°, and
753°; find the fourth angle.
8. In a quadrilateral ABCD, the angles at B, C, and D are re-
spectively equal to 2A, 3A, and 4A; find all the angles.
9. Four angles of an irregular pentagon (5 sides) are 40°, 78°, 122°,
and 135°; find the fifth angle. '
on
10. In any regular polygon of n sides, each angle contains — 2)
right angles. d
(i) Deduce this result from the Enunciation of Corollary 1.
(ii) Prove it independently by joining one vertex A to each of the
others (except the two immediately adjacent to A), thus dividing the
polygon into n —2 triangles.
11. How many sides have the regular polygons each of whose
angles is (i) 108°, (1i) 156°?

12. Shew that the only regular figures which may be fitted together
so as to form a plane surface are (i) equilateral triangles, (ii) squares,
(iii) regular hexagons.
46 GEOMETRY.

CorouLary 2. If the sides of a rectilineal figure, which has

no re-entrant angle, ave produced im order, then all the exterior
ancles so formed are together equal to four right angles.

1st Proof. Suppose, as before, that the figure has m sides;

and consequently n vertices.
Now at each vertex
the interior 2+ the exterior 2=2 rt. 2;
and there are n vertices,
.'. the sum of the izt. 2°+ the sum of the ext. 2° =2n rt. c*.
But by Corollary 1,
the sum of the int. 2°+ 4 rt. L" 2M Tt. £75
.. the sum of the ext. 2°=4 rt. 2°.
Q.E.D.
2nd Proof.

Take any point O, and suppose Oa, Ob, Oc, Od, and Oe, are
lines parallel to the sides marked, A, B, C, D, E (and drawn
from O in the sense in which those sides were produced).
Then the exterior 2 between the sides A and B =the 2aOb.
And the other exterior 2*=the 2*bOc, cOd, dOe, eOa,
respectively.
*, the sum of the ext. 2*’=the sum of the 2’ at O
=4rt. 2’.
 THE ANGLES OF RECTILINEAL FIGURES. 47

EXERCISES.

1. If one side of a regular hexagon is produced, shew that the

exterior angle is equal to the interior angle of an equilateral triangle.
2. Express in degrees the magnitude of each exterior angle of
(i) a regular octagon, (11) a regular decagon.
3. How many sides has a regular polygon if each exterior angle is
(i) 30°, (ii) 24°?
4. Ifa straight line meets two parallel straight lines, and the two
interior angles on the same side are bisected, shew that the bisectors
meet at right angles.
5. If the base of any triangle is produced both ways, shew that the
~

sum of the two exterior angles minus the vertical angle is equal to two
right angles,
6. In the triangle ABC the base angles at B and C are bisected by
BO and CO respectively. Shew that the angle BOC=90° +5.
7. In the triangle ABC, the sides AB, AC are produccd, and the
exterior angles are bisected by BO and CO. Shew that the angle
eat
BOC=90 “5

8. The angle contained by the bisectors of two adjacent angles of

a quadrilateral is equal to half the sum of the remaining angles.
9. Ais the vertex of an isosceles triangle ABC, and BA is produced
to D, so that AD is equal to BA; if DC is drawn, shew that BCD isa
right angle. /
10. The straight line joining the middle point of the hypotenuse
of a right-angled triangle to the right angle is equal to half the
hypotenuse.

EXPERIMENTAL PROOF OF THEOREM 16. [A+B+C=180°.]

In the A ABC, AD is perp. to BC the
greatest side. AD is bisected at right
angles by ZY; and YP, ZQ are perp*. on
BC.
If now the A is folded about the three
dotted lines, the Z£8A, B, and C will coin-
cide with the £4:ZDY, ZDQ, YDP;
*, their sum is 180°.
48 - GEOMETKY.

THEOREM 17. (Euclid I. 26.]

If two triangles have two angles of one equal to two angles of the
other, each to each, and any side of the first equal to the corresponding
side of the other, the triangles are equal in all respects.
2)

B C E F
Let ABC, DEF be twe triangles in which
the LA=the 2D,
the .B=the zE,
also let the side BC =the corresponding side EF.
It is reqiived to prove that the A*ABC, DEF are equal i all
respects,

Proof. The sum of the z°A, B, and C

= 2 rt. 2° Theor. 16.
=the sum of the z"D, E, and F;
and the 2°A and B=the 2*D and E respectively,
., the 2C=the zF.
Apply the AABC to the ADEF, so that B falls on E, and
BC along EF.
Then because BC = EF,
.. C must coincide with F.
And because the 2B=the 2 E, F
*. BA must fall along ED.
And because the 2C =the ZF,
.. CA must fall along FD.
*, the point A, which falls both on ED and on FD, must coin-
cide with D, the point in which these lines intersect.
‘*. the AABC coincides with the A DEF,
and is therefore equal to it in all respects.
So that AB=DE, and AC=DF;
and the AABC =the A DEF in area. Q.E.D.
 CONGRUENT ‘TRIANGLES. 49

EXERCISES.

Ow THE IDENTICAL EquaLity OF TRIANGLES.

1. Shew that the perpendiculars drawn from the extremities of the

base of an isosceles triangle to the opposite sides are equal.
2. Any point on the bisector of an angle is equidistant from the arms
of the angle.
3. Through O, the middle point of a straight line AB, any straight
line is drawn, and perpendiculars AX and BY are dropped upon it from
A and B: shew that AX is equal to BY.
4. If the bisector of the vertical angle of a triangle is at right
angles to the base, the triangle is isosceles.
5. If ina triangle the perpendicular from the vertex on the base
bisects the base, then the triangle is isosceles.
6. If the bisector of the vertical angle of a triangle also bisects the
base, the triangle is isosceles.
[Produce the bisector, and complete the construction after the
manner of Theorem 8. ]

7. ‘The middle point of any straight line which meets two parallel
straight. lines, and is terminated by them, is equidistant from the
parallels.
8. A straight line drawn between two parallels and terminated by
them, is bisected ;shew that any other straight line passing through
the middle point and terminated by the parallels, is also bisected at.
that point.
9. If through a point equidistant from two parallel straight lines,
two straight lines are drawn cutting the parallels, the portions of the
latter thus intercepted are equal.
10. Ina quadrilateral, ABCD, if AB=AD, and BC=DC: shew that
the diagonal AC bisects each of the angles which it joins; and that A
is perpendicular to BD.
ll. A surveyor wishes to ascertain the breadth of a river which he
cannot cross. Standing at a point A near the bank, he notes an object B
immediately opposite on the other bank. He lays down a line AC ofany
length at right angles to AB, fixing a mark at O the middle point of AC.
From C he walks along a line perpendicular to AC until he reaches a point
D from which O and B are seen in the same direction. He now measures
CD: prove that the result gives him the width of the river.

H.S.G. »
50 GEOMETRY.

ON THE IDENTICAL EQUALITY OF TRIANGLES.

Three cases of the congruence of triangles have been dealt
with in Theorems 4, 7, 17, the results of which may be
summarised as follows :
Two triangles are equal in all respects when the following
three parts in each are severally equal:
1. Two sides, and the included angle. Theorem 4.
2. The three sides. Theorem 7.
3. Two angles and one side, the side given in one triangle
CORRESPONDING fo that given in the other. Theorem 17.
Two triangles are not, however, necessarily equal in all
respects when any three parts of one are equal to the corre-
sponding parts of the other.
Tor example :
(i) When the three angles of one are
equal to the three angles of the other,
each to each, the adjoining diagram
shews that the triangles need not be
equal in all respects.
(ii) When two sides and one angle in one are equal to two
sides and one angle of the other, the given angles being opposite
to equal sides, the diagram below shews that the triangles
need not be equal in all respects.
A D

we
BCc E
wee
F’ F

For if AB=DE, and AC=DF, and the 2ABC=the 2 DEF, it

will be seen that the shorter of the given sides in the
triangle DEF may lie in either of the positions DF or DF’.
Notre. From these data it may be shewn that the angles opposite
to the equal sides AB, DE are either equal (as for instance the £* ACB,
DF’E) or supplementary (as the L* ACB. DFE); and that in the former
ease the triangles are equal in all respects. This is called the
ambiguous case in the congruence of eee, [See Problem 9, p. 82.]
If the given angles at B and E are right angles, the ambiguity dis-
appears. This exception is proved in the following Theorem.
 CONGRUENT TRIANGLES. 51

THEOREM 18.
Two right-angled triangles which have their hypotenuses equal,
and one side of one equal to one side of the other, are equal in all
respects.

ee ine
B (CX aot E F

Let ABC, DEF be two right-angled triangles, in which

the L°ABC, DEF are right angles,
the hypotenuse AC =the hypotenuse DF,
and AB = DE.
It is required to prove that the A*ABC, DEF are equal in all
respects.
Proof. Apply the AABC to the ADEF, so that AB falls
on the equal line DE, and C on the side of DE opposite to F.
Let C’ be the point on which C falls.
Then DEC’ represents the A ABC in its new position.
Since each of the 2° DEF, DEC’ is a right angle,
.. EF and EC’ are in one straight line.
- And in the AC'DF, because DF = DC’ (i.e. AC),
, the:2 DFC =the 2 DCF. Theor. 5.
Hence in the A°* DEF, DEC’,
the L DEF =the £2 DEO, being right angles ;
because - the 2 DFE =the 2 DC'E, Proved.
and the side DE is common.
... the A*DEF, DEC’ are equal in all respects; Theor. 17.
that is, the A*DEF, ABC are equal in all respects.
Q.E.D.
52 GEOMETRY.

*THEOREM 19. [Euclid I. 24.]

Tf two triangles have two sides of the one equal to two sides of the
other, cach to each, but the angle included by the two sides of one
greater than the angle included by the corresponding sides of
the other; then the base of that which has the greater angle is
greater than the base of the other.
A D

B GY -E KG
F

Let ABC, DEF be two triangles, in which

BA=ED,
and AC = DF,
but the 2 BAC is greater than the 2 EDF.
It is required to prove that the base BC is greater than the
base EF.
Proof. Apply the AABC to the A DEF, so that A falls on D,
and AB along DE.
Then because AB= DE, B must coincide with E.
Let DG, GE represent AC, CB in their new position.
Then if EG passes through F, EG is greater than EF ;
that is, BC is greater than EF.
But if EG does not pass through F, suppose that DK bisects
the LFDG, and meets EGin K. Join FK.
Then in the A*FDK, GDK,
FD=GD,
because DK is common to both,
and the included 2 FDK =the included 2 GDK;
*, FK=QK. Theor. 4.
Now the two sides EK, KF are greater than EF ;
that is, EK, KG are greater than EF.
‘, EG (or BC) is greater than EF. Q.E.D,
 CONVERSE OF THEORLM 19, 53

Conversely, if two triangles have two sides of the one equal to two
sides of the other, each to each, but the base of one greater than the
base of the other ; then the angle contained by the sides of that which
has the greater base, is greater than the angle contained by the
corresponding sides of the other.

A D

B C e
rm

Let ABC, DEF be two triangles in which.

BA=ED,
and AC= DF,
but the base BC is greater than the base EF.
It ts required to prove that the L BAC 1s greater than the L EDF.
Proof. If the 2 BAC is not greater than the 2 EDF,
it must be either equal to, or less than the 2 EDF.
Now if the 2 BAC were equal to the z EDF,
then the base BC would be equal to the base EF; Theor. 4.
but, by hypothesis, it is not.
Again, if the 2 BAC were less than the 2 EDF,
then the base BC would be less than the base EF; Theor. 19.
but, by hypothesis, it is not.
That is, the 2 BAC is neither equal to, nor less than the 2 EDF;
*, the 2 BAC is greater than the 2 EDF.
Q.E.D.
* Theorems marked with an asterisk may be omitted or postponed at the
discretion of the teacher.
54 GEOMETRY.

REVISION LESSON ON TRIANGLES.

1. State the properties of a triangle relating to

(1) the sum of its interior angles;
(ii) the sum of its exterior angles,
What property corresponds to (i) in a polygon of n sides? With
what other figures does a triangle share the property (ii)?

2. Classify triangles with regard to their angles. Enunciate any

Theorem or Corollary assumed in the classification.

3. Enunciate two Theorems in which from data relating to the sides

a conclusion is drawn relating to the angles.
In the triangle ABC, if a=3°6 em., b=2°8 em., c=3°6 em., arrange
the angles in order of their sizes (before measurement) ; and prove that
the triangle is acute-angled.

4. Enunciate two Theorems in which from data relating to the

angles a conclusion is drawn relating to the sides.
In the triangle ABC, if
(i) A=48° and B=5I’, find the third angle, and name the greatest
side.
(ii) A=B=62}°, find the third angle, and arrange the sides in order
of their lengths,

5. From which of the conditions given below may we conclude that

the triangles ABC, A’B’C’ are identically equal? Point out where
ambiguity arises ; and draw the triangle ABC in each case.
A=A'=71°. a= a'=4'2 cm. A=A’=36".
(i) 4B=B’=46", (ii) 1b= b'=2-40em. (iii) 1B=B’=121°.
| a=a' =3°7 em. C=C’=81*. C=C’=23".

@= a'=3:0 om B=B’=53°. C=C’=90°.

(iv) < b= b'=5°2 om (v) 4 b=0’ =4:3.0m. (vi) 4 c= c'=6cm,
c= c'=4'5 cm c=c’ =5'0 om. a=a'=38ecm,

6. Summarise the results of the last question by stating generally

under what conditions two triangles
(i) are necessarily congruent ;
(ii) may or may not be congruent.

7. If two triangles have their angles equal, each to each, the triangles
are not necessarily equal in all respects, because the three data are not
independent. Carefully explain this statement.
 EXERCISES ON TRIANGLES. 5D

(Miscellaneous Examples.)

_ 8. (i) The perpendicular is the shortest line that can be drawn to a

given straight line from a given point.
(ii) Obliques which make equal angles with the perpendicular are
equal.
(ili) Of two obliques the less is that which makes the smaller anyle with
the perpendicular.

9. If two triangles have two sides of the one equal to two sides of the
other, each to each, and have likewise the angles opposite to one pair of
equal sides equal, then the angles opposite to the other pair of equal sides
are either equal or supplementary, and in the former case the triangles are
equal in all respects.

10. PQ is a perpendicular (4 cm. in length) to a straight line XY.

Draw through P a series of obliques making with PQ the angles 15°,
30°, 45°, 60°, 75°. Measure the lengths of these obliques, and tabulate
the results.
11. PAB is a triangle in which AB and AP have constant lengths
4 cm. and 3 cm. If AB is fixed, and AP rotates about A, trace the
changes in PB, as the angle A increases from 0° to 180°.
Answer this question by drawing a series of figures, increasing A by
increments of 30°. Measure PB in each case, and tabulate the results.

12. From B the foot of a flagstaff AB a horizontal line is drawn

passing two points C and D which are 27 feet apart. The angles BCA’
and BDA are 65° and 40° respectively. Represent this on a diagram
(scale 1 em. to 10 ft.), and find by measurement the approximate height
of the flagstaff.

13. From P, the top of a lighthouse PQ, two boats A and B are
seen at anchor ina line due south of the lighthouse. It is known that
PQ =126 ft., LPAQ=57°, LPBQ=33°; hence draw a plan in which
1” represents 100 ft., and find by measurement the distance between A
and B to the nearest foot.

14. From a lighthouse L two ships A and B, which are 600 yards
apart, are observed in directions §.W. and 15° East of South respec-
tively. At the same time B is observed from A in a 8.E. direction.
Draw a plan (scale 1” to 200 yds.), and find by measurement the distance
of the lighthouse from each ship.
56 GEOMETRY.

PARALLELOGRAMS.

DEFINITIONS,

1. A quadrilateral is a plane figure bounded

by four straight lines.
The straight line which joins opposite angular
points in a quadrilateral is called a diagonal.

2. A parallelogram is a quadrilateral
whose opposite sides are parallel.
{It will be proved hereafter that the opposite ee oe
sides of a parallelogram are equal, and that its
opposite angles are equal. }

3. A rectangle is a parallelogram which

has one of its angles a right angle. cose
[It will be proved hereafter that a// the angles of ‘
a rectangle are right angles. See page 59.]

4, A square is a rectangle which has two

adjacent sides equal.
[It will be proved that all the sides of a square are
equal and all its angles right angles. See page 59.]

5. A rhombus is a quadrilateral which

has all its sides equal, but its angles are
not right angles.

6. A trapezium is a quadrilateral which has

one pair of parallel sides.
 PARALLELOGRAMS. 57

THEOREM 20. [Euclid I. 33.]

The straight lines which join the extremities of two equal and
parallel straight lines towards the same parts are themselves equal
and parallel.
A B

Cc D

Let AB and CD be equal and parallel straight lines; and let

them be joined towards the same parts by the straight lines
AC and BD.
It is required to prove that AC and BD are equal and parallel.
Join BC.
Proof. Then because AB and CD are parallel, and BC meets
them,
.. the ABC =the alternate 2 DCB.
Now in the A* ABC, DCB,
AB=DC,
because + BC is common. to both ;
and the .ABC=the 2 DCB; Proved.
—— :, the triangles are equal in all respects ;
sor that AC = DES Satta Nauin peer cert: (i)
and the 4 ACB=Z DBC.
But these are alternate angles ;
set AG. and BDrarOWarallel, Jee. cseson. sense (11)
That is, AC and BD are both equal and parallel. |
Q.E.D.
58 GEOMETRY.

THEOREM 21. [Euclid I. 34.]

The opposite sides and angles of a parallelogram are equal to one
another, and each diagonal bisects the parallelogram.
A B

D Cc
Let ABCD be a parallelogram, of which BD is a diagonal.
It is required to prove that
(i) AB=CD, and AD=CB,
(ii) the L BAD =the 2 DCB,
(ili) the LADC =the L CBA,
(iv) the AABD=the ACDB tn area.

Proof. Because AB and DC are parallel, and BD meets them,

*. the ABD = the alternate 2 CDB.
Again, because AD and BC are parallel, and BD meets them,
.. the ADB =the alternate 2 CBD.
Hence in the A* ABD, CDB,
the ABD =the 2 CDB,
. because |the 2 ADB =the 2 CBD, Proved.
lana BD is common to both ;
.. the triangles are equal in all respects; = Theor. 17,
so that AB= OD, etd AD SOBs viicescscercceecvsees (i)
and tho LBAD =the ZLDOB; ...t.cccccsssevsecneases (ii)
and the AABD =the ACDB in area. .............5 (iv)
And because the 2 ADB=the 2 CBD, Proved.
and the 2 CDB =the 2 ABD,
.. the whole ADC =the whole 2 CBA.......... (iii)
Q.E.D.
 PARALLELOGRAMS. 59

CoroLuary 1. Jf one angle of a parallelogram is a right

angle, all ats angles are right angles.
In other words :
All the angles of a rectangle are right angles.
For the sum of two consecutive £§=2 rt. £*; (Theor, 14.)
“., if one of these is a rt. angle, the other must be a rt. angle.
And the opposite angles of the par™ are equal ;
‘. all the angles are right angles.

CoROLLARY 2. All the sides of a square are equal; and all

us angles are right angles.
CoROLLARY 3. The diagonals of a parallelogram bisect one
another. D C
‘Let the diagonals AC, BD of the par™
ABCD intersect at O.
To prove AO=OC, and BO=OD. ! /
In the AS AOB, COD, A B
the 2OAB=the alt. 2OCD,
because + the ~AOB=vert. opp. 2 COD,
and AB=the opp. side CD;
TLOA—OC sand OOD) Theor. 17

EXERCISES.

1. If the opposite sides ofa quadrilateral are equal, the figure is a

parallelogram.
2. If the opposite angles of a quadrilateral are equal, the figure is a
parallelogram.
3. If the diagonals of a quadrilateral bisect each other, the figure is a
parallelogram.
4. The diagonals of a rhombus bisect one another at right angles.
5. If the diagonals of a parallelogram are equal, all its angles are
right angles.
6. In a parallelogram which is not rectangular the diagonals are
unequal.
 60 GEOMETRY,

EXERCISES ON PARALLELS AND PARALLELOGRAMS.

(Symmetry and Swperposition.)

1. Shew that by folding a rhombus about one of its diagonals the
triangles on opposite sides of the crease may be made to coincide.
That is to say, prove that a rhombus is symmetrical about either
diagonal.

2. Prove that the diagonals of a square are axes of symmetry. Name

two other lines about which a square is symmetrical.

3. The diagonals of a rectangle divide the figure into two congruent

triangles: is the diagonal, therefore, an axis of symmetry? About
what two lines is a rectangle symmetrical?

4, Is there any axis about which an oblique parallelogram is sym-

metrical? Give reasons for your answer.

5. Ina quadrilateral ABCD, AB=AD and CB=CD;; but the sides

are not all equal. Which of the diagonals (if either) is an axis of
symmetry ? ;

6. Prove by the method of superposition that

(i) Two parallelograms are identically equal if two adjacent sides of
one are equal to two adjacent sides of the other, each to each, and one
angle of one equal to one angle of the other.
(ii) Two rectangies are equal if two adjacent sides of one are equal to
two adjacent sides of the other, each to each.

7. Two quadrilaterals ABCD, EFGH have the sides AB, BC, CD, DA
equal respectively to the sides EF, FG, GH, HE, and have also the
angle BAD equal to the angle FEH. Shew that the figures may be
made to coincide with one another.

(Miscellaneous Theoretical Examples.)

8. Any straight line drawn through the middle point of a diagonal

of a parallelogram and terminated by a pair of opposite sides, is bisected
at that point.

9. In a parallelogram the perpendiculars drawn from one pair of

opposite unites to the diagonal which joins the other pair are equal.

10. If ABCD is a parallelogram, and X, Y respectively the middle

points of the sides AD, BC; shew that the figure AYCX is a paral-
lelogram,
 PARALLELS AND PARALLELOGRAMS. 61

11. ABC and DEF are two triangles such that AB, BC are respec-
tively equal to and parallel to DE, EF; shew that AC is equal and
parallel to DF.
12. ABCD is a quadrilateral in which AB is parallel to DC, and AD
equal but not parallel to BC; shew that
(i) the 2A+the LC=180°=the LB+the 2D;
(ii) the diagonal AC=the diagonal BD ;
(iii) the quadrilateral is symmetrical-about the straight line joining
the middle points of AB and DC.
13. AP, BQ are straight rods of equal length, turning at equal
rates (both clockwise) about two fixed pivots A and B respectively. If
the rods start parallel but pointing in opposite senses, shew that
(i) they will always be parallel ;
(ii) the line joining PQ will always pass through a certain fixed
point.

(Miscellaneous Numerical and Graphical Luanyples.)

14. Calculate the angles of the triangle ABC, having given:

int. LA=# of ext. LA; 3B=4C.
15. A yacht sailing due East changes her course successively by 63°,
by 78°, by 119°, and by 64°, with a view to sailing round an island.
What further change must be made to set her once more on an Easterly
course ?
16. If the sum of the interior angles of a rectilineal figure is equal
to the sum of the exterior angles, how many sides has it, and why?
17. Draw, using your protractor, any five-sided figure ABCDE,
in which
ABE, -fC=lIb>, LD=93), LE=152-.
Verify by a construction with ruler and compasses that AE is parallel
to BC, and account theoretically for this fact.
18. A and B are two fixed points, and two straight lines AP, BQ,
unlimited towards P and Q, are pivoted at Aand B. AP, starting from
the direction AB, turns about A clockwise at the uniform rate of 74° a
second ; and BQ, starting simultaneously from the direction BA, turns
about B counter-clockwise at the rate of 32° a second.
(i) How many seconds will elapse before AP and BQ are parallel?
(ii) Find graphically and by calculation the angle between AP and
BQ twelve seconds from the start.
(iii) At what rate does this angle decrease?
62 GEOMETRY.

THEOREM 22. .
If there are three or more parallel straight lines, and the intercepts
made by them on any transversal are equal, then the corresponding
mtercepts on any other transversal are also equal.

Let the parallels AB, CD, EF cut off equal intercepts PQ, QR
from the transversal PQR; and let XY, YZ be the corresponding
intercepts cut off from any other transversal XYZ.

It is required to prove that XY = YZ.

Through X and Y let XM and YN be drawn parallel to PR.
Proof. Since CD and EF are parallel, and XZ meets them,
.. the .XYM=the corresponding 2 YZN.
And since XM, YN are parallel, each being parallel to PR,
‘, the 2 MXY =the corresponding 2 NYZ.
Now the figures PM, QN are parallelograms,
.. XM=the opp. side PQ, and YN=.the opp. side QR;
and since by hypothesis PQ = QR,
. XM=YN.,

Then in the A*XMY, YNZ,

the 2.XYM=the 2 YZN,
because {the 2 MXY =the 2 NYZ,
and XM = YN ;
.‘, the triangles are identically equal; Theor, 17.
“. XY=YZ.
Q.E.D.
 PARALLELS AND PARALLELOGRAMS, 63

COROLLARY. Ina triangle ABC, if a set of lines Pp, Q@, Rr, ...,
drawn parallel to the base, divide one side AB into equal parts, they
also divide the other side AC into equal parts.

The lengths of the parallels Pp, Qq, Rr, ..., may thus be expressed
in terms of the base BC.
Through p, qg, and r let pl, q2, 73 be drawn par! to AB.
Then, by Theorem 22, these par’ divide BC into four equal parts, of
which Pp evidently contains one, Qq two, and Rr three.
In other words,
1 2 3
Ep BC; Qq=7- BC; ee BC.

Similarly if the given par’ divide AB into x equal parts,

Pp=--- BC, Qq=" - BC, Rr=2 - BC; and so on.

** Problem 7, p. 78, should now be worked.

DEFINITION.

If from the extremities of a straight line AB perpendiculars

Ax, BY are drawn to a straight line PQ of indefinite length,
then XY is said to be the orthogonal projection of AB on PQ.
B
A
 64 ’ GEOMETRY.

EXERCISES ON PARALLELS AND PARALLELOGRAMS.

/ 1. The straight line drawn through the middle point of a side of a

|triangle, parallel to the base, bisects the remaining side.

| [This is an important particular case of A

| Theorem 22.
In the AABC, if Z is the middle point of
_ AB, and ZY is drawn par! to BC, we have to Z iY
prove that AY=YC. ‘
Draw YX par! to AB, and then prove the
As ZAY, XYC congruent. ] B x Cc
2. The straight line which joins the A
middle points of two sides of a triangle is
parallel to the third side.
{In the A ABC, if Z, Y are the middle Z Y
points of AB, AC, we have to prove ZY
par! to BC.
Produce ZY to V, making YV equal to
ZY, and join CV. Prove the A*AYZ, 8B re)
CYV congruent ; the rest follows at once.]

3. The straight line which joins the middle points of two sides of a
triangle is equal to half the third side.

4. Shew that the three straight lines which join the middle points
of the sides of a triangle, divide it into four triangles which are wdenti-
cally equal.

5. Any straight line drawn from the vertex of a triangle to the base
is bisected by the straight line which joins the middle points of the other
sides of the triangle.

6. ABCD is a parallelogram, and X, Y are the middle points of

a opposite sides AD, BC: shew that BX and DY trisect the diagonal
AC. ;

7. If the middle points of adjacent sides of any quadrilateral are

joined, the figure thus formed is a parallelogram.

8. Shew that the straight lines which join the middle points of
opposite sides of a quadrilateral, biseet one another.
 PARALLELS AND PARALLELOGRAMS. 65

9. From two points A and B, and from O the mid-point between

them, perpendiculars AP, BQ, OX are drawn to a straight line CD.
If AP, BQ measure respectively 4:2 cm. and 5’8 cm., deduce the length
of OX, and verify your result by measurement.
Shew that OX=3(AP+BQ) or 4(AP~BQ), according as A and B
are on the same side, or on opposite sides of CD.

10. When three parallels cut off equal intercepts from two trans-
versals, shew that of the three parallel lengths between the two
transversals the middle one is the Arithmetic Mean of the other two.

ll. The parallel sides of a trapezium are a centimetres and b centt-

metres in length. Prove that the line joining the middle points of the
oblique sides is parallel to the parallel sides, and that its length is
(a+b) centimetres.

12. OX and OY are two straight lines, and along OX five points
1, 2, 3, 4, 5 are marked at equal distances. Through these points
parallels are drawn in any direction to meet OY. Measure the lengths
of these parallels: take their average, and compare it with the length
of the third parallel. Prove geometrically that the 3'4 parallel is the
mean of all five.
State the corresponding theorem for any odd number (2n+1) of
parallels so drawn.

13. From the angular points of a parallelogram perpendiculars

are drawn to any straight line which is outside the parallelogram:
shew that the sum of the perpendiculars drawn from one pair of
opposite angles is equal to the sum of those drawn from the other pair.
[Draw the diagonals, and from their point of intersection suppose
a perpendicular drawn to the given straight line.]

14. The sum of the perpendiculars drawn from any point in the
base of an isosceles triangle to the equal sides is equal to the perpen-
dicular drawn from either extremity of the base to the opposite side.
[It follows that the sum of the distances of any point in the base
of an isosceles triangle from the equal sides is constant, that is, the
same whatever point in the base is taken. ]
How would this property be modified if the given point were taken
in the base produced?

15. The sum of the perpendiculars drawn from any point within
an equilateral triangle to the three sides is equal to the perpendicular
drawn from any one of the angular points to the opposite side, and is
therefore constant.

16. Equal and parallel lines have equal projections on any other
straight line.
H.S.G. u
66 GEOMETRY.

DIAGONAL SCALES.

Diagonal scales form an important application of Theorem 22.

We shall illustrate their construction and use by describing a -
Decimal Diagonal Scale to shew Inches, Tenths, and Hundredths.
A straight line AB is divided (from A) into inches, and the
points of division marked 0, 1, 2,..... The primary division
0A is subdivided into tenths, these secondary divisions being
numbered (from 0) 1, 2, 3,...9. We may now read on AB
mches and tenths of an inch.

A98765438210 : 1 2 as

In order to read hundredths, ten lines are taken at any equal

intervals parallel to AB; and perpendiculars are drawn through
o£ |...
The primary (or inch) division corresponding to 0A on the
tenth parallel is now subdivided into fen equal parts; and
diagonal lines are drawn, as in the diagram,
joining 0 to the first point of subdivision on the 10" parallel.
» 1 to the second ,, ” 9 ’ ;
» 2tothe third ,, 8 » ” ;
and so on.
The scale is now complete, and its use is shewn in the
following example.
Example. To take from the scale a length of 2°47 inches.
(i) Place one point of the dividers at 2 in AB, and extend them till
the other point reaches 4 in the subdivided inch 0A. We have now
2°4 inches in the dividers.
(ii) To get the remaining 7 hundredths, move the right-hand point
up the perpendicular through 2 till it reaches the 7" parallel. Then
extend the dividers till the left point reaches the diagonal 4 also on the
7 parallel. We have now 2°47 inches in the dividers.
 DIAGONAL SCALES.-~ 67

REASON FOR THE ABOVE PROCESS,

The first step needs no explanation. The reason of the

second is found in the Corollary of Theorem 22.
Joining the point 4 to the corresponding point _5 4
on the tenth parallel, we have a triangle 4,4,5,
of which one side 4,4 is divided into ten equal
parts by a set of lines parallel to the base 4,5.
Therefore the lengths of the parallels between
4,4, and the diagonal 4,5 are =y, 3%, 33;,--. of the
base, which is ‘1 inch. :
Hence these lengths are respectively
Ol Gz, 208,2.o8 | inch, a aa
Similarly, by means of this scale, the length of a given
straight line may be measured to the nearest hundredth of an
inch. .

Again, if one inch-division on the scale is taken to represent

10 feet, then 2°47 inches on the scale will represent 24°7 fect.
And if one inch-division on the scale represents 100 links, then
2°47 inches will represent 247 links. Thus a diagonal scale is
of service in preparing plans of enclosures, buildings, or field-
works, where it is necessary that every dimension of the actual
object must be represented by a line of proportional length on
‘the plan.

NOTE.

The subdivision of a diagonal scale need not be decimal.

For instance we might construct a diagonal scale to read centimetres,
millimetres, and quarters of a millimetre; in which case we should take
four parallels to the line AB.

[For Exercises on Linear Measurements see the following page.]

68 GEOMETRY.

EXERCISES QN LINEAR MEASUREMENTS

1. Draw straight lines whose lengths are 1:25 inches, 2°72 inches,
3°08 inches.
2. Draw a line 2°68 inches long, and measure its length in centi-
metres and the nearest millimetre.
3. Draw a line 5°7 cm. in length, and measure it in inches (to the
nearest hundredth). Check your result by calculation, given that
1 em. =0°3937 inch.
4. Find by measurement the equivalent of 3°15 inches in centi-
metres and millimetres. Hence calculate (correct to two decimal
places) the value of 1 em. in inches.
5. Draw lines 29 em. and 6:2 cm. in length, and measure them in
inches. Use each equivalent to find the value of 1 inch in centimetres
and millimetres, and take the average of your results.
6. A distance of 100 miles is represented on a map by 1 inch,
Draw lines to represent distances of 336 miles and 408 miles.
7. If 1 inch on a map represents 1 kilometre, draw lines to represent
850 metres, 2980 metres, and 1010 metres.

8. A plan is drawn to the scale of 1 inch to 100 links. Measure in

centimetres and millimetres a line representing 417 links.
9. Find to the nearest hundredth of an inch the length of a line
which will represent 42°500 kilometres in a map drawn to the scale of
1 centimetre to 5 kilometres.
10. The distance from London to Oxford (in a direct line) is
55 miles. If this distance is represented on a map by 2°75 inches, to
what scale is the map drawn? That is, how many miles will be
represented by 1 inch? How many kilometres by 1 centimetre?
{1 em. =0°3937 inch; 1 km. =§ mile, nearly.]

11. On a map of France drawn to the scale 1 inch to 35 miles, the

distance from Paris to Calais is represented by 42 inches. Find the
distance accurately in miles, and approximately in kilometres, and
express the scale in metric measure. [1 km.=§ mile, nearly.]
12. The distance from Exeter to Plymoyth is 874 miles, and
appears on a certain map to be 24”; and the distance from Lincoln to
York is 88 km., and appears on another map to be 7 em. Compare the
scales of these maps in miles to the inch.
13. Draw a diagonal scale, 2 centimetres to represent 1 yard,
shewing yards, feet, and inches.
 PRACTICAL GEOMETRY. 69

PRACTICAL GEOMETRY.

PROBLEMS.

The following problems are to be solved with ruler and

compasses only. No step requires the actual measurement
of any line or angle; that is to say, the constructions are to be
made without using either a graduated scale of length, or a
protractor.

The problems: are not merely to be studied as propositions ;

but the construction in every case is to be actually performed
by the learner, great care being given to accuracy of drawing.
Each problem is followed by a theoretical proof; but the
results of the work should always be verified by measurement,
as a test of correct drawing. Accurate measurement is also
required in applications of the problems.
In the diagrams of the problems lines which are inserted
only for purposes of proof are dotted, to distinguish them
from lines necessary to the construction.
For practical applications of the problems the student
should be provided with the following instruments:
1. A flat ruler, one edge being graduated in centimetres
and millimetres, and the other in inches and tenths.
2. Two set squares; one with angles of 45°, and the other
with angles of 60° and 30°.
3. <A pair of pencil compasses.
4. <A pair of dividers, preferably with screw adjustment.
5. A semi-circular protractor.
70 GEOMETRY.

PROBLEM 1.

To bisect a given angle.

Let BAC be the given angle to be bisected.

Construction. With centre A, and any radius, draw an
are of a circle cutting AB, AC at P and Q.
With centres P and Q, and radius PQ, draw two ares cutting
at O. -Join AO.
Then the 2 BAC is bisected by AO.

Proof. Join PO, QO.

In the A* APO, AQO,
AP=AQ, being radii of a circle,
because PO=Q0, , ss equal circles,
and AO is common ;
. the triangles are equal in all respects; Theor. 7.
so that the 2 PAO =the 2 QAO ;
that is, the 2 BAC is bisected by AO.
Notre. PQ has been taken as the radius of the ares drawn from the
centres P and Q, and the intersection of these ares determines the
point O. Any radius, however, may be used instead of PQ, provided
that it is great enough to secure the intersection of the ares.
 PROBLEMS ON LINES AND ANGLES. 71

PROBLEM 2.

Lo bisect a given straight line.

Let AB be the line to be bisected.

Construction. With centre A, and radius AB, draw twe
ares, one on each side of AB.
With centre B, and radius BA, draw two arcs, one on each
side of AB, cutting the first arcs at P and Q.
Join PQ, cutting AB at O.
Then AB is bisected at O.
Proof. Join AP, AQ, BP, BQ.
In the A* APQ, BPQ,
[ AP =BP, being radii of equal circles,
because AQ= BQ, for the same reason,
lana PQ is common ;
-, the ,APQ=the z BPQ. Theor. 7.
Again in the A* APO, BPO,
AP = BP,
because PO is common,
and the ~APO=the 2 BPO;
nO OB. Theor. 4.
that is, AB is bisected at O.
Notes. (i) AB was taken as the radius of the arcs drawn from the
centres A and B, but any radius may be used provided that it is great
enough to secure the intersection of the arcs which determine the points
P and Q.
(ii) From the congruence of the AsAPO, BPO it follows that the
LAOP=the CBOP. As these are adjacent angles, it follows that PQ
bisects AB at right angles.
72 GEOMETRY.

PROBLEM 3.

To draw a straight line perpendicular to a given straight line at

a given point in tt.

Let AB be the straight line, and X the point in it at which

a perpendicular is to be drawn.
Construction. With centre X cut off from AB any two
equal parts XP, XQ.
With centres P and Q, and radius PQ, draw two arcs cutting
at O. 4
Join XO.
Then XO is perp. to AB.

Proof. Join OP, OQ.

In the A* OXP, OXQ,
XP = XQ, by construction,
because OX is common,
and PO = QO, being radii of equal circles;
*, the 2OXP=the 2 OXQ. Theor. 7.
And these being adjacent angles, each is a right angle ;
that is, XO is perp. to AB.

Obs. If the point X is near one end of AB, one or other of

the alternative constructions on the next page should be used
 PROBLEMS ON LINES AND ANGLES, 73

PROBLEM 3. SECOND METHOD.

Construction. Take any point C ©)

outside AB.
With centre C, and radius CX, draw
a circle cutting AB at D. C.
Join DC,-and produce it to meet
the circumference of the circle at O.
Join XO. x
WO
Then XO is perp. to AB. Bae a

Proof. Join CX.

Because CO=CX; ..:.. the £CXO=the 2 COX;
and because CD—=CX; ..°. the ~CXD=the 2CDxX.
*, the whole 2DXO=the 2 XOD + the 2 XDO
= of 180°
= SI,
*, XO is perp. to AB.

PropteEM 3. Turrp METHOD.

Construction. With centre X

and any radius, draw the arc CDE,
cutting AB at C.
With centre C, and with the D E
same radius, draw an are, cutting
the first arc at D.
With centre D, and with the
same radius, draw an arc, cut-
ting the first are at E. BY Ge xX B
Bisect the 2 DXE by XO. Prob. 1.
Then XO is perp. to AB.
Proof. Hach of the 2*CXD, DXE may be proved to be 60°;
and the 2 DXO is half of the 2 DXE;
the L CXO is 90°.
That is, XO is perp. to AB.
74 GEOMETRY.

PROBLEM 4.

To draw a straight line perpendicular to a given straight line

from a given external point.

Let X be the given external point from which a perpen-

dicular is to be drawn at AB.
Construction. ‘Take any point C on the side of AB remote
from X.
With centre X, and radius XC, draw an arc to cut AB at P
and Q.
With centres P and Q, and radius PX, draw ares cutting at
Y, on the side of AB opposite to X.
Join XY cutting ‘AB at O.
Then XO is perp. to AB.

Proof. Join PX, QX, PY, QY,

In the A* PXY, QXY,
PX = QX, being radii of a cirele,
because +PY = QY, for the same reason,
and XY is common ; a
.. the 2 PXY =the 2 QXY. Theor. 7.
Again, in the A* PXO, QXO,
PX = QX,
because XO is common,
and the 2 PXO=the 2 Qx0O;
. the 2XOP=the 2 XOQ. Theor. 4:
And these being adjacent angles, each is a right angle,
that is, XO is perp. to AB.
 PROBLEMS ON LINES AND ANGLES. 15

Obs. When the point X is nearly opposite one end of AB,

one or other of the alternative constructions given below
should be used.

PROBLEM 4. SECOND METHOD.

Construction. Take any point D in x

AB. Join DX, and bisect it at C. cC
With centre C, and radius CX, draw
a circle cutting AB at D and O.
Join XO. AD Ons
Then XO is perp. to AB.
_ For, as in Problem 3, Second Method, the 2 XOD is a right
angie.

Propuem 4. TuHirpd METHOD.

Construction. Take any two points

D and E in AB.
With centre D, and radius DX, draw
an arc of a circle, on the side of AB
opposite to X.
With centre E, and radius EX, draw
another are cutting the former at Y.
Join XY, cutting AB at O.
Then XO is perp. to AB.
(i) Prove the A*XDE, YDE equal
in all respects by Theorem 7,
so that the 2XDE =the ZYDE.
(ii) Hence prove the A*XDO, YDO equal in all respects
by Theorem 4, so that the adjacent L* DOX, DOY are equal.
That is, XO is perp. to AB.
76 GEOMETRY.

PROBLEM 5.

At a given Len in a given straight line to make an angle equal

fo a given angle.

ae
Let BAC be the given angle, and FG the given straight line;
and let O be the point at which an angle is to be made equal
to the 2 BAC.
Construction. With centre A, and with any radius, draw
an are cutting AB and AC at D and E.
With centre O, and with the same radius, draw an arc
cutting FG at Q.
With centre Q, and with radius DE, draw an are cutting the
former are at P.
Join OP.
Then POQ is the required angle.

Proof. Join ED, PQ.

In the A*POQ, EAD,
OP =AE, being radii of equal circles,
because {oa=n0 for the same reason,
PQ =ED, by construction ;
*. the triangles are equal in all respects;
so that the 2 POQ =the 2 EAD, Theor, 7
 PROBLEMS ON LINES AND ANGLES. 77

PROBLEM 6

Through a given point to draw a straight line parallel to a given

straight line.

B fe)

on

Let XY be the given straight line, and O the given point,

through which a straight line is to be drawn par! to XY.
Construction. In XY take any point A, and join OA.
Using the construction of Problem 5, at the point O in
the line AO make the LAOP equal to the Z OAY and alternate
to it.
Then OP is parallel to XY.

Proof. Because AO, meeting the straight lines OP, XY,

makes the alternate L* POA, OAY equal ;
2. OP is par to XY.

** ~The constructions of Problems 3, 4, and 6 are not usually

followed im practical applications. Parallels and perpendiculars
may be more quickly drawn by the aid of set squares, (See LESSONS
IN EXPERIMENTAL GEOMETRY, pp. 36, 42.)
78 GEOMETRY.

PROBLEM 1,
To divide a given straight line into any number of equal parts.

Let AB be the given straight line, and suppose it is required

to divide it into jive equal parts.
Construction. From A draw AC, a straight line of unlimited
Jength, making any angle with AB.
From AC mark off five equal parts of any length, AP, PQ,
QR, RS, ST.
Join TB; and through P, Q, R, S draw par" to TB, meeting
AB in p, 4, 7; 8
Then since the par™ Pp, Qy, Rr, Ss, TB cut off five equal parts
from AT, they also cut off five equal parts from AB.
(Theorem 22.)

SECOND METILOD.

From A draw AC at any angle with C

AB, and on it mark off fowr equal parts
AP, PQ, QR, RS, of any length.
From B draw BD par' to AC, and on
it mark off BS’, S’R’, R’Q’, Q’P’, each
equal to the parts marked on AC. A?
Join PP’, QQ’, RR’, SS’ meeting AB
in p, 7, 7, 8 Then AB is divided into
five equal parts at these points.
[Prove by Theorems 20 and 22.] D
 PROBLEMS. 79

EXERCISES ON LINES AND ANGLES.

(Graphical Haxercises.)
1. Construct (with ruler and compasses only) an angle of 60°.
By repeated bisection divide this angle into fowr equal parts.
2. By means of Exercise 1, trisect a right angle; that is, divide it
into three equal parts.
Bisect each part, and hence shew how to trisect an angle of 45°.
[No construction is known for exactly trisecting any angle.]
3. Draw a line 6°7 cm. long, and divide it into five equal parts.
Measure one of the parts in inches (to the nearest hundredth), and verify
your work by calculation. [1 em. =0°3937 inch.]
4. From a straight line .3°72’ long, cut off one seventh. Measure
the part in centimetres and the nearest millimetre, and verify your
work by calculation. :
5. Ata point X in a straight line AB draw XP perpendicular to AB,
making XP 1°8” in length. From P draw an oblique PQ, 3:0” long,
to meet AB in Q. Measure XQ.

(Problems. State your construction, and give a theoretical proof.)

6. In a straight line XY find a point which is equidistant from two.

given points A and B.
When is this impossible”

7. Inastraight line XY find a point which is equidistant from two,

intersecting lines AB, AC. —
When is this impossible?

8. From a given point P draw a straight line PQ, making with a

given straight line AB an angle of given magnitude.

9. From two given points P and Q on the same side of a straight

line AB, draw two lines which meet in AB and make equal angles
with it.
[Construction. From P draw PH perp. to AB, and produce PH to P’,
making HP’ equal to PH. Join P’Q cutting AB at K. Join PK.
Prove that PK, QK are the required lines.]

10. Through a given point P draw a straight line such that the
perpendiculars drawn to it from two points A and B may be equal.
Is this always possible?
80 GEOMETRY.

THE CONSTRUCTION OF TRIANGLES.

PROBLEM 8.

To draw a triangle having given the lengths of the three sides.

Let a, 6, ¢ be the lengths to which the sides of the required

triangle are to be equal. ;
Construction. Draw any straight line BX, and cut off from
it a part BC equal to a.
With centre B, and radius c, draw an are of a circle.
With centre C, and radius }, draw a second are cutting the
first at A.
Join AB, AC.
Then ABC is the required triangle, for by construction the
sides BC, CA, AB are equal to a, b, ¢ respectively.

Obs. The three data a, b, c may be understood in two

ways: either as three actual lines to which the sides of the
triangle are to be equal, or as three numbers expressing the
lengths of those lines in terms of inches, centimetres, or some
other linear unit.

Norrs. (i) In order that the construction may be possible it is

necessary that any two of the given sides should be together greater
than thé third side (Theorem 11); for otherwise the ares drawn from
the centres B and C would not eut.
(ii) The ares which cut at A would, if continued, cut again on the
other side of BC, ‘Thus the construetion gives two triangles on opposite
sides of a common base.
 THE CONSTRUCTION OF TRIANGLES.” 8]

“ON THE CONSTRUCTION OF TRIANGLES. —

It has been scen (page 50) that to prove two triangles

identically equal, three parts of one must be given equal to
the corresponding parts of the other (though any three parts
do not necessarily serve the purpose). This amounts to saying
that to determine the shape and size of a triangle we must know
three of its parts: or, in other words,
To construct a triangle three independent data are required.
For example, we may construct a triangle
(i) When two sides (b, c) and the included angle (A) are given.
The method of construction in this case is obvious.

(ii) When two angles (A, B) and one side (a) are given.
Here, since A and B are given, we at once know C;
for A+B+C=180°.
fence we have only to draw the base equal
to a, and at its ends make angles equal to Nee
B and C; for we know that the remaining B | iC
angle must necessarily be equal to A. a

(iii) If the three angles A, B, C are given (and no side), the

problem is indeterminate, that is, the number of solutions
is unlimited.
For if at the ends of any base we make angles equal to
B and ©, the third angle is equal to A.
This construction is indeterminate, because the three data
are not independent, the third following necessarily from the
other two.

H.S.G. RP
82 GEOMETRY.

PROBLEM 9.

To construct a triangle having given two sides and an angle

opposite to one of them.

Let /, ¢ be the given sides and B the given angle.

Construction. Take any straight line BX, and at B make

the 2 XBY equal to the given 2B.
From BY cut off BA equal to c.
With centre A, and radius b, draw an are of a circle.
If this are cuts BX in two points C, and C,, both on the
same side of B, both the A*ABC,, ABC, satisfy the given con-
ditions.
This double solution is known as the Ambiguous Case, and
will occur when 3 is less than ¢ but greater than the perp.
from A on BX.

EXERCISE.

Draw figures to illustrate the nature and number of solutions in the

following cases :
(i) When 0 is greater than c.
(ii) When 6 is equal to c,
(iii) When b is equal to the perpendicular from A on BX,
(iv) When 6 is less than this perpendicular,
 THE CONSTRUCTION OF TRIANGLES. 83

PROBLEM 10.

To construct a vight-angled triangle having given the hypotenuse

and one side.
C

A O B

Let AB be the hypotenuse and P the given side.

Construction. Bisect AB at O; and with centre O, and
radius OA, draw a semicircle.
With centre A, and radius P, draw an arc to cut the semi-
circle at C.
Join AC, BC.
Then ABC is the required triangle.

Proof. Join OC.

Because OA=OC;
.. the 2OCA=the z OAC.
And because OB=OC;
.. the 2OCB=the 2 OBC.
.. the whole LACB=the 2 OAC + the LOBC
= +4 of 180° Theor. 16.
= 90°.
84 GEOMETRY.

ON THE CONSTRUCTION OF TRIANGLES,

‘Graphical Exercises.)

1. Draw a triangle whose sides are 7°5 em., 6°2 cm., and 5°3 cm.
Draw and measure the perpendiculars dropped on these sides from
the opposite vertices.
[N.B. The perpendiculars, if correctly drawn, will meet at a point,
as will be seen later. See page 207.]

2. Draw a triangle ABC, having given a=3:00", b=2°50", c=2°75".

Bisect the angle A by a line which meets the base at X. Measure
BX and XC (to the nearest hundredth of an inch); and hence calculate
the value of a to two places of decimals. Compare your result with

the value of 5

3. Two sides of a triangular field are 315 yards and 260 yards, and
the included angle is known to be 39°. Draw a plan (1 inch to
100 yards) and find by measurement the length of the remaining side
of the field.
4. ABC is a triangular plot of ground, of which the base BC is
75 metres, and the angles at Pret C are 47° and 68° respectively. Draw
a plan (scale 1 em. to 10 metres). Write down without measurement
the size of the angle A; and by measuring the plan, obtain the approxi-
mate lengths of the other sides of the field; also the perpendicular
drawn from A to BC.
5. A yacht on leaving harbour steers N.E. sailing 9 knots an hour,
After 20 minutes she goes about, steering N.W. for 35 minutes and
making the same average speed as*before. How far is she now from
the harbour, and what course (approximately) must she set for the
run home? Obtain your results from a chart of the whole course,
scale 2 cm. to 1 knot.
6. Draw a right-angled triangle, given that the hypotenuse
c=10°6 em. and one side a=5‘6 cm. Measure the third side b; and
find the value of Nc?-a*. Compare the two results.
7. Construct a triangle, having given the following parts: B=34°,
b=5°5em., c=8'5em. Shew that there are two solutions. Measure
the two values of a, and ‘also of C, and shew that the latter are
supplementary.
8. Ina triangle ABC, the angle A=50°, and b=6'5 em. Illustrate
by figures the cases which arise in constructing the triangle, when
(i) a=7em, (ii) a=6em. (iii) a=5em. (iv) a=4 cm.
 THE CONSTRUCTION OF TRIANGLES. 85

9. Two straight roads, which cross at right angles at A, are carried

over a straight canal by bridges at Band C. The distance between the
bridges is 461 yards, and the distance from the crossing A to the bridge
B is 261 yards. Draw a plan, and by measurement of it ascertain the
distance from A to C.

(Problems. State your construction, and give a theoretical proof.)

10. Draw an isosceles triangle on a base of 4 em., and having an

altitude of 62 cm. Prove the two sides equal, and measure them to
the nearest millimetre.

11. Draw an isosceles triangle having its vertical angle equal to a

given angle, and the perpendicular from the vertex on the base equal to
a given straight line.
Hence-draw an equilateral triangle in which the perpendicular from
one vertex on the opposite side is 6cm. Measure the length of a side
to the nearest millimetre.

12. Construct a triangle ABC in which the perpendicular from A on

BC is 5:0 cm., and the sides AB, AC are 5'8 cm. and 9:0 cm. respectively.
Measure BC.

13. Construct a triangle ABC having the angles at B and C equal

to two given angles L and M, and the perpendicular from A on BC
equal to a given line P.

14. Construct a triangle ABC (without protractor) having given two

angles B and C and the side b.
15. On a given base construct an isosceles triangle having its
vertical angle equal to the given angle L.

16. Construct a right-angled triangle, having given the length of

the hypotenuse c, and the sum of the remaining sides a and b.
If c=5'3 cm., and a+b=7'3 cm., find a and b graphically ; and
calculate the value of Va?+b?.

17. Construct a triangle having given the perimeter and the angles
at the base. For example, a+b+c=12cm., B='70°, C=80°

18. Construct a triangle ABC from the following data :

¢=6 em, .6--c—10em., and B=60".
Measure the lengths of b and c.
19. Construct a triangle ABC from the following data :
@=7 em, c=b=1¢m., and B=55°.
Measure the lengths of 6 and c.
86 GEOMETRY.

THE CONSTRUCTION OF QUADRILATERALS.

It has been shewn that the shape and size of a triangle are
completely determined when the lengths of its three sides are
given. A quadrilateral, however, is not completely determined
by the lengths of its four sides. From what follows it will
appear that five independent data are required to construct a
quadrilateral.

PROBLEM 11.

To construct a quadrilateral, given the lengths of the four sides,

and one angle.

Let a, b, ¢, d be the given lengths of the sides, and A the

angle between the sides equal to @ and d.

Construction. Take any straight line AX, and cut off from
it AB equal to a.
Make the 2 BAY equal to the 2A.
From AY cut off AD equal to d.
With centre D, and radius c, draw an are of a circle.
With centre B and radius }, draw another are to cut the
former at C.
Join DC, BC.
Then ABCD is the required quadrilateral ; for by construction
the sides are equal to a, }, c, d, and the 2 DAB is equal to the
given angle.
 CONSTRUCTION OF QUADRILATERALS., 87

PROBLEM 12.

To construct a parallelogram having given two adjacent sides and

the included angle.

D Cc ea ae
0
©

A
A B

Let P and Q be the two given sides, and A the given angle.
Construction 1. (With ruler and compasses.) Take a line
AB equal to P; and at A make the 2 BAD equal to the ZA, and
make AD equal to Q.
With centre D, and radius P, draw an arc of a circle.
With centre B, and radius Q, draw another are to cut the
former at C.
Then ABCD is the required par™.
Proof. Join DB.
In the A* DCB, BAD,
DC =BA,
oecause - CB=AD,
land DB is common :
‘, the ~CDB=the ZABD; Theor. 7.
and these are alternate angles,
[2 DC iS pat’ oVAB:
Also DC =AB;
. DA and BC are also equal and parallel. Theor. 20.
*, ABCD is a par™.

Construction 2. (With set squares.) Draw AB and AD as

before ; then with set squares through D draw DC par’ to AB,
and through B draw BC par' to AD.
By construction, ABCD is a par™ having the required parts.
88 GEOMETRY.

PROBLEM 13.

To construct a square on a given side,

.
X
6 Cc

A B

Let AB be the given side.

Construction 1. (With ruler and compasses.) At A draw AX
perp. to AB, and cut off from it AD equal to AB.
With B and D as centres, and with radius AB, draw two arcs
cutting at C.
Join BC, DC.
Then ABCD is the required square.
Proof. As in Problem 12, ABCD may be shewn to be a par”.
And since the 2 BAD is a right angle, the figure is a rectangle.
Also, by construction all its sides are equal.
.. ABCD is a square.

Construction 2. (With set squares.) At A draw AX perp. to

AB, and cut off from it AD equal to AB.
Through D draw DC par' to AB, and through B draw BC
par' to AD meeting DC in C.
Then, by construction, ABCD isa rectangle. [Def. 3, page 56.]
Also it has the two adjacent sides AB, AD equal.
. it is a square.
 CONSTRUCTION OF QUADRILATERALS. 89

EXERCISES.

ON THE CONSTRUCTION OF QUADRILATERALS.

1. Draw a rhombus each of whose sides is equal to a given straight
line PQ, which is also to be one diagonal of the figure.
Ascertain (without measurement) the number of degrees in each
angle, giving a reason for your answer.

2. Draw a square on a side of 2°5 inches. Prove theoretically that

its diagonals are equal ; and by measuring the diagonals to the nearest
hundredth of an inch test the correctness of your drawing.

3. Construct a square on a diagonal of 3:0’, and measure the lengths

of each side. Obtain the average of your results.

4. Draw a parallelogram ABCD, having given that one side

AB=5'5 em., and the diagonals AC, BD are 8 cm., and 6 cm. respectively.
Measure AD.

5. The diagonals of a certain quadrilateral are equal, (each 6:0 cm.),

and they bisect one another at an angle of 60°. Shew that jive inde-
pendent data are here given.
Construct the quadrilateral. Name its species; and give a formal
proof of your answer. Measure the perimeter. If the angle between
the diagonals were increased to 90°, by how much per cent. would the
perimeter be increased?

6. Ina quadrilateral ABCD,

AB—»5'6 om, SC=]2-5 cm., CD=4'0 cm., and DA=3'3 em:
Shew that the shape of the quadrilateral is not settled by these data.
Draw the quadrilateral when (i) A= 30°, (ii) A=60°. Why does the
construction fail when A=100° ?
Determine graphically the least value of A for which the con-
- struction fails.

7. Shew how to construct a quadrilateral, having given the lengths

of the four sides and of one diagonal. What conditions must hold
among the data in order that the problem may be possible?
Illustrate your method by constructing a quadrilateral ABCD, when
Gy AB=305 BC=1*7", CD=2:5", DA=2'8", and the diagonal
BD=2°6”. Measure AC.
(ii) AB=3°6 ecm., BC=7°7 cm., CD=6'8 cm., DA=5:1 cm., and the
diagonal AC=8°5 cm. Measure the angles at B and D.
 90 GEOMETRY.

LOCI.

_ DEFINITION. The locus of a point is the path traced out

by it when it moves in accordance with some given law.

Example 1. Suppose the point P to move so

that its distance from a fixed point O is constant
(say 1 centimetre).
Then the locus of P is evidently the cireum-
ference of a circle whose centre is O and radius
1 cm.

Example 2. Suppose the point P :

moves at a constant distance (say 1 em.) :
from a fixed straight line AB. )
Then the locus of P is one or other of ~=——___»__
two straight lines parallel to AB, on :
either side, and at a distance of 1 cm. :
from it. '

Thus the locus of a point, moving under some given con-

dition, consists of the line or lines to which the point is
thereby restricted ; provided that the condition is satisfied by
every point on such line or lines, and by no other.
When we find a series of points which satisfy the given
law, and through which therefore the moving point must pass,
we are said to plot the locus of the point.
 LOC 91

Prosirem 14.
To find the locus of a point P which moves so that its distances
from two tized points A and B are always equal to one another.

Here the point P moves through all positions in which PA= PB ;

*, one position of the moving point is at O the middle point
of AB.
Suppose P to be any other position of the moving point:
that is, let PA= PB.
Join OP.
Then in the A* POA, POB,
PO is common,
because OA= OB,
and PA=PB, by hypothesis;
*, the £ POA=the 2 POB. Theor. 7.
Hence PO is perpendicular to AB.
That is, every point P which ts equidistant from A and B lies on
the straight line bisecting AB at right angles.
Likewise it may be proved that every point on the perpen- |
dicular through O is equidistant from A and B.
This line is therefore the required locus.
 92 GEOMETRY.

PROBLEM 15.

To find the locus of a point P which moves so that its perpen-

dicular distances from two given straight lines AB, CD are equal to
one another.

Let P be any point such that the perp. PM=the perp. PN.
Join P to O, the intersection of AB, CD.
Then in the A* PMO, PNO,
the 2* PMO, PNO are right angles,
because {the hypotenuse OP is common, .
and one side PM=one side PN ;
. the triangles are equal in all respects; Theor. 18.
so that the 2 POM=the 2 PON.
Hence, if P lies within the 2 BOD, it must be on the bisector
of that angle;
and, if P is within the ZAOD, it must be on the bisector of
that angle.
_ It follows that the required locus is the pair of lines which bisect
the angles between AB and CD.
 LOCI. 93

INTERSECTION OF LOCI.

The method of Loci may be used to find the position of a

point which is subject to two conditions. For corresponding
to each condition there will be a locus on which the required
point must lie. Hence all points which are common to these
two loci, that is, all the points of intersection of the loci, will
satisfy both the given conditions.

ExampLe l. To find a point equidistant from three given points

A, B, C, which are not in the same straight line.
(i) The locus of points equidistant from
A and B is the straight line PQ, which
bisects AB at right angles.
(ii) Similarly, the locus of points equi-
distant from B and C is the straight line
RS which bisects BC at right angles.
Hence the point common to PQ and RS
must satisfy both conditions: that is to. . -__
say, X the point of intersection of PQ@and A
RS will be equidistant from A, B, and C.

EXAMPLE 2. Zo construct. a triangle, having given the base, the

altitude, and the length of the median which bisects the base.
Let AB be the given base, and P and
Q the lengths of the altitude aud median
respectively. CE ‘an )
Then the triangle is known if its vertex
is known.
(i) Draw a straight line CD parallel to
AB, and at a distance from it equal to P:
then the required vertex must lie on CD. coi ES eal
(ii) Again, from O the middle point of —_—__
AB as centre, with radius equal to Q, Q
describe a circle:
then the required vertex must lie on this circle.
Hence any points which are common to CD and the circle, satisfy
both the given conditions: that is to say, if CD intersect the circle in
E, F, each of the points of intersection might be the vertex of the
required triangle. This supposes the length of the median Q to be
greater than the altitude.
It may happen that the data of the problem are so related to one
another that the resulting loci do not intersect. In this case the
problem is impossible.
94 GEOMETRY.

Obs. In examples on the Intersection of Loci the student

should make a point of investigating the relations which must
exist among the data, in order that the problem may be
possible ;and he must observe that if under certain relations
two solutions are possible, and under other relations no solu-
tion exists, there will always be some intermediate relation
under which the two solutions combine in a single solution.

EXAMPLES ON LOCI.

1. Find the locus of a point which moves so that its distance

(measured radially) from the circumference of a given circle is constant.
2. <A point P moves along a straight line RQ; find the position in
which it is equidistant from two given points A and B.
3. A and B are two fixed points within a circle: find points on the
circumference equidistant from A and B. How many such points are
there?
4. A point P moves along a straight line RQ; find the position in
which it is equidistant from two given straight lines AB and CD.
5. Aand B are two fixed points 6 om. apart. Find by the method
of loci two points which are 4 em. distant from A, and 5 em. from B.
6. AB and CD are two given straight lines. Find points 3 em.
distant from AB, and 4 cm. from CD. How many solutions are there?
7. A straight rod of given length slides between two straight
rulers placed at right angles to one another.
*/ot the locus of its middle point; and shew that this locus is the
fourth part of the circumference of a circle. [See Problem 10.]
8. Onagiven base as hypotenuse right angled triangles are described.
Find the locus of their vertices.
9. A is a fixed point, and the point X moves on a fixed straight
line BC. :
Plot the locus of P, the middle point of AX ; and prove the locus to
be a straight line parallel to BC.
10. A is a fixed point, and the point X moves on the circumference
of a given circle,
Plot the locus of P, the middle point of AX; and prove that this
locus is a cirele. [See Ex. 3, p. 64]
 EXAMPLES ON LOCI. OD.

11. ABisa given straight line, and AX is the perpendicular drawn.

from A to any straight line |passing through B. If BX revolve about B,
find the locus of the middle point of AX.
12. Two straight lines OX, OY cut at right angles, and from P, a
point within the ‘angle XOY, perpendiculars PM, PN are drawn to.
OX, OY respectively. Plot the locus of P when
(i) PM+PN is constant (=6 em., say) :
(ii) PM —PN is constant (=3 cm., say).
And in each ease give a theoretical proof of the result you arrive at.
experimentally.
13. Two straight lines OX, OY intersect at right angles at O; and
from a movable point P perpendiculars PM, PN are drawn to OX, OY..
Plot (without proof) the locus of P, when
(i) PM=2PN;
(ii) PM=3PN.
14. Find a point which is at a given distance from a given point,.
and is equidistant from two given parallel straight lines.
When does this problem admit of two solutions, when of one only,.
and when is it impossible?
15. S is a fixed point 2 inches distant from a given straight line-
MX. Find two points which are 2% inches distant from S, and also-
23 inches distant from MX.
16. Find a series of points equidistant from a given point S and
a given straight line MX. Draw a curve freehand passing through all.
the points so found.
17. On a given base construct a triangle of given altitude, having:
its vertex on a given straight line.

18. Find a point equidistant from the three sides of a triangle.

19. Two straight lines OX, OY cut at right angles; and Q and R
are points in OX ‘and OY respectively. Plot the locus of the middle-
point of QR, when
(i) OQ+OR=constant.
(ii) OQ -OR=constant.
20. Sand S$’ are two fixed points, Find a series of points P such.
that
(i) SP +S’P=constant (say 3°5 inches).
(ii) SP - S’P=constant (say 1°5 inch).
In each case draw a curve freehand passing through all the points.
so found.
96 GEOMETRY.

ON THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.

I. The perpendicuars drawn to the sides of a triangle from their

middle points are concurrent.
Tet ABC bow Ay and MZ iemiddle A
points of its sides.
From Z and Y draw perps. to AB, AC,
meeting at O. Join OX
. . \g

It is required to prove that OX is perp. geo.

to BC. Lae ine
Join OA, OB, OC. B Xx Cc
Proof. Because YO bisects AC at right angles,
. it is the locus of points equidistant from A and C;
.. OR=OR,
Again, because ZO bisects AB at right angles,
*, it is the locus of points equidistant from A and B;
."OA=0B:
Hence OB=OC.
*. O is on the locus of points equidistant from B and C;
that is, bx is perp. to BC.
Hence the perpendiculars from the mid-points of the sides meet at O.
Q.E.D.

II. The visectors of the angles of a triangle are concurrent.

Let ABC be a A. Bisect the L* ABC,
BCA by straight lines which meet at O.
Join AO
It is required to prove that AO bisects the
LBAC.
From O draw OP, OQ, OR perp. to the
sides of the A. B rs) Cc
Proof. Because BO bisects the L ABC,
', it is the locus of points equidistant from BA and BC;
OP=OR.
Similarly CO is the locus of points equidistant from BC and CA;
OP=O0Q.
Hence OR=OQ.
. O is on the locus of points equidistant from AB and AG:
that is, OA is the bisector of the 2 BAC.
Hence the bisectors of the angles meet at O. Q.E.D.
 THE CONCURRENCE OF LINES IN A TRIANGLE. 97

Ill. The medians of a triangle are concurrent.

Let ABC bea A. A
Let BY and CZ be two of its medians, and let
them intersect at O.
Join AO, , 2
and produce it to meet BC in X.
It is required to shew that AX is the remaining VE
median of the A.
Through C draw CK parallel to BY ; ee es
produce AX to mect CK at K.
Join BK. ; K
Proof. In the AAKC,
oecause Y is the middle point of AC, and YO is parallel to CK,
“. O is the middle point of AK. Theor. 22.
Again in the AABK, -
since Z and O are the middle points of AB, AK,
*. ZO is parallel to BK,
that is, OC is parallel to BK,
*. the figure BKCO is a par™.
But the diagonals of a par™ bisect one another;
.. X is the middle point of BC.
That is, AX is a median of the A.
Hence the three medians meet at the point O. (API 1D),

DeFrIniTion. The point of intersection of the medians is called the

centroid of the triangle.

CoRoLLaRy. The three medians of a triangle cut one another at a

point of trisection, the greater segment in each being towards the angular
point.
For in the above figure it has been proved that

also that OX is half of OK;

*, OX is half of OA:
that is, OX is one third of AX.
Similarly OY is one third of BY,
and OZ is one third of CZ. Q.E.D.

By means of this Corollary it may be shewn that in any triangle

the shorter median bisects the greater side.

Nors. It will be proved hereafter that the perpendiculars drawn

from the vertices of a triangle to the opposite sides are concurrent.
HW S.G. ie
98 GEOMETRY.

MISCELLANEOUS PROBLEMS.

(A theoretical proof is to be given in each case.)

1. Aisa given point, and BC a given straight line. From A draw

a straight line to make with BC an angle equal to a given angle X.
How many such lines can be drawn?

2. Draw the bisector of an angle AOB, without using the vertex O

in your construction.

3. P is a given point within the angle AOB. Draw through P a

straight line terminated by OA and OB, and bisected at P.

4. OA, OB, OC are three straight lines meeting at O. Draw a

transversal terminated by OA and OC, and bisected by OB.

5. Through a given point A draw a straight line so that the part

intercepted between two given parallels may be of given length.
When does this problem admit of two solutions? When of only one?
And when is it impossible?

6. In a triangle ABC inscribe a rhombus having one of its angles

coinciding with the angle A.
7. Use the properties of an equilateral triangle to trisect a given
straight line.

(Construction of Triangles.)

8. Construct a triangle, having given

(i) The middle points of the three sides.
(ii) The lengths of two sides and of the median which bisects the
third side.
(iii) The lengths of one side and the medians which bisect the other
two sides.
(iv) The lengths of the three medians,
 AREAS. 99

PART IL

ON AREAS.

DEFINITIONS.

1. The altitude (or height) of a parallelogram with refer-

ence to a given side as base, is the perpendicular distance
between the base and the opposite side.
2. The altitude (or height) of a triangle with reference to
a given side as base, is the perpendicular distance of the
opposite vertex from the base.
Nore. It is clear that parallelograms or triangles which are between
the same parallels have the same altitude.

For let AP and DQ be the alti- =— —

tudes of the A®*ABC, DEF, which
are between the same parallels BF,
1

Then the fig. APQD is evidently

a rectangle ; Z !

3. The area of a figure is the amount of surface contained

within its bounding lines.

4. A square inch is the area of a .

square drawn on a side one inch in MAKE
“ length. inch

5. Similarly a square centimetre is the area of Sq.

a square drawn on a side one centimetre in length. cm.

The terms square yard, square foot, square metre are to be understood
in the same sense.
6. Thus the unit of area is the area of a square on a side
of unit length.
100 GEOMETRY.

THEOREM 23.
Area of a rectangle. Jf the number of units in the length of a
rectangle is multiplied by the number of wnits in its breadth, the
product gives the number of square units in the area.

Let ABCD represent a rectangle whose length AB is 5 feet,

and whose breadth AD is 4 feet.
Divide AB into 5 equal parts, and BC into 4 equal parts, and
through the points of division of each line ‘draw parallels to
the other. “-
The rectangle ABCD is now divided into compartments,
each of which represents one square foot.
Now there are 4 rows, each containing 5 squares,
. the rectangle contains 5 x 4 square feet.
Similarly, if the length=a linear units, and the breadth
=b
linear units
the rectangle contains ab units of area.
And if each side of a square =a linear units,
the square contains a? wnits of area.
These statements may be thus abridged:
the area of a rectangle =length x breadth ........00 (i),
the area of a squire = (side)? ......00ssverseseees (ii).

CorOLLARIES, (i) Rectangles which have equal lengths and

equal breadths have equal areas.
(ii) Rectangles which have equal areas and equal lengths have
also equal breadths.
 AREA OF A TRIANGLE. 10]

NOTATION.

The rectangle ABCD is said to be contained by AB, AD; for

these adjacent sides fix its size and shape.
A rectangle whose adjacent sides are AB, AD is denoted by
rect. AB, AD, or simply AB x AD.
A square drawn on the side AB is denoted by sq. on AB, or AB.

EXERCISES.

(On Tables of Length and Area.)

1. Draw a figure to shew why
(i) 1 sq. yard =3? sq. feet.
(ii) 1 sq. foot: =12? sq. inchies.
(ia) Lesq. em. 10" sq2 mm.
2. Draw a figure to shew that the square on a straight line is four
times the square on half the line.
3. Use squared paper to shew that the square on 1”=10? times the
square on 0'1”.
4, If 1” represents 5 miles, what docs an area of 6 square inches
represent ? =

EXTENSION OF THEOREM 23.

The proof of Theorem 23 here given supposes that the length and
breadth of the given rectangle are expressed by whole numbers ; but the
formula holds good when the length and breadth are fractional.
This may be illustrated thus :
Suppose the length and breadth are 3:2 cm. and 2°4 cm.; we shal}
shew that the area is (3°2 x 24) sq. em.
For length =3°2 cm. =32 mm.
breadth =2°4 em. =24 mm.
32 x 24
“. area=(32 x 24) sq. mm. = sq. cm.
10?
=(3°2 x 2°4) sq. cm.
102 GEOMETRY.

EXERCISES.

(On the Area of a Rectangle.)

Draw on squared paper the rectangles of which the length (a) and
breadth (b) are given below. Calculate the areas, and verify by the
actual counting of squares.
l. a=2", b=3". 2. a=1°5", b=4".
3. a=0°8’, b=3°5". 4. a=2:5", b=1°4".
5. a=2-2’, b=1°5", 6. a=1°6", b=2°1",
Calculate the areas of the rectangles in which
7. a=18 metres, b=11 metres. 8. O=/ it. V=72in.
9, a=2°5km., b=4 metres. 10. a= mile, b=1 inch.

11. The area of a rectangle is 30 sq. cm., and its length is 6 em.
Find the breadth. Draw the rectangle on squared paper; and verify
your work by counting the squares.
12. Find the length of a rectangle whose area is 3°9 sq. in., and
breadth 1°5’. Draw the rectangle on squared paper; and verify your
work by counting the squares.
13. (i) When you treble the length of a rectangle without altering
its breadth, how many times do you multiply the area?
(ii) When you treble both length and breadth, how many times do
you multiply the area?
Draw a figure to illustrate your answers ; and state a general rule.
14. In a plan of a rectangular garden the length and breadth
are 3°6” and 2°5”, one inch standing for 10 yards. Find the area of the
garden.
If the area is increased by 300 sq. yds., the breadth remaining the
same, what will the new length be? And how many inches will repre.
sent it on your plan?
15. Find the area of a rectangular enclosure of which a plan
(scale 1 cm. to 20 metres) measures 6°5 em. by 4°5 em,
16. The area of a rectangle is 1440 sq. yds. If in a plan the sides
of the rectangle are 3°2 em. and 4°5 cm., on what scale is the plan
drawn ?
17. The area of a rectangular field is 52000 sq. ft. On a plan of
this, drawn to the scale of 1” to 100 ft., the length is 3°25”. hat is
the breadth ?
 EXERCISES ON RECTANGLES. 103

a ae

Calculate the areas r epresented by the shaded parts of the following

The dimensions are mar ked in feet.

21
a

at.
ey.

See 2
a

<+

BS

eee
-3-.>
|

ae
WN es) :
2
' ’
a

rh

aea
>
104 GEOMETRY.

THEOREM 24. [Euclid I. 35.}

Parallelograms on the same base and between the same parallels
are equal in area.
A E D F

B Cc

Let the par™ ABCD, EBCF be on the same base BC, and
between the same par" BC, AF.
It is required to prove that
the par" ABCD = the par” EBCF in area,
Proof. In the A* FDC, EAB,
DC =the opp. side AB; Theor. 21.
because }the ext. 2 FDC =the int. opp. ©EAB; Theor. 14.
(the int. 2DFC=the ext. 2AEB;
. the AFDC =the A EAB. Theor. 17.
Now, if from the whole fig. ABCF the AFDC is taken, the
remainder is the par™ ABCD.
And if from the whole fig. ABCF the AEAB is taken, the
remainder is the par™ EBCF.
.. these remainders are equal ;
that is, the par" ABCD=the par™ EBCF. = Q.E.D.

EXERCISE.

In the above diagram the sides AD, EF overlap. Draw diagrams in

which (i) these sides do not overlap ; (ii) the ends E and D coincide.
Go through the proof with these diagrams, and ascertain if it applies
to them without change.
 AREAS. 105

THE AREA OF A PARALLELOGRAM.

Let ABCD be a _ parallelogram, ane Ea’

and ABEF the rectangle on the
same base AB and of the same
altitude BE. Then by Theorem 24,
area of par” ABCD =area of rect. ABEF A ° 8B
= AB x BE
= base x altitude.

CoROLLARY. Since the area of a parallelogram depends

only on its base and altitude, it follows that
_ Parallelograms on equal bases and of equal altitudes are equal
m area.

EXERCISES.

(Numerical and Graphical.)

i, Find the area of parallelograms in which

(i) the base=5°5 cm., and the height
=4 em.
(ii) the base =2°4”, and the height =1°5”.

2. Draw a parallelogram ABCD having given AB=23”, AD=1}",

and the LA=65°. Draw and measure the perpendicular from D on AB,
and hence calculate the approximate area. Why approximate?
Again calculate the area from the length of AD and the perpendicular
on it from B. Obtain the average of the two results.

3. Two adjacent sides of a parallelogram are 30 metres and 25 metres,

and the included angle is 50°. Draw a plan, 1 cm. representing
5 metres; and by measuring each altitude, make two independent
ealculations of the area. Give the average result.

4, The arca of a parallelogram ABCD is 4:2 sq. in., and the base
AB is 2°8”. Find the height. If AD=2’, draw the parallelogram.

5. Hach side of a rhombus is 2”, and its area is 3°86 sq. in. Calculate
an altitude. Hence draw the rhombus, and measure one of its acute
angles.
106 GEOMETRY.

‘THEOREM 25,

The Area of a Triangle. The area of a triangle is half the area

of the rectangle on the same base and having the same altitude.

D A E D E A
i

B F Cc B Cc F
Fig. 1. Pip. 2,

Let ABC be a triangle, and BDEC a rectangle on the same

base BC and with the same altitude AF.
It is required to prove that the 4 ABC is half the rectangle BDEG.

Proof. Since AF is perp. to BC, each of the figures DF, EF

is a rectangle.
Because the diagonal AB bisects the rectangle DF,
. the A ABF is half the rectangle DF.
Similarly, the AAFC is half the rectangle FE.
*, adding these results in ig. 1, and taking the difference in
Fig.2
‘the A ABC is half the rectangle BDEC.
Q.E.D,

CoroLuaRy. A triangleis half any parallelogram on the same

base and between the same parallels.
For the A ABC is half the rect. BCED. G HO AE
And the rect. BCED=any par™ BCHG
on the same base and between the same
par”.
*, the A ABC is half the par™ BCHG.,
 AREAS. 107

THE AREA OF A TRIANGLE,

_ [f BC and AF respectively contain a units andp units of

length, the rectangle BDEC contains ap units of area,
‘, the area of the AABC=4ap units of area.
This result may be stated thus:
Area of a Triangle=4. base x altitude.

EXERCISES ON THE AREA OF A TRIANGLE.

(Numerical and Graphical.)

1. Calculate the areas of the triangles in which
(i) the base =24 ft., the height=15 ft.
(ii) the base=4°8", the height=3'5".
(iii) the base=160 metres, the height =125 metres,

2. Draw triangles from the following data. In each case draw and
measure the altitude with reference to a given side as base: hence cal-
culate the approximate area.
(i) a=8°4 cm., b=6°8 cm., c=4°0 cm.
(ie b=s0'em., ¢=6°8 cm. A=65-.
(iit) ¢=6°5 cm., B=52°; C=76".
3. ABCisa triangle right-angled at C ; shew that its area=4BC x CA.
Given a=6 cm., b=5 cm., calculate the area.
Draw the triangle and measure the hypotenuse ¢ ; draw and measure
the perpendicular from C on the hypotenuse; hence calculate the
approximate area.
Note the error in your approximate result, and express it as a per-
centage of the true value.
4, Repeat the whole process of the last question for a right-angled
triangle ABC, in which a=2°8” and b=4°5"; C being the right angle
as before.
5. Ina triangle, given
(i) Area=80 sq. in., base=1 ft. 8 in.; calculate the altitude.
(ii) Area=10°4 sq. em., altitude=1°6 cm.; calculate the base.
6. Construct a triangle ABC, having given a=3:0", b=2°8’, c=2'6”,
Draw and measure the perpendicular from A on BC; hence calculate
the approximate area. :
108 GEOMETRY.

THEOREM 26. [Euclid I. 37.]

Triangles on the same base and between the same parallels
(hence, of the same altitude) are equal in
area. D ae G
Let the A*ABC, GBC be on the
same base BC and between the same
par’ BC, AG.
It is required to prove that
the A ABC =the A GBC in area. B C

Proof. If BCED is the rectangle on the base BC, and

between the same parallels as the given triangles,
the A ABC is half the rect. BCED ; Theor. 25.
also the A GBC is half the rect. BCED ;
.. the AABC =the A GBC. Q.E.D.
Similarly, triangles on equal bases and of equal altitudes are
equal in area.

THEOREM 27. [Euclid I. 39.]

If two triangles are equal in area, and stand on the same base
and on the same side of it, they are between the same parallels.
Let the A* ABC, GBC, standing on A G
the same base BC, be equal in area;
and let AF and GH be their altitudes.
It is required to prove that AG and
BC are par’. B FC H
Proof. The AABC is half the rectangle contained by BC
and AF ;
and the AGBC is half the rectangle contained by BC
and GH;
the rect. BC, AF =the rect. BC, GH;
.. AF =GH, Theor. 23, Cor. 2.
Also AF and GH are par';
hence AG and FH, that is BC, are par. =. E.D
 AREAS, 109

EXERCISES ON THE AREA OF A TRIANGLE,

(Theoretical.)

1. ABC is a triangle and XY is drawn parallel to the base BC,

cutting the other sides at X and Y. Join BY and CX; and shew that
(i) the A XBC=the A YBC;
(ii) the A BKY=the A CXY;
(iii) the A ABY=the A ACX.
if BY and CX cut at K, shew that
(iv) the A BKX=the A CKY.
2. Shew that a median of a triangle divides it into two parts of
equal area.
How would you divide a triangle into three equal parts by straight
lines drawn from its vertex ?
3. Prove that a parallelogram is divided by its diagonals into four
triangles of equal area.
4. ABC is a triangle whose base BC is bisected at X. If Y is any
point in the median AX, shew that
the A ABY =the A ACY in area.

5. ABCD is a parallelogram, and BP, DQ are the perpendiculars

from B and D on the diagonal AC.
Shew that BP=DQ.
Hence if X is any point in AC, or AC produced,
prove (i) the A ADX=the A ABX;
(ii) the A CDX=the A CBX.

6. Prove by means of Theorems 26 and 27 that the straight line

joining the middle points of two sides of a triangle is parallel to the third
side.
7. The straight line which joins the middle points of the oblique
sides of a trapezium is parallel to each of the parallel sides.

8. ABCD is a parallelogram, and X, Y are the middle points of the

sides AD, BC; if Z is any point in XY, or XY produced, shew that the
triangle AZB is one quarter of the parallelogram ABCD.
9. If ABCD isa parallelogram, and X, Y any points in DC and AD
respectively : shew that the triangles AXB, BYC are equal in area.
10. ABCD is a parallelogram, and P is any point within it ; shew
that the sum of the triangles PAB, PCD is equal to half the parallelo-
gram.
110 GEOMETRY.

EXERCISES ON THE AREA OF A TRIANGLE.

(Numerical and Graphtcal.)

1, The sides of a triangular field are 370 yds., 200 yds., and 196
yds. Draw a plan (scale 1” to 100 yards). Draw and measure an
altitude ; hence calculate the approximate area of the field in square
yards. :
2. Two sides of a triangular enclosure are 124 metres and 144
metres respectively, and the included angle is observed to be 45°.
Draw a plan (scale 1 cm. to 20 metres). Make any necessary measure-
ment, and calculate the approximate area.
3. Ina triangle ABC, given that the area=6°6 sq. cm., and the base
BC=5'5 em., find the altitude. Hence determine the locus of the
vertex A.
If in addition to the above data, BA=2°6 cm., construct the tri-
angle ; and measure CA.
4. In a triangle ABC, given area=3-06 sq. in., and a=3°0’. Find
the altitude, and the locus of A, Given C=68°, construct the triangle;
and measure b,
5. ABC is a triangle in which BC, BA have constant lengths 6 em.
and 5em. If BC is fixed, and BA revolves about B, trace the changes
in the area of the triangle as the angle B increases from 0° to 180°.
Answer this question by drawing a series of triangles, increasing
B by increments of 30°. Find the area in each case and tabulate the
results.

(Theoretical. )
6. If two triangles have two sides of one respectively equal to two
sides of the other, and the angles contained by thdse sides supple-
mentary, shew that the triangles are equal in area. Can such triangles
ever be identically equal ?
7. Shew how to draw on the base of a given triangle an isosceles
triangle of equal area.
8. If the middle points of the sides of oF cig are joined in
order, prove that the parallelogram so formed [see Ex. 7, p. 64), is half
the quadrilateral.
9. ABC is a triangle, and R, Q the middle points of the sides
AB, AC; shew that if BQ and CR intersect in X, the triangle BXC
is equal to the quadrilateral AQXR.
10. Two triangles of ee area stand on the same base but on
opposite sides of it: shew that the straight line joining their vertices
is bisected by the base, or by the base produced.
 THK AREA OF A TRIANGLE. kil

[The method given below may be omitted from a first course. In

any case it must be postponed till Theorem 29 has been read.]

The Area of a Triangle. (Given the three sides of a triangle,

to calculate the area.
Exameire. Find the area of a triangle whose sides measure 21 m.,
17 m., and 10 m.
Let ABC represent the given
triangle. A
Draw AD perp. to BC, and
denote AD by p.
We shall first find the length 10 p i?
of BD.
Let BD=2 metres; then DC
=21— «x metres.
Bae D 21-x C
From the right-angled AADB,
we have by Theorem 29
AD?= AB? - BD?= 10? - 2?.
And from the right-angled AADC,
AD?=AC?— DC?=17?- (21-2);
3, 10 =F =P - Ol =x)?
or 100 — x? = 289 — 441 +442 —2?;
whence p= (0.
Again, AD?=AB?- BD?;
or p= 107 —6=64;
os

New Area of triangle =4. base x altitude

=($x 21 x 8) sq. m.=84 sq. m.

EXERCISES.

Find by the above method the area of the triangles, whose sides
are as follows: ~
Teme Onto enitessl lett, 2. 15 yds., 14 yds., 13 yds.
Sao lene 20) ie, eho. A> 30icm., Zo\em,, lem:
Dy, Ol Luss OO fb; lots 6. 6) mi, 37 m.,.20 m,
7. If the given sides are a, b and c units in length, prove
Sa Se eee rece len
(i) ie aa are (Git) yore |} 3

(iii) A=4N(a+b+0)(—a+b+c)(a—b+c)(a+b—c).
112 GEOMETRY.

THE AREA OF QUADRILATERALS.

THEOREM 28,
To find the area of (i) a@ trapezium.
(ii) any quadrilateral.
(i) Let ABCD be a trapezium, having
the sides AB, CD parallel. Join BD, D Cc
and from C and D draw perpendiculars
CF, DE to AB.
Let the parallel sides AB, CD measure
a and 6 units of length, and let the AE EB
height CF contain / units.
Then the area of ABCD =A ABD +A DBC
1 ]

1 1 h
= gh + gh = 34 + b).

That is,
the area of a trapezium ms height x (the swm of the parallel sides).

(ii) Let ABCD be any quadrilateral. C

Draw a diagonal AC; and from B D
and D draw perpendiculars BX, DY to
AC. These perpendiculars are called
offsets.
If AC contains d units of length, and 4
BX, DY p and q units respectively,
the area of the quad' ABCD =A ABC + AADC
= JAC. BX +5AC. DY
l ]
[—4 5p “+ sly = Lup +. q):

That is to say, 2 9
Ti
the area of a quadrilateral = = diagonal x (sum of offsets).
~
 EXERCISES ON QUADRILATERALS, ups}

EXERCISES.

(Numerical and Graphical.)

1. Find the area of the trapezium in which the two parallel sides
are 4-7” and 3:3”, and the height 1°5”.

2. In a quadrilateral ABCD, the diagonal AC=17 feet; and the

offsets from it to B and D are 11 feet and 9 feet. ind the area.

3. In a plan ABCD of a quadrilateral enclosure, the diagonal AC

measures 8°2 em., and the offsets from it to B and D are 3°4 cm. and
2°6 cm. respectively. If 1 cm. in the plan represents 5 metres, find
the area of the enclosure.
4. Draw a quadrilateral ABCD from the ad-
joining rough plan, the dimensions being given in
inches.
Draw and measure the offsets to A and C from
the diagonal BD; and hence calculate the area of
the quadrilateral.

5. Draw a quadrilateral ABCD from the

details given in the adjoining plan. The
dimensions are to be in centimetres.
Make any necessary measurements of your
figure, and calculate its area.
A TU B

6. Draw a trapezium ABCD from the following data: AB and CD

are the parallel sides) AB=4"; AD=BC=2’; the LA=the LB=60°.
Make any necessary measurements, and calculate the area.

7. Draw a trapezium ABCD in which AB and CD are the parallel

sides; and AB=9 cm., CD=3 em., and AD=BC=5 cm.
Make any necessary measurement, and calculate the area.

8. From the formula area of quad'=4 diag. x (sum of offsets) shew

that, if the diagenals are at right angles,
area =4(product of diagonals).

9. Given the lengths of the diagonals of a quadrilateral, and the

angle between them, prove that the area is the same wherever they
intersect.

H.S.G. ; H
114 GEOMETRY.

THE AREA OF ANY RECTILINEAL FIGURE.

1*Mernop. A rectilineal figure

may be divided into triangles whose
areas can be separately calculated
from suitable measurements. The
sum of these areas will be the area
of the given figure.
Example. The measurements re-
uired to find the area of the figure
BCDE are AC, AD, and the offsets BX,
DY, EZ.

2"* MerHop. The area of a rectilineal figure is also found

by taking a base-line (AD in the diagram below) and offsets
from it. These divide the figure into right-angled triangles
and right-angled trapeziums, whose areas may be found after
measuring the offsets and the various sections of the base-line.
Example. Y¥ind the area of the enclosure ABCDEF from the plan
and measurements tabulated below.
YARDS.
AD =56
VC=12} AV=50
AZ=40 | ZE=18
¥B=20) sAV=18
AX=10 | XF=15

The measurements are made from

A along the base line to the points
from which the offsets spring.

Sq. yds. Sq. yds.

Here A AXF =4.AX x XF =¢xlOxlb= 76
4 AYB=4.AYx YB =x 18ix 20 180
A DZE=t. Dex Ze =4x16xI18= 144
A DVC=4.DVxVC =t% Ox12= $6
trap" XFEZ=4.XZ x (XF+ZE)=4x 30x 33= 495
trap" YBCV=4.YV x (YB+VC)=4 x 32x 32= 512
., by addition, the fig. ABCDEF = 1442 sq. yds.
 EXERCISES ON RECTILINEAL FIGURES. 116

EXERCISES.
1. Calculate the areas of the figures (i) and (ii) from the plans and
dimensions (in cms.) given below.

(ii)

A Xx B

AC=6cm., AD=5 cm. AB=BD=DA=6 cm.

Lengths of offsets figured BY=CZ— ikem:
in diagram. DX —=52 em,

2. Draw full size the figures whose plans and dimensions are given
below ; and calculate the area in each case.

D
(1) (ii) C

A x \¢ B
The fig. is equilateral : Ae OY =a
each side to be 23”. : YB=12"

3. Find the area of the figure ABCDEF from the following measure:
ments and draw a plan in which |] em. represents 20 metres.
THE Pian.
METRES. C
to C
180 2 B
80toD| 150 E
40 toE 120 50 to B
60 to F 50 7
From A
116 GEOMETRY.

EXERCISES ON QUADRILATERALS,

(Tleoretical.)

1. ABCD is a rectangle, and PQRS the figure formed by joining in

order the middle points of the sides.
Prove (i) that PQRS is a rhombus ;
(ii) that the area of PQRS is half that of ABCD.
Hence shew that the area of a rhombus is half the product of tts
diagonals.
Is this true of any quadrilateral whose diagonals cut at right angles?
Illustrate your answer by a diagram.

2, Prove that a parallelogram is bisected by any straight line which

passes through the middle point of one of its diagonals.
Hence shew how a parallelogram ABCD may be bisected by a
straight line drawn
(i) through a given point P;
(ii) perpendicular to the side AB ;
(iii) parallel to a given line QR.

In the trapezium ABCD, AB is parallel to DC; and X is the

midlale point of BC. Through X draw PQ parallel to AD to meet AB
and DC prodaced at Pand Q. ‘Then prove
(i) trapezium ABCD=par™ APQD.
(ii) trapezium ABCD =twice the AAXD.

(Graphical. )

4. The diagonals of a quadrilateral ABCD cut at right angles, and

measure 3°0’ and 2°2” respectively. Find the area.
Shew by a figure that the area is the same wherever the diagonals
cut, so long as they are at right angles.

5. In the parallelogram ABCD, AB=8:0 cm., AD=3°2 cm., and the

perpendicular distance between AB and DC=3°0 cm. Draw the par-
allelogram. Calculate the distance between AD and BC; and check
your result by measurement.

6. One side of a parallelogram is 2°5”, and its diagonals are 3:4”

and 2°4”. Construct the parallelogram ; and, after making any neces-
sary measurement, calculate the area.

7. ABCD is a parallelogram on a fixed base AB and of constant

area, Find the locus of the intersection of its diagonals.
 EXPERIMENTAL EXERCISES. V7

EXERCISES LEADING TO THEOREM 29,

In the adjoining diagram, ABC is a triangle

right-angled at C; and squares are drawn on the
three sides. Let us compare the area of the square
on the hypotenuse AB with the swm of the squares
on the sides AC, CB which contain the right angle.

1. Draw the above diagram, making AC=3 cm., and BC=4 cm. ;
Then the area of the square on AC=3?, or 9 sq. cm.
Me acg tds anes Saxe the square on BC=4?, or 16 sq. cm.
*. the sum of the squares on AC, BC= 25 sq. em.
Now measure AB ; hence calculate the area of the square on AB, and
compare the result with the swm already obtained.

a Repeat the process of the last exercise, making AC=1°0’, and

DA

3. If a=15, b=8, c=17, shew arithmetically that c?=a?+ 62.

Now draw on squared paper a triangle ABC, whose sides a, b, and ¢
are 15, 8, and 17 units of length ; and measure the angle ACB.

4. Take any triangle ABC, right-

angled at C; and draw squares on AC,
CB, and on the hypotenuse AB.
Through the mid-point of the square
on CB (2.e. the intersection of the dia-
gonals) draw lines parallel and perpen-
dicular to the hypotenuse, thus dividing
the square into four congruent quadri-
laterals. These, together with thesquare
on AC, will be found exactly to fit into
the square on AB, in the way indicated
by corresponding numbers.

These experiments point to the conclusion that:

In any right-angled triangle the square on the hypotenuse ts equal
to the sum of the squares on the other two sides.
A formal proof of this theorem is given on the next page.
118 : GEOMETRY.

THEOREM 29. [Euclid I. 47.1

In a right-angled triangle the square described on the hypotenuse
is equal to the sum of the squares described on the other two sides.

Let ABC be a right-angled A, having the angle ACB a rt. cz.

It is required to prove that the square on the hypotenuse AB =the
sum of the squares on AC, CB.
On AB describe the sq. ADEB; and on AC, CB describe the
sqq. ACGF, CBKH.
Through C draw CL par' to AD or BE.
Join CD, FB.

Proof. Because each of the 2* ACB, ACG is a rt. 2,

‘, BC and CG are in the same st. line.
Now the rt. 2 BAD=the rt. 2 FAC;
add to each the 2 CAB:
then the whole 2 CAD=the whole 2 FAB.
Then in the A*CAD, FAB,
CA= FA,
because AD = AB,
and the included 2 CAD =the included 2 FAB;
“. the ACAD=the A FAB Theor. 4.
 THEOREM OF PYTHAGORAS. 119

Now the rect. AL is double of the ACAD, being on the

same base AD, and between the same par’ AD, CL.
And the sq. GA is double of the A FAB, being on the same
base FA, and between the same par’ FA, GB.
.. the rect. AL=the sq. GA.
Similarly by joining CE, AK, it can be shewn that
the rect. BL=the sq. HB.
*. the whole sq. AE=the sum of the sqq. GA, HB:
that is, the square on the hypotenuse AB=the sum of the
squares on the two sides AC, CB.
Q.E.D.

Obs. This is known as the Theorem of Pythagoras. The

result established may be stated as follows:
AB? = BC? + CA?.
That is, if a and 6 denote the lengths of the sides containing
the right angle ; and if ¢ denotes the hypotenuse,
=a? + 0%,
Hence a?—c?—0?; and 0?=c —a?,

Nott 1. The following important results should be noticed.

If CL and AB intersect in O, it has been shewn in the course of the
proof that
the sq. GA=the rect. AL;
that is, AC?=the rect. contained by AB, AO. ............ (i)
Also the sq. HB=the rect. BL;
that is, BC?=the rect. contained by BA, BO............. (ii)

Nore 2. It can be proved by superposition that squares standing on

equal sides are equal in area.
Hence we conclude, conversely,
Iftwo squares are equal in area they stand on equal sides.
120 GEOMETRY.

EXPERIMENTAL PROOFS OF PYTHAGORAS'S THEOREM.

I. Here ABC is the given ee eal a eh)

rt.-angled A; and ABED is ewe ped eet Id rer)
fhe square on the hypotenuse | HRS HH as
B. |
/
a
By drawing lines par! to the a y, LT
sides BC, CA, it is easily seen e =
SE
that the sq. BD is divided i S ever
ae
oe
into 4 rt.-angled A’, each Y| be
identically equal to ABC, to- = WEa
gether with a central square. || eS
Z
Hence os Le
ea
sq. on hypotenuse c=4rt. £4 A®
Ay
iz
oa
i
+the central square > ac
he
3S
=4.4ab+(a—b)? | fay
pn cone 2408 aun
=2ab + a? —2ab + b? | SEeoeanette
i “SPR ee
aoe SPER
aon
Sarance Se SS SRE
=u? +b? = eI
SRSRET
SRR
PSH
tt SSSSUUSeEE |i128 US
ESTE
ERRARE

Il. Here ABC is the Ahn

ar-angled A, and the figs
F. HK are the nae on C
CA ‘placed side by si
FE is made equal to DH
or CA; and the two sqq. Ree
a°
CF, HK are cut along the ry
VT
TTT
SESReR@
SERRE
lines BE, ED. ee | a.
eo WH
Bae 2
42RN
4S

tie
Then it will be found that Gant
epee)
the ADHE may be placed
a

FECES
so as to fill up the space ACB;
and the ABFE may be made >
asmie
HUSkR
SSRE
to fill the space AK
_ aoa

coy
Hence the two sqq. CF,
HK may be fitted to ether - HH
so as to form the single fig.
ABED, which will be found
to be a perfect square, namely JDUSERRSSRPES
SERT
oPSeen 8¢!
the square on the hypotenuse
 THEOREM OF PYTHAGORAS. 121

EXERCISES.

(Numerical and Graphical.)

1. Draw a triangle ABC, right-angled at C, having given:

(a= orem. u 0 —=4acmats
(Giita=2-5em-,, 0=6"0)em.s
(in) sa 127 b=3°5".
In each case calculate the length of the hypotenuse c, and verify
your result by measurement.
2. Draw a triangle ABC, right-angled at C, having given:
(i) c=3°4", a=3:°0"; [See Problem 10]
(it)"c=5'3 em., b=4:5 cm.
In each case calculate the remaining side, and verify your result by
measurement.

(The following examples are to be solved by calculation; but in each

case a plan should be drawn on some suitable scale, and the calculated
result verified by measurement.)
3. A ladder whose foot is 9 feet from the front of a house reaches
to a window-sill 40 feet above the ground. What is the length of the.
ladder ?

4. A ship sails 33 miles due South, and then 56 miles due West.
How far is it then from its starting point?

5. Two ships are observed from a signal station to bear respectively

N.E. 6:0 km. distant, and N.W. 1:1 km. distant. How far are they
apart ?

6. A ladder 65 feet long reaches to a point in the face of a house

63 feet above the ground. How far is the foot from the house?
7. Bis due Kast of A, but at an unknown distance. C is due South
of B, and distant 55 metres. AC is known to be 73 metres. Find AB.

8. A man travels 27 miles due South; then 24 miles due West;

finally 20 miles due North. How far is he from his starting point?
9. From A go West 25 metres, then North 60 metres, then East
80 metres, finally South 12 metres. How far are you then from A?

10. A ladder 50 feet long is placed so as to reach a window 48 feet

high ; and on turning the ladder over to the other side of the street, it
reaches a point 14 feet high. Find the breadth of the street.
122 GEOMETRY.

THEOREM 30. [Euclid I. 48.]

If the square described on one side of a triangle is equal to the
swm of the squares described on the other two sides, then the angle
contained by these two sides is a right angle.

A D

B Cc E F

Let ABC be a triangle in which

the sq. on AB=the sum of the sqq. on BO, CA,
It is required to prove that ACB is a right angle.
Make EF equal to BC.
Draw FD perp" to EF, and make FD equal to CA.
Join ED.
Proof. Because EF = BC,
*. the sq. on EF =the sq. on BC.
And because FD =CA,
.. the sq. on FD=the sq. on CA.
Hence the sum of the sqq. on EF, FD=the sum of the sqq.
on BO, CA.
But since EFD is a rt. Z,
. the sum of the sqq. on EF, FD=the sq. on DE: Theor. 29.
ha by hypothesis, the sqq. on BC, CA=the sq. on AB,
*, the sq. on DE=the sq. on AB.
*, DE=AB.
Then in the A* ACB, DFE,
AC = DF,
because | CB=FE,
and AB=DE;
.. the .ACB=the 2DFE. Theor. 7.
But, by construction, DFE is a right angle;
*. the 2 ACB is a right angle.
Q.E.D.
 THEOREM OF PYTHAGORAS AND ITS CONVERSE. 123

EXERCISES ON THEOREMS 29, 30.

(Theoretical. )
1. Shew that the square on the diagonal of a given square is double
of the given square.
2. In the AABC, AD is drawn perpendicular to the base BC. If
the side c is greater than ),
shew that c?- 6?=BD?- DC?.
3. If from any point O within a triangle ABC, perpendiculars OX,
OY, OZ are drawn to BC, CA, AB respectively : shew that
AZ? + BX?+ CY?=AY?+ CX?+ BZ?.
4. ABC is a triangle right-angled at A; and the sides AB, AC are
intersected by a straight line PQ, and BQ, PC are joined. Prove that
BQ?+ PC?=BC?+ PQ?.
5. Ina right-angled triangle four times the sum of the squares on
the medians drawn from the acute angles is equal to five times the
square on the hypotenuse.
6. Describe a square equal to the sum of two given squares.
7. Describe a square equal to the difference between two given
squares.
8. Divide a straight line into two parts so that the square on one
part may be twice the square on the other.
9. Divide a straight line into two parts such that the sum of their
squares shall be equal to a given square.

(Numerical and Graphical.)

10. Determine which of the following triangles are right-angled :
(ee eC I — 4 SGins, C—O UnemiE:
(i) a=40 cm., b=10cm., c= 41 cm.;
(ii1)) @=20 cm., b=99 cm., c=10) cm.
11. ABC is an isosceles triangle right-angled at C; deduce from
Theorem 29 that
AB?=2AC?.
Illustrate this result graphically by drawing both diagonals of the
square on AB, and one diagonal of the square on AC.
If AC=BC=2", find AB to the nearest hundredth of an inch, and
verify your calculation by actual construction and measurement.
12. Draw a square on a diagonal of 6 cm. Calculate, and also
measure, the length of a side. Find the area.
124 : GEOMETRY.

ProbiEM 16.
To draw squares whose areas shall be respectively twice, three-times 1
four-times, ..., that of a given square.
Hence find graphically approximate values of /2, 3, \/4, V5, ....
¢

Take OX, OY at right angles P

to one another, and from them
mark off OA, OP, each one
unit of length. Join PA.
fe) kK & Oo oS
Then PA2= OP? + OA2=14+1=2.
. PA=,/2.
From OX mark off OB equal to PA, and join PB;
then PB? = OP? + OB?=1+2=3.
* PB=V3.
From OX mark off OC equal to PB, and join PC;
then PC?=OP?+00?=1+3=4.
“. PC=,/4,
The iengths of PA, PB, PC may now be found by measurement;
and by continuing the process we may find /5, /6, V7, ....

EXERCISES ON THEOREMS 29, 30 (Continued).

13. Prove the following formula :
Diagonal of square
= side x V2.
Hence find to the nearest centimetre the diagonal of a square on a
side of 50 metres.
Draw a plan (scale 1 em. to 10 metres) and obtain the result as
nearly as you can by measurement.
14. ABC is an equilateral triangle of which each side=2m units,
and the perpendicular from any vertex to the opposite side
= p.
Prove that p=mN3. -
Test this result graphically, when each side=8 cm.
 THEOREM OF PYTHAGORAS. 125

15. Ifina triangle a=m?-1?, b=2mn, c=m?+n?; prove algebraic-

ally that c?=a?+6?
Hence by giving various numerical values to m and n, find sets of
numbers representing the sides of right-angled triangles.

_16. In a triangle ABC, AD is drawn perpendicular to BC. Let p

denote the length of AD.
(i) If a=25 cm., p=12 cm., BD=9 cm.; find b and c
(u) If b=41", c=50", BD=380"; find » and a.
And prove that Vb? — p? +c? -—p?=a.

17. In the triangle ABC, AD is drawn perpendicular to BC.

Prove that
c? - BD?=b?
- CD?.
ika—aliemeo—20ieme. cs cmas: tind BID;
Thence find p, the length of AD, and the area of the triangle ABC.
18. Find by the method of the last example the areas of the triangles
whose sides are as follows:
fijasi7 ,0—107, c=9": (Oo) Paitin, WIM aitig, CHD ie.
(ili) a=41 cm., b=28cm.,c=1l5cem. (iv) a=40 yd.,b=37 yd.,c=13 yd._

19. A straight rod PQ slides between two straight rulers OX, OY

laced at right angles to one another. In one position of the rod
OP—56 em., and OQ=3°3 cm. If in another position OP=4:0 cm.,
find OQ@ graphically; and test the accuracy of your drawing by
calculation.

20. ABC is a triangle right-angled at C, and p is the length of the

perpendicular from C on AB. By expressing the area of the triangle in
two ways, shew that
peo=ab.
ieee le Furl
Hence deduce oFaa 4 be
126 GEOMETRY.

PROBLEMS ON AREAS.

PROBLEM 17.
To describe a parallelogram equal to a given triangle, and having
one of its angles equal to a given angle.

Let ABC be the given triangle, and D the given angle.

It is required to describe a parallelogram equal to ABC, and
having one of its angles equal to D.

Construction. Bisect BC at E.
At E in CE, make the -CEF equal to D;
through A draw AFG par' to BC ;
and through C draw CG par' to EF.
Then FECG is the required par”.

Proof. Join AE.

Now the A*ABE, AEC are on equal bases BE, EC, and of
the same altitude ; ;
.. the AABE =the A AEC.
‘, the A ABC is double of the A AEC.
But FECG is a par™ by construction ;
and it is double of the A AEC,
being on the same base EC, and between the same par™ EC
and AG.
.. the par™ FECG= the A ABC;
and one of its angles, namely CEF, =the given 2D.
 PROBLEMS ON AREAS. 127

EXERCISES,

(Graphical.)
1. Draw a square on a side of 5 cm., and make a parallelogram of
equal area on the same base, and having an angle of 45°.
Find (i) by calculation, (ii) by measurement the length of an oblique
side of the parallelogram.

2. Draw any parallelogram ABCD in which AB=23” and AD=2’;

and on the base AB draw a rhombus of equal area.

DEFINITION. In a_ parallelogram
ABCD, if through any point K in the
diagonal AC parallels EF, HG are drawn
to the sides, then the figures EH, GF
are called parallelograms about AC, and
the figures EG, HF are said to be their
complements.

3. In the diagram of the preceding definition shew by Theorem 21 that

the complements EG, HF are equal in area.
Hence, given a parallelogram EG, and a straight line HK, deduce a
construction for drawing on HK as one side a parallelogram equal and
equiangular to the parallelogram EG.

4. Construct a rectangle equal in area to a given rectangle CDEF,

and having one side equal to a given line AB.
If AB=6 cm., CD=8 cm., CF=3 em., find by measurement the
remaining side of the constructed rectangle.

5. Given a parallelogram ABCD, in which AB=2°-4", AD =1°8", and

the LA=55°. Construct an equiangular parallelogram of equal area,
the greater side measuring 2°7”.. Measure the shorter side.
Repeat the process giving to A any other value; and compare your
results. What conclusion do you draw?

6. Draw a rectangle on a side of 5 cm. equal in area to an

equilateral triangle on a side of 6 cm.
Measure the remaining side of the rectangle, and calculate its
approximate area.
. 128 GEOMETRY.

PROBLEM 18.

To draw a triangle equal m area to a given quadrilateral.

A 5 Xx

Let ABCD be the given quadrilateral.

It is required to describe a triangle equal to ABCD in area.

Construction. Join DB.

Through C draw CX par' to DB, meeting AB produced in X.
Join DX.
Then DAX is the required triangle.

Proof. Now the A* XDB, CDB are on the same base DB and
between the same par™ DB, CX ;
... the A XDB =the ACDB in area.
To each of these equals add the A ADB;
then the A DAX = the fig. ABCD.

CoroLiAry. In the same way it is always possible to draw

a rectilineal figure equal to a given rectilineal figure, and
having fewer sides by one than the given figure; and thus
step by step, any rectilineal figure may be reduced to a
triangle of equal area. 5
For example, in the adjoining dia-
gram the five-sided fig. EDCBA is equal Cc
in area to the four-sided fig. EDXA.
The fig. EDXA may now be reduced
to an equal A DXY. Y A B Xx
 PROBLEMS ON AREAS. 13%

PROBLEM 19.

To draw a paralleiogram equal in area to a gwen rectilineal

fisure, and having an angle equal to a given angle.

K H

A G@ B
i
ist ABCD be the given rectil. fig., and E the given angle.
Ti is required to draw a par” equal to ABCD and having an,
angie equal to E

Construction. Join DB.

Through C draw CF par’ to DB, and meeting AB produced
in F.
Join DF.
Then the A DAF = the fig. ABCD. Prob. 18.
Draw the par™ AGHK equal to the AADF, and having the
4 KAG equal to the 2 E. Prob. 17
Then the par™ KG= the A ADF
=the fig. ABCD ;
| and it has the 2 KAG equal to the 2 E.

Nors. If the given rectilineal figure has more than four sides, it
|must first be reduced, step by step, until it is replaced by an equivalent
‘triangle,
130 GEOMETRY.

EXERCISES.

(Reduction of a Rectilineal Figure to an equivalent Triangle.)

1. Draw a quadrilateral ABCD from the following data :
AB=BC=5'5 em. ; CD=DA=4'5 em.; the LA=75°.
Reduce the quadrilateral to a triangle of equal area. Measure the
base and altitude of the triangle; and hence calculate the approximata
area of the given figure.
2. Draw a quadrilateral ABCD having given :
AB=2'8", BC=3-2’, CD=3°3’, DA=3'6", and the diagonal BD=3°0.
Construct an equivalent triangle; and hence find the approximate
area of the quadrilateral.

3. Ona base AB, 4 cm. in length, describe an equilateral pentagon

{5 sides), having each of the angles at A and B 108°.
Reduce the figure to a triangle of equal area; and by measuring ite
base and altitude, calculate the approximate area of the pentagon.

4. A quadrilateral field ABCD has the following measurements :

AB=450 metres, BC=380 m., CD=330 m., AD=390 m.,
wad the diagonal AC=660 m.
Draw a plan (scale 1 cm. to 50 metres). Reduce your plan to an
equivalent triangle, and measure its base and altitude. Hence estimate
the area of the field.

(Problems. State your construction, and give a theoretical proof.)

5. Reduce a triangle ABC to a triangle of equal area having ite

pase BD of given length. (D lies in BC, or BC produced.)

6. Construct a triangle equal in area to a given triangle, and

having a given altitude.

7. ABC isa given triangle, and X a given point. Draw a triangle

equal in area to ABC, having its vertex at X, and its base in the same
straight line as BC.

8. Construct a triangle equal in area to the quadrilateral ABCD,

having its vertex at a given point X in DC, and its base in the same
straight line as AB.

9. Shew how a triangle may be divided into n equal parts by straight

lines drawn through one of its angular points,
 PROBLEMS ON AREAS, 13}

X10. Bisect a triangle by a straight line drawn through a given point

im one of its sides.
[Let ABC be the given A, and P the A
given point in the side AB. Pp
Bisect AB at Z; and join CZ, CP. Z
Through Z draw ZQ parallel to CP.
Join PQ.
Then PQ bisects the A.]
B Q Cc

ll. Trisect a triangle by straight lines drawn from a given pont an

one of its sides,
[Let ABC be the given A, and X the A
given point in the side BC.
Trisect BC at the points P, Q. p00: be H K
Join AX, and through P and Q draw PH and
QK parallel to AX.
Join XH, XK. :
These straight lines trisect the A; as ma. :
be shewn by joining AP, AQ.] B Pie C

XZ, Cut off from a given triangle a fourth, fifth, sixth, or any part
required by a straight line drawn from a given point in one of its sides,

13. Bisect a quadrilateral by a straight line drawn through an angular

goint.
[Reduce the quadrilateral to a triangle of equal area, and join the
vertex to the middle point of the base.]

14. Cut off from a given quadrilateral a third, a fourth, a fifth, or

any part required, by a straight line drawn through a given angular
pot.
132 GEOMETRY.

AXES OF REFERENCE. COORDINATES.

EXERCISES FOR SQUARED PAPER.
If we take two fixed straight lines XOX’, YOY’ cutting ons
another at right angles at O, the position of any point P with
reference to these lines is known when we know its distances
from each of them. Such lines are called axes of reference,
XOX’ being known as the axis of x, and YOY’ as the axis of y.
Their point of intersection O is called the origin.

The lines XOX’, YOY’ are usually drawn horizontally and

vertically.
In practice the distances of P from the axes are estimated
thus:
From P, PM is drawn perpendicular to X’X; and OM and
PM are measured.
OM is called the abscissa of the point P, and is denoted by 2.
PM _ ordinate oy iy iy ¥.
The abscissa and ordinate taken together are called the
coordinates of the point P, and are denoted by (a, y).
We may thus find the position of a point if its coordinater
are known.
ExampLeE. Plot the point whose coordinates are (5, 4).
Along OX mark off OM, 5 units in length.
At M draw MP perp. to OX, making MP 4 units in length.
Then P is the point whose coordinates are (5, 4).

The axes of reference divide the plane into four regions XOY,
yYOx’, X’‘OY’, Y‘OX, known respectively as the first, second,
third, and fourth quadrants,
 AXES. COORDINATES. 833

It is clear that in each quadrant there is a point whose

distances from the axes are equal to those of P in the above
diagram, namely, 5 units and 4 units.
The coordinates of these points are distinguished by the use
of the positive and negative signs, according to the following
system:
Abscisse measured along the a-axis to the right of the
origin are positive, those measured to the left of the origin
are negative. Ordinates which lie above the «z-axis (that is, in
the first and second quadrants) are positive ; those which lie
below the z-axis (that is, in the third and fourth quadrants)
are negative.
Thus the coordinates of the points Q, R, S are
(-5, 4), (-5, -4), and (5, —4) respectively.

Norg. The coordinates of the origin are (0, 0).

In practice it is convenient to use squared paper. Twa

intersecting lines should be chosen as axes, and slightly
thickened to aid the eye, then one or more of the length-
divisions may be taken as the linear unit. The paper used in
the following examples is ruled to tenths of an inch.

ExamMeLe 1. The coordinates of the points A and B are (7, 8) and

{-—5, 3): plot the points and find the distance between tiem.
After plotting the points as
in the diagram, we may find
AS approximately by direct
measurement.
Or we may proceed thus:
Draw through B a line par!
to XX’ to meet the ordinate
of A at C. Then ACB is a
rt.-angled A in which BC=12,
and AC=5.
Now AB?=BC?+ AC?
= 12?
+ 5?
= 144425
= 169,
- AB=12%
134 GEOMETRY.

EXAMPLE 2. The coordinates of A, B, and C are (5, 7), (-8, 2), and
(3, —5); plot these points and find the area of the triangle of which thess
are the vertices.
na
, : : BEASSHAVSE MBER.
Having plotted the points as in EAL
the diagram, we may measure AB,
and draw and measure the perp.
from Con AB. Hence the approxi-.
mate area may be calculated.

Or we may proceed thus:

Sagi A and B draw AP, BQ
par! to YY’.
oe C draw PQ par’ to
FN 0 Sa'S
RSE
eRit
SP

‘Then the AABC=the trap" APQB -the two rt.-angled At APC, BQG
—}PQ(AP+BQ)-4.AP.PC-4.BQ.QC
=4}x13x19 -4x12x2 -4x7xU
=73 units of area,

EXERCISES.

i, Pilot the following sets of points:

(i) (6, 4), (-6, 4), (-6, - 4), (6, - 4);
(ii) (8, 0), (0, 8), (-8, 0), (0, - 8);
(iii) (12, 5), (5, 12), (-12, 5), (—5, 12).
2. Plot the following points, and shew experimentally that each sed
lie in one straight line,
(i) (9, 7), (0, 0), (-9, —7); (ii) (-9, 7), (0, 0), (9, 7).
Explain these results theoretically.
3.
Ey
Plot the following pairs of points; join the points in each oaas,
and measure the coordinates of the mid-point of the joining line.
(i) (4, 3), (12, 7); (ii) (5, 4), (15, 16).
Shew why in each case the coordinates of the mid-point are respeo-
tively half the sum of the abscissa and half the sum of the ordinates of
the given points.
4, Plot the following pairs of points; and find the coordinates of
the mid-point of their joining lines.
(i) (0, 0), (8, 10); (ii) (8, 0), (0, 10);
(iti) (0, 0), (-8, - 10); {iv) (-8, 0), (0, — 10),
 EXERCISES FOR SQUARED PAPER. 135

5. Find the coordinates of the points of trisection of the line joining

(0, 0) to (18, 15). :
6. Plot the two following series of points:
(i) (5, 0), (5, 2), (5, 5), (5, =)5 (5, a)
(ii) (=4, 8), (= 1, 8), (0, 8), (3, 8), (6, 8).
Shew that they lie on two lines respectively parallel to the axis of y,
and the axis of x Find the coordinates of the point in which they
intersect.
7. Plot the following points, and calculate their distances from the
origin.
(7) (15, 8); (il) (— 15, - 8); (itt) (2°4", 7") sv) (- 7"", 2°4").
Check your results by measurement.
8. Plot the following pairs of points, and in each case calculate the
distance between them.
(i) (4, 0), (0, 3); (ii) (9, 8), (5, 5)5
(aii) (15, 0), (0, 8); (iv) (10, 4), (—5, 12);
(v) (20, 12), (-15, 0); (vi) (20, 9), (—15, — 3).
Verify your calculation by measurement.
9. Shew that the points (-—3, 2), (3, 10), (7, 2) are the angular
points of an isosceles triangle. Calculate and measure the lengths of
the equal sides.
10. Plot the eight points (0, 5), (3, 4), (5, 0), (4, -3), (—5, 0),
(0, —5), (—4, 3), (-4, -3), and shew that they all lie on a circle whose
sentre is the origin.
il. Explain by a diagram why the distances between the following
pairs of points are all equal.
(i) (a, 0), (0, b) ; (ii) (6, 0), (0, a); (iii) (0, 0), (a, 6).
12. Draw the straight lines joining
(i) (a, 0) and (0, a) ; (ii) (0, 0) and (a, a) ;
and prove that these lines bisect each other at right angles.
13. Shew that (0, 4), (12, 9), (12, —4) are the vertices of an isosceles
triangle whose base is bisected by the axis of z.
14. Three vertices of a rectangle are (14, 0), (14, 10), and (0, i9);
find the coordinates of the fourth vertex, and of the intersection of the
diagonals.
15. Prove that the four points (0, 0), (13, 0), (18, 12), (5, 12) are the
angular points of a rhombus. Find the length of each side, and the
coordinates of the intersection of the diagonals.
16. Plot the locus of a point which moves so that its distances from
the points (0, 0) and (4, —4) are always equal to one another. Where
does the locus cut the axes?
136 GEOMETRY.

17. Shew that the following groups of points are the vertices of
rectangles. Draw the figures, and calculate their areas.
(i) (4, 3), (17, 3), (17, 12), (4, 12);
(ii) (3, 2), (3, 15), (-6, 15), (-6, 2) ;

(iii) (5, 1), (-8, 1), (-8, —8), (5, =).

18. Join in order the points (1”, 0), (0, 1”), (-1", 0), (Q, -1”). Ot
what kind is the quadrilateral so formed? Find its area.
If a second figure is formed by joining the middle points of the first,
find its area.

19. Plot the triangles given by the following sets of points ; and find
their areas.
(i) (10, 10), (4, 0), (81, 0); (ii) (10, = 10), (4, 0), (18, 0);
(iii) (- 10, 10), (-4, 0), te 18, 0) > (iv) (= 10, vax 10), (-4, 0), (- 18, 0}.

20. Draw the triangles given by the points

(i) (9, 0), (5, 3), (6, 0); (ii) (0, 0), (3, 0), (0, 6).
Find their areas ; and measure the angles of the first triangle.

21. Plot the triangles given by the following sets of points. Shew
that in each case one side is parallel to one of the axes. ence find the
area.
(i) (0, 0), (12, 10), (12, — 6); (ii) (0, 0), (5, 8), (—15, 8);
(iii) (0, 0), (-12, 12), (-12, - )5 (iv) (0, 0), (-6, —8), (20, -8).
“2. In the following triangles shew that two sides of each are
parallel to the axes. Find their areas,
(i) (5, 5), (15, 5), (15, 15); (ii) (8, 3), (8, 18), (0, 18);
(iii) (4, 8), (—16, —4), (4, -4); (iv) (1, 15), (—11, 15), (1, -7).
23. Shew that (-5, 5), (7, 10), (10, 6), (—2, 1) are the angular
points of a parallelogram. Find its sides and area.

24. Shew that each of the following sets of points gives a trapezium.
Find the area of each. 5
(i) (3, 0), (3, 3), (9, 0), (9, 6) ; (ii) (0, 3), (hms 5, 3), ( _ 2, = 3), (0, = 3);
(iii) (8, 4), (4, 4), (11, -1), (3,-1)5 (iv) (0, 0), (1, 5), (-4, 5), (—8, 0).
25. Find the area of the triangles given by the following points :
(i) (5, 5), (20, 10), (12, 14) ; (ii) (7, 6), (= 10, 4), (- 4, — 3);
(iii) (0, ~ 6), (0, - 3), (14, 5); (iv) (6, 4), (-7, - 6), (-2, - 15).

26. Shew that (-5, 0), (7, 5), (19, 0), (7, —5) are the angular
points of a rhombus. Find its sides and its area.
 EXERCISES FOR SQUARED PAPER. Pou

27, Join the points (0, —5), (12, 0), (4, 6), (—8, —3), in the order
“Saad Calculate the lengths of the first three sides and measure the
urth. Find the areas of the portions of the figure lying in the first
snd fourth quadrants.

28. The coordinates of four points A, B, C, D are respectively

(-—4, —4), (-10, 4), (—10, 18), (5, 5).
Caleulate the lengths of AB, BC, CD, and measure AD. Also calculate
the area of ABCD by considering it as the difference of two triangles.

29. Draw the figure whose angular points are given by

(0, — 3), (8, 3), (=4; 8), (-4, 3), (0, 0).

Find the lengths of its sides, taking the points in the above order.
Also divide it into three right-angled triangles, and hence find its area.

30. A plan of a triangular field ABC is drawn on squared paper™

jscale 1”=100 yds.). On the plan the coordinates of A, B, & are
— 1”, —2”), (3”, 4”), (- 5”, —2”) respectively. Find the area of the field,
the length of the side represented by BOC, and the distance from this
side of the opposite corner of the field.

31. Shew that the points (6, 0), (20, 6), (14, 20), (0, 14) are the
vertices of a square. Measure a side and hence find the approximate
area. Calculate the area exactly (i) by drawing a circumscribing
square through its vertices; (ii) by subdividing the given square as in
the first figure on page 120.
138 GEOMETRY.

MISCELLANEOUS EXERCISES.

1. AB and AC are unequal sides of a triangle ABC; AX is ths

median through A, AP bisects the angle BAC, and AD is the perpen-
dicular from A to BC. Prove that AP is intermediate in position and
magnitude to AX and AD.
2. In a triangle if a perpendicular is drawn from one extremity of
the base to the bisector of the vertical angle, (i) it will make with
either of the sides containing the vertical angle an angle equal to half
the sum of the angles at the base; (ii) it will make with the base an
angle equal to half the difference of the angles at the base.
3. In any triangle the angle contained by the bisector of the
vertical angle and the perpendicular from the vertex to the base is
equal to half the difference of the angles at the base.
4. Construct a right-angled triangle having given the hypotenuse
and the difference of the other sides.
5. Construct a triangle, having given the base, the difference of the
angles at the base, and (i) the difference, (ii) the sum of the remaining
sides.
6. Construct an isosceles triangle, having given the base and the
sum of one of the equal sides and the perpendicular from the vertex te
the base.
7. Shew how to divide a given straight line so that the square on
one part may be double of the square on the other.
8. ABCD is a parallelogram, and O is any point without the angle
BAD or its n goon vertical a shew that the triangle OAC is equal
to the sum of the triangles OAD, OAB.
If O is within the angle BAD or its opposite vertical angle, shew
that the triangle OAC is equal to the difference of the triangles
OAD, OAB.
9. The area of a quadrilateral is equal to the area of a triangle
having two of its sides equal to the diagonals of the given figure, and
the included angle equal to either of the angles between the diagonals.
10. Find the locus of the intersection of the medians of triangles
described on a given base and of given area.
11, On the base of a given triangle construct a second triangle
_ in area to the first, and having its vertex in a given straight
ne.
12. ABCD isa parenologas made of rods connected by hinges, If
AB is fixed, find the loous of the middle point of CD.
 PART Tit.

THE CIRCLE.

DEFINITIONS AND First PRINCIPLES.

1. Acircle is a plane figure contained by a line traced out

by a point which moves so that its distance from a certain
fixed point is always the same.
The fixed point is called the centre, and the bounding line
is called the circumference.
Notr. According to this definition the term circle strictly applies
to the figure contained by the circumference ; it is often used however
for the circumference itself when no confusion is likely to arise.

2. A radius of a circle is a straight line drawn from the

centre to the circumference. It follows that all radii of a
eircle are equal.
3. A diameter of a circle is a straight line drawn through
the centre and terminated both ways by the circumference.

4. A semi-circle is the figure bounded by a diameter of

a circle and the part of the circumference cut off by the
diameter.
It will be proved on page 142 that a diameter divides a circle into
two identically equal parts.

5. Circles that have the same centre are said to be

eoncentric.
 —
£40 GEOMETRY.

From these definitions we draw the following inferences:

(i) A circle is a closed curve; so that if the circumference
is crossed by a straight line, this line if produced will cross
the circumference at a second point.
(ii) The distance of a point from the centre of a circle
is greater or less than the radius according as the point is
without or within the circumference.
(iii) A point is outside or inside a circle according as its
distance from the centre is greater or less than the radius.
(iv) Circles of equal radii are identically equal. For by
superposition of one centre on the other the circumferences
must coincide at every point.
(v) Concentric circles of unequal radii cannot intersect, for
the distance from the centre of every point on the smaller
circle is less than the radius of the larger.
(vi) If the circumferences of two circles have a common
point they cannot have the same centre, unless they coincide
altogether.

6. An arc of a circle is any part of the circumference.

7. A chord of a circle is a straight line joining any two

points on the circumference.
Nots. From these definitions it may be seen
that a chord of a circle, which does not pass through
the centre, divides the circumference into two un-
equal ares ; of these, the greater is called the major
are, and the less the minor are. ‘Thus the major
are is greater, and the minor are /ess than the semi-
circumference.
The major and minor ares, into which a cir- — *s ’
cumference is divided by a chord, are said to be “s .
conjugate to one another.
 * SYMMETRY OF A CIRCLE, ; 141

SYMMETRY.
Some elementary properties of circles are easily proved by
considerations of symmetry. For convenience the definition
given on page 21 is here repeated.
DEFINITION 1. A figure is said to be symmetrical about a
line when, on being folded about that line, the parts of the
figure on each side of it can be brought into coincidence.
The straight line is called an axis of symmetry.
That this may be possible, it is clear that the two parts of the figure
must have the same size and shape, and must be similarly placed with
regard to the axis. ;

DEFINITION 2. Let AB be a straight line and P a point

outside it.
P

From P draw PM perp. to AB, and produce it to Q, making

MQ equal to PM.
Then if the figure is folded about AB, the point P may be
made to coincide with Q, for the .AMP=the ZAMQ, and
MP= MQ.

The points P and Q are said to be symmetrically opposite

with regard to the axis AB, and each point is said to be the
image of the other in the axis.

Note. A point and its image are equidistant from every point on
the axis. See Prob. 14, page 91.
142 GEOMETRY.

SoME SYMMETRICAL PROPERTIES OF CIRCLES.

I. A circle is symmetrical about any diameter.

Let APBQ be a circle of which O is the centre, and AB any

diameter.
It is required to prove that the circle is symmetrical about AB.
Proof. Let OP and OQ be two radii making any equal
4*AOP, AOQ on opposite sides of OA.
Then if the figure is folded about AB, OP may be made to
fall along OQ, since the 2 AOP=the 2 AOQ.
And thus P will coincide with Q, since OP=0Q.
Thus every point in the arc APB must coincide with some
eo in the are AQB; that is, the two parts of the circum-
erence on each side of AB can be made to coincide.
. the cirele is symmetrical about the diameter AB.
CoroLuary. If PQ is drawn cutting AB at M, then on
folding the figure about AB, since P falls on Q, MP will
coincide with MQ,
“. MP=MQ;
and the 2 OMP will coincide with the 2 OM
‘. these angles, being adjacent, are rt. 2";
. the points P and Q are symmetrically opposite with
regard to AB.
Hence, conversely, if a circle passes through a given point P,
it also passes through the symmetrically opposite point with regard
to any diameter.
DerinitTion. The straight line passing through the centres
of two circles is called the line of centres.
 SYMMETRICAL PROPERTIES. 143

I. Two circles are divided symmetrically by their line of centres.

Vet O, O’ be the centres of two circles, and let the st. line
through O, O’ cut the O° at A, B and A, B’.. Then AB and
A’B’ are diameters and therefore axes of symmetry of their
respective circles. That is, the line of centres divides each
circle symmetrically.

ill. Jf two circles cut at one point, they must also cut at a
second point ; and the common chord is bisected at right angles by
the line of centres.

Let the circles whose centres are O, O’ cut at the point P.

Draw PR perp. to OO’, and produce it to Q, so that
RQ = RP.
Then P and Q@ are symmetrically opposite points with
regard to the line of centres OO’ ;
.. since P is on the O® of both circles, it follows that Q is
also on the O° of both. [I. Cor.]
And, by construction, the common chord PQ is bisected at
right angles by OO’.
144 GEOMETRY.

ON CHORDS.

THEOREM 31. [Euclid III. 3.]

If a straight line drawn from the centre of a circle bisects a
chord which does not pass through the centre, ti cuts the chord at
right angles.
Conversely, if it cuts the chord at right angles, ti bisects it.

<a
Let ABC be a circle whose centre is O; and let OD bisect
a chord AB which does not pass through the centre.
It is required to prove that OD is perp. to AB.
Join OA, OB.
Proof. Then in the A* ADO, BDO,
AD = BD, by hypothesis,
because OD is common,
and O4= OB, being radii of the circle;
. the .ADO=the 2 BDO; Theor. 7.
and these are adjacent angles,
.. OD is perp. to AB. Q.E.D.
Conversely. Let OD be perp. to the chord AB.
It is required to prove that OD bisects AB.
Proof. In the A* ODA, ODB,
the 2" ODA, ODB are right angles,
because 4 the hypotenuse OA = the hypotenuse OB,
lend
‘ OD is common ;
*, DA=DB; Theor. 18,
that is, OD bisects AB at D. Q.E.D,
 CHORD PROPERTIES. 145

CoroLuaRY 1. The straight line which bisects a chord at

right angles passes through the centre.

COROLLARY 2. A straight line cannot meet a circle at more

than two points.
For suppose a st. line meets a O
circle whose centre is O at the points
A and B.
Draw OC perp. to AB.
Then AC=CB A C BD
Now if the circle were to cut AB in a third point D, AC
would also be equal to CD, which is impossible.

Corotuary 3. A chord of a circle lies wholly within it. .

EXERCISES.

(Numerical and Graphical.)

1. In the figure of Theorem 31, if AB=8 cm., and OD=3 em., find
OB. Draw the figure, and verify your result by measurement.
2. Calculate the length of a chord which stands at a distance 5”
from the centre of a circle whose radius is 13”.

3. In a circle of 1” radius draw two chords 1:6” and 1:2” in length.

Calculate and measure the distance of each from the centre.

4. Draw a circle whose diameter is 8°0 cm. and place in it a chord

6:0 cm. in length. Calculate to the nearest millimetre the distance of
the chord from the centre ; and verify your result by measurement,
5. Find the distance from the centre of a chord 5 ft. 10 in. in length
in a circle whose diameter is 2 yds. 2in. Verify the result graphically
by drawing a figure in which 1 em. represents 10’.
6. AB is a chord 2:4” long in a circle whose centre is O and whose
radius is 1°3”; find the area of the triangle OAB in square inches.
7. Two points P and Q are 8” apart. Draw a circle with radius 1°7”
to pass through P and Q. Calculate the distance of its centre from the
chord PQ, and verify by measurement.
H.S.G. K
146 GEOMETRY.

THEOREM 32.
One circle, and only one, can pass through any three points not
on the same straight line.

Let A, B, C be three points not in the same straight line.

It is required to prove that one circle, and only one, can pass
through A, B, and C.
Join AB, BC,
Let AB and BC be bisected at right angles by the lines
DF, EG.
Then since AB and BC are not in the same st. line, DF and
EG are not par’.
Let DF and EG meet in O.

Proof. Because DF bisects AB at right angles,

. every point on DF is equidistant from A and B.
Prob. 14,
Similarly every point on EG is equidistant from B and C.
‘. O, the only point common to DF and EG, is equidistant
from A, B, and C;
and there is no other point equidistant from A, B, and C.
. a circle having its centre at O and radius OA will pass
through B and C; and this is the only circle which will pass
through the three given points. Q.E.D.
 CHORD PROPERTIES. 147

CoROLLaRY 1. The size and position of a circle are fully

determined if it is known to pass through three given points ; for
then the position of the centre and length of the radius can
be found.
COROLLARY 2. Two circles cannot cut one another in more
than two points without coinciding entirely; for if they cut at
three points they would have the same centre and radius.
HYPOTHETICAL ConstRucTION. From Theorem 32 it appears
that we may suppose a circle to be drawn through any three points
not in the same straight line.
For example, a circle can be assumed tu pass through the vertices of
any triangle. :

DrFINITION. The circle which passes through the vertices

of a triangle is called its circum-circle, and is said to be
circumscribed about the triangle. The centre of the circle is
called the circum-centre of the triangle, and the radius is called
the circum-radius.

EXERCISES ON THEOREMS 31 AND 32.

(Theoretical. )
1, The parts of a straight line intercepted between the circum-
ferences of two concentric circles are equal.
2. Two circles, whose centres are at Aand B, intersect at C, D; and
M is the middle point of the common chord. Shew that AM and BM
are in the same straight line.
Hence prove that the line of centres bisects the common chord at right
anyles.
3. AB, AC are two equal chords of a circle ;shew that the straight
line which bisects the angle BAC passes through the centre.
4. Find the locus of the centres of all circles which pass through two
given points.
5. Describe a circle that shall pass through two given points and have
its centre in a given straight line.
When is this impossible?
6. Describe a circle of given radius to pass through two given points.
When is this impossible?
148 GEOMETRY.

* THEOREM 33. [Euclid III. 9.]

If from a point within a circle more than two equal straight
lines can be drawn to the circumference, that point is the centre
of the circle.

DT
Let ABC be a circle, and O a point within it from which
more than two equal st. lines are drawn to the O*, namely
OA, OB, OC.
It is required to prove that O is the centre of the circle ABC.
Join AB, BC.
Let D and E be the middle points of AB and BC respectively.
Join OD, OE. :

Proof. In the A* ODA, ODB,

{ DA= DB,
because DO is common,
land OA= OB, by hypothesis;
*, the .ODA=the 2 ODB; Theor. 7.
.. these angles, being adjacent, are rt. 2°.
Hence DO bisects the chord AB at right angles, and therefore
passes through the centre. Theor. 31, Cor. 1.
Similarly it may be shewn that EO passes through the
centre,
.*. O, which is the only point common to DO and EO, must
be the centre. Q.E.D.
 CHORD PROPERTIES. 149

EXERCISES ON CHORDS.

(Numerical and Graphical.)

1. AB and BC are lines at right angles, and their lengths are 1-6”
and 3:0” respectively. Draw the circle through the points A, B, and C;
find the length of its radius, and verify your result by measurement.
2. Draw a circle in which a chord 6 cm, in length stands at a
distance of 3 cm. from the centre.
Calculate (to the nearest millimetre) the length of the radius, and
verify your result by measurement.
3. Draw a circle on a diameter of 8 cm., and place in it a chord
equal to the radius.
Calculate (to the nearest millimetre) the distance of the chord from
the centre, and verify by measurement. 3
4. Two circles, whose radii are respectively 26 inches and 25
inches, intersect at two points which are 4 feet apart. Find the
distances between their centres.
Draw the figure (scale 1 cm. to 10”), and verify your result by
measurement.

5.. Two parallel chords of a circle whose diameter is 13” are re-
spectively 5’ and 12” in length: shew that the distance between them
is either 8°5’ or 3°5".
6. Two parallel chords of a circle on the same side of the centre are
6 cm. and 8 cm. in length respectively, and the perpendicular distance
between them is lem. Calculate and measure the radius.
7. Shew on squared paper that if a circle has its centre at any point
on the z-axis and passes through the point (6, 5), it also passes through
the point (6, - 5). [See page 132.]

(Theoretical.)

8. The line joining the middle points of two parallel chords of a

circle passes through the centre.
9. Find the locus of the middle points of parallel chords in a circle.
10. Two intersecting chords of a circle cannot bisect each other
unless each is a diameter.
11. Ifa parallelogram can be inscribed in a circle, the point of inter~
section of its diagonals must be at the centre of the circle.
12. Shew that rectangles are the only parallelograms that can be:
inscribed in a circle.
150 GEOMETRY.

THEOREM 34. [Euclid III. 14.]

Equal chords of a circle are equidistant from the centre.
Conversely, chords which are equidistant from the centre are
equal.

Let AB, CD be chords of a circle whose centre is O, and let

OF, OG be perpendiculars on them from O.
First. Let AB=CD.
It is required to prove that AB and CD are equidistant from O.
Join OA, OC.

Proof. Because OF is perp. to the chord AB,

.. OF bisects AB; Theor. 31.
.. AF is half of AB.
Similarly CG is half of CD.
But, by hypothesis, AB =CD,
“, AF=CG.

Now in the A* OFA, OGC,

the 2" OFA, OGC are right angles,
because ,the hypotenuse OA=the hypotenuse OC,
| and AF=CG;
.”, the triangles are equal in all respects; Theor. 18,
so that OF = OG ;
that is, AB and CD are equidistant from O.
Q.E.D,
 CHORD PROPERTIES. 151

Conversely. Let OF = OG,

It is required to prove that AB=CD.
Proof.. As before it may be shewn that AF is half of AB,
and CG half of CD.
Then in the A* OFA, OGC,
ee L* OFA, OGC are right angles,
because {the hypotenuse OA=the hypotenuse OC,
| and OF = OG;
Ar —CGs Theor. 18.
.. the doubles of these are equal ;
that is, AB=CD.
Q.E.D.

EXERCISES.

(Theoretical.)
1. Find the locus of the middle points of equal chords of a circle.
2. If two chords of a circle cut one another, and make equal
angles with the straight line which joins their point of intersection to
the centre, they are equal.
3. If two equal chords of a circle intersect, shew that the segments
of the one are equal respectively to the segments of the other.
4. In a given circle draw a chord which shall be equal to one given
straight line (not greater than the diameter) and parallel to another.
5. PQ is a fixed chord in a circle, and AB is any diameter: shew
that the sum or difference of the perpendiculars let fall from A and B
on PQ is constant, that is, the same for all positions of AB.
[See Ex. 9, p. 65.]

(Graphical.)
6. In a circle of radius 4:1 em. any number of chords are drawn
each 18 cm. in length. Shew that the middle points of these chords
all lie on a circle. Calculate and measure the length of its radius, and
draw the circle.
7. he centres of two circles are 4” apart, their common chord is
2°4” in length, and the radius of the larger circle is 3°7’.. Give a con-
struction for finding the points of intersection of the two circles, and
find the radius of the smaller circle.
152 GEOMETRY.

THEOREM 35. [Euclid III. 15.]

Of any two chords of a circle, that which is nearer to the centre
is greater than one more remote.
Conversely, the greater of two chords is nearer to the centre than
the less.

A Cc

Let AB, CD be chords of a circle whose centre is O, and let

OF, OG be perpendiculars on them from O.
It is required to prove that
(i) if OF ts less than OG, then AB is greater than CD ;
(ii) if AB is greater than CD, then OF is less than OG.
Join OA, OC.

Proof. Because OF is perp. to the chord AB,

. OF bisects AB ;
*, AF is half of AB.
Similarly CG is half of CD.
Now OA=00C,
*. the sq. on OA=the sq. on OC,
But since the 2 OFA is a rt. angle,
*. the sq. on OA=the sqq. on OF, FA.
Similarly the sq. on OC =the sqq. on OG, GC.
.. the sqq. on OF, FA=the sqq. on OG, GC.
 CHORD PROPERTIES. 153

(i) Hence if OF is given less than OG;

the sq. on OF is less than the sq. on OG.
.*. the sq. on FA is greater than the sq. on GC;
*, FA is greater than GC:
‘. AB is greater than CD.

(ii) But if AB is given greater than CD,

that is, if FA is greater than GC ;
then the sq. on FA is greater than the sq. on GC.
*, the sq. on OF is less than the sq. on OG;
‘, OF is less than OG. Q.E.D.

COROLLARY. The greatest chord in a circle is a diameter.

EXERCISES.

(Miscellaneous. )

1. Through a given point within a circle draw the least possible

chord.
2. Draw a triangle ABC in which a=3°5", b=1°2", c=3°7".. Through
the ends of the side a draw a circle with its Centre on the side ec.
Caleulate and measure the radius.

38. Draw the circum-circle of a triangle whose sides are 2°6’, 2°8”,
and 3:0”. Measure its radius.

4, AB isa fixed chord of a circle, and XY any other chord having

its middle point Z on AB; what is the greatest, and what the least
length that XY may have ?
Shew that XY increases, as Z approaches the middle point of AB.
5. Shew on squared paper that a circle whose centre is at the
origin, and whose radius is 3:0’, passes through the points (2°4”, 1°8”),
(iss 240) e
Find (i) the length of the chord joining these points, (ii) the co-
ordinates of its middle point, (iii) its perpendicular distance from the
origin.
154 GEOMETRY.

*THEOREM 36. [Euclid III. 7.]

If from any internal point, not the centre, straight lines are
drawn to the circumference of a circle, then the greatest is that
which passes through the centre, and the least is the remaining part
of that diameter.
And of any other two such lines the greater is that which sub-
tends the greater angle at the centre.
A

Let ACDB be a circle, and from P any internal point, which

is not the centre, let PA, PB, PC, PD be drawn to the O%,
so that PA passes through the centre O, and PB is the remain-
ing part of that diameter. Also let the 2 POC at the centre
subtended by PC be greater than the 2 POD subtended by PD.
It is vequired to prove that of these st. lines
(i) PA is the greatest,
(ii) PB is the least,
(iii) PC 1s greater than PD.
Join OC, OD.
Proof. (i) In the APOC, the sides PO, OC are together
greater than PC. Theor. 11.
But OC =OA, being radii ;
.. PO, OA are together greater than PC ;
that is, PA is greater than PC.
Similarly PA may be shewn to be greater than any other
st. line drawn from P to the O* ;
*, PA is the greatest of all such lines.
 DISTANCE OF A POINT TO THE CIRCUMFERENCE. 155

(ii) In the AOPD, the sides OP, PD are together greater

than OD.
But OD = OB, being radii ;
.. OP, PD are together greater than OB.
Take away the common part OP;
then PD is greater than PB.
Similarly any other st. line drawn from P to the O* may
be shewn to be greater than PB ;
‘. PB is the least of all such lines.

(iii) In the A‘ POC, POD,

PO is common,
because OC = OD, being radii,
but the 2 POC is greater than the 2 POD ;
. PC is greater than PD. Theor. 19.
Q.E.D.

EXERCISES.

(Miscellaneous. )
1. All circles which pass through a fixed point, and have their centres
on a given straight line, pass also through a second fixed point.
2. If two circles which intersect are cut by a straight line parallel
to the common chord, shew that the parts of it intercepted between
the circumferences are equal.
3. If two circles cut one another, any two parallel straight lines
drawn through the points of intersection to cut the circles are equal.
4. If two eireles cut one another, any two straight lines drawn
through a point of section, making equal angles with the common
. chord, and terminated by the circumferences, are equal.
5. Two circles of diameters 74 and 40 inches respectively have a
common chord 2 feet in length: find the distance between their centres.
Draw the figure (1 em. to represent 10”) and verify your result by
measurement,
6. Draw two circles of radii 1°0’ and 1°7’, and with their centres
271’ apart. Find by calculation, and by measurement, the length of
the common chord, and its distance from the two centres.
156 GEOMETRY.

*THEOREM 37. [Euclid IL 8.]

If from any external point straight lines are drawn to the
circumference of a circle, the greatest is that which passes through
the centre, and the least is that which when produced passes through
the centre.
And of any other two such lines, the greater is that which sub-
tends the greater angle at the centre.

Ss.
2)

Let ACDB be a circle, and from any external point P let the
lines PBA, PC, PD be drawn to the O*, so that PBA passes
through the centre O, and so that the 2 POC subtended by PC
at the centre is greater than the 2 POD subtended by PD.
It is required to prove that of these st. lines
(i) PA is the greatest,
(ii) PB is the least,
(iii) PC is greater than PD.
Join OC, OD.

Proof. (i) In the APOC, the sides PO, OC are together

greater than PC.
But OC = OA, being radii ;
‘, PO, OA are together greater than PC;
that is, PA is greater than PC.
Similarly PA may be shewn to be greater than any other st.
line drawn from P to the O°;
that is, PA is the greatest of all such lines.
 DISTANCE OF A POINT TO THE CIRCUMFERENCE. 157

(ii) In the APOD, the sides PD, DO are together greater

than PO.
But OD = OB, being radii ;
*, the remainder PD is greater than the remainder PB.
Similarly any other st. line drawn from P to the O* may be
shewn to be greater than PB ;
that is, PB is the least of all such lines.

(tii) In the A* POC, POD,

fee is common,
because ;OC = OD, being radii ;
\but the 4 POC is greater than the 2 POD;
*. PC is greater than PD. Theor. 19.
Q.E.D.

EXERCISES.

(Miscellaneous. )
1. Find the greatest and least straight lines which have one ex-
tremity on each of two given circles which do not intersect.
2. If from any point on the circumference of a circle straight lines
are drawn to the circumference, the greatest is that which passes
through the centre; and of any two such lines the greater is that
which subtends the greater angle at the centre.
3. Of all straight lines drawn through a point of intersection of two
circles, and terminated by the circumferences, the greatest is that which
is parallel to the line of centres.
4. Draw on squared paper any two circles which have their centres
on the x-axis, and cut at the point (8, —11). Find the coordinates of
their other point of intersection.
5. Draw on squared paper two circles with centres at the points
(15, 0) and (—6, 0) respectively, and cutting at the point (0, 8). Find
the lengths of their radii, and the coordinates of their other point of
intersection.
6. Draw an isosceles triangle OAB with an angle of 80° at its vertex
O. With centre O and radius OA draw a circle, and on its circum-
ference take any number of points P, Q, R, ... on the same side of
AB as the centre. Measure the angles subtended by the chord AB at
the points P, Q, R, .... Repeat the same exercise with any other given
angle at O. What inference do you draw?
158 GEOMETRY.

ON ANGLES IN SEGMENTS, AND ANGLES AT THE

CENTRES AND CIRCUMFERENCES OF CIRCLES.

THEOREM 38. [Euclid III. 20.]

The angle at the centre of a circle is double of an angle at the
circumference standing on the same are.
A

Fig. 1. Figs

Let ABC be a circle, of which O is the centre; and let BOC

be the angle at the centre, and BAC an angle at the O*,
standing on the same are BC.
It is required to prove that the L BOC ts twice the 2 BAC.
Join AO, and produce it to D.
Proof. In the A OAB, because OB= OA,
*. the 2 OAB=the 2 OBA.
*. the sum of the 2'OAB, OBA= twice the 2 OAB.
‘But the ext. 2 BOD =the sum of the -* OAB, OBA;
. the BOD =twice the 2 OAB.
Similarly the 2 DOC =twice the 2 OAC.
.., adding these results in Fig. 1, and taking the difference
in Fig. 2, it follows in each case that
the 2 BOC =twice the 2 BAC. Q.E.D.
 ANGLE PROPERTIES. 159

x S,
E
Fig. 3. Fig. 4.

Obs. If the are BEC, on which the angles stand, is a semi-

circumference, as in Fig. 3, the 2 BOC at the centre is a
straight angle; and if the are BEC is greater than a semi-
circumference, as in Fig. 4, the 2 BOC at the centre. is reflex.
But the proof for Fig. 1 applies without change to both these
cases, shewing that whether the given arc is greater than,
equal to, or less than a semi-circumference,
the L BOC = twice the 2 BAC, on the same arc BEC.

DEFINITIONS.

A segment of a circle is the figure bounded

by a chord and one of the two ares into which
. the chord divides the cireumference.
Notr. The chord of asegment is sometimes called . 7
its base.

An angle in a segment is one formed by two

straight lines drawn from any point in the arc
of the segment to the extremities of its chord.

We have seen in Theorem 32 that a circle may be drawn

through any three points not in a straight line. But it is
only under certain conditions that a circle can be drawn
through more than three points.
DEFINITION. If four or more points are so placed that a
circle may be drawn through them, they are said to be
concyclic.
160 GEOMETRY.

THEOREM 39. (Euclid III. 21.j

Angles in the same segment of a circle are equal.

Let BAC, BDC be angles in the same segment BADC of a

circle, whose centre is O.
It is required to prove that the L BAC =the 2 BDC.
Join BO, OC.

Proof. Because the 2 BOC is at the centre, and the 2 BAC —

at the O", standing on the same are BC,
‘. the 2 BOC = twice the z BAC. Theor. 38.

Similarly the 2 BOC = twice the 2 BDC.

.. the 2 BAC=the 2 BDC. Q.E.D.

Nort. The given segment may be greater than a semicirele ag in

Fig. 1, or less than a semicircle as in Fig. 2: in the latter case the angle
BOC will be reflex. But by virtue of the extension of Theorem 38.
"ewe on the preceding page, the above proof applies equally to both
igures.
 ANGLE PROPERTIES. 161

CoNVERSE OF THEOREM 39.

Equal angles standing on the same base, and on the same side of tt,
have their vertices on an arc of a circle, of which the given base is the
chord.
Let BAC, BDC be two equal angles standing on A ED
the same base BC, and on the same side of it.
It is required to prove that A and D lie on an are
of a circle having BC as tts chord.
Let ABC be the circle which passes through
the three points A, B, C; and suppose it cuts BD B C
or BD produced at the point E.
Join EC.

Proof. Then the 2BAC=the 2 BEC in the same segment.

But, by hypothesis, the 2 BAC=the BDC;
sethes“BEC=the ZBDO-:
which is impossible unless E coincides with D ;
*. the circle through B, A, C must pass through D.

Corotuary. The locus of the vertices of triangles drawn on the same

side of a given base, and with equal vertical angles, 1s an arc of a circle.

EXERCISES ON THEOREM 39.

1. In Fig. 1, if the angle BDC is 74°, find the number of degrees in

each of the angles BAC, BOC, OBC.
2. In Fig. 2, let BD and CA intersect at X. If the angle DXC=40°,
and the angle XCD =25°, find the number of degrees in the angle BAC
and in the reflex angle BOC.
3. In Fig. 1, if the angles CBD, BCD are respectively 43° and 82°,
find the number of degrees in the angles BAC, OBD, OCD.
4. Shew that in ‘Fig. 2 the angle OBC is always less than the angle
BAC by a right angle.

[For further Exercises on Theorem 39 see page 170.]

H.S.G. L
162 GEOMETRY.

THEOREM 40. [Euclid III. 22.]

The opposite angles of any quadrilateral inscribed in a circle
are together equal to two right angles.

Let ABCD be a quadrilateral inscribed in the © ABC,

It is required to prove that
(i) the L'ADC, ABC together = two rt. angles.
(ii) the £8 BAD, BCD together= two rt. engles.
Suppose O is the centre of the circle.
Join OA, OC.

Proof. Since the ~ADC at the O*=half the 2 AOC at the

centre, standing on the same arc ABC;
and the 2 ABC at the O°=half the reflex 2 AOC at the centre,
standing on the same are ADC;
. the L*ADC, ABC together =half the sum of the 2 AOC and
the reflex 2 AOC.
But these angles make up four rt. angles.
‘. the L*ADC, ABC together= two rt. angles.
Similarly the 2* BAD, BCD together= two rt. angles.
Q.E.D.
Notre. The results of Theorems 39 and 40 should be carefully
compared,
From Theorem 39 we learn that angles in the same segment are
equal.
From Theorem 40 we learn that angles in conjugate segments are
supplementary.
DeFINITION. A quadrilateral is called cyclic when a circle
can be drawn through its four vertices.
 ANGLE PROPERTIES. 163

CONVERSE OF THEOREM 40.

If a pair of opposite angles of a quadrilateral are supplementary, its

vertices are concyclic.
Let ABCD be a quadrilateral in which the A
opposite angles at B and D are supplementary. E D
It is required to prove that the points A, B, C, D
are concyclic.
Let ABC be the circle which passes through
the three points A, B, C; and suppose it cutsAD Bp C
or AD produced in the point E.
Join EC.

Proof. Then since ABCE is a cyclic quadrilateral,

“. the LAEC is the supplement of the LABC.
But, by hypothesis, the L ADC is the supplement of the LABC;
: the LAEC=the ZLADC;
which is impossible unless E coincides with D.
*. the circle which passes through A, B, C must pass through D:
that is, A, B, C, D are concyclie. Q.E.D.

EXERCISES ON THEOREM 40.

1. Ina circle of 1°6” radius inscribe a quadrilateral ABCD, making
the angle ABC equal to 126’. Measure the remaining angles, and
hence verify in this case that opposite angles are supplementary.
2. Prove Theorem 40 by the aid of Theorems 39 and 16, after first
joining the opposite vertices of the quadrilateral.
3. Ifa circle can be described about a parallelogram, the parailelo
gram must be rectangular.
4. ABC is an isosceles triangle, and XY is drawn parallel to the
base BC cutting the sides in X and Y: shew that the four points B, C,
X, Y lie on a circle.
5. If one side of a cyclic quadrilateral is produced, the exterior angle
ts equal to the opposite interior angle of the quadrilateral.
164 GEOMETRY.

THEOREM 41. [Euclid III. 31.]

The angle in a semi-circle is a right angle.

Let ADB be a circle of which AB is a diameter and O the

centre ; and let C be any point on the semi-circumference ACB.
It is required to prove that the 2 ACB is a rt. angle.

1st Proof. The 2ACB at the O” is half the straight angle

AOB at the centre, standing on the same are ADB;
and a straight angle = two rt. angles:
‘. the 2 ACB is a rt. angle. Q.E.D.

2nd Proof. Join OC.

Then because OA= OC,
.. the 2 OCA=the z OAC. Theor. 5.
And because OB = OC,
.. the 2 OCB=the z OBC.
.. the whole .ACB=the 2 OAC + the 2 OBC.
But the three angles of the A ACB together = two rt. angles;
‘, the .ACB= one-half of two rt. angles
=one rt. angle. Q.E.D.
 ANGLE PROPERTIES. 165

CoRoLuLARY. The angle in a segment greater than a semi-circle

is acute ; and the angle in a segment less than a semi-circle is obtuse.

The 2 ACB at the O* is half the 2 AOB at the centre, on the

same are ADB.
(i) If the segment ACB is greater than a semi-circle, then
_ADB is a minor arc ;
*, the 2 AOB is less than two rt. angles;
*, the 2 ACB is less than one rt. angle.
(11) If the segment ACB is less than a semi-circle, then ADB
is a major arc ;
.. the LAOB is greater than two rt. angles;
*, the 2 ACB is greater than one rt. angle.

EXERCISES ON THEOREM 41.

1. A circle described on the hypotenuse of a right-angled triangle as
diameter, passes through the opposite angwar point.
2. Two circles intersect at A and B; and through A two diameters
AP, AQ are drawn, one in each circle: shew that the points P, B, Q
are collinear.
3. A circle is described on one of the equal sides of an isosceles
triangle as diameter. Shew that it passes through the middle point of
the base.
4, Circles described on any two sides of a triangle as diameters
intersect on the third side, or the third side produced.
5. A straight rod of given length slides between two straight rulers
placed at right angles to one another ; find the locus of its middle point.
6. Find the locus of the middle points of chords of a circle drawn
through a fixed point. Distingwish between the cases when the given point
is within, on, or without the circumference. ectmse

DeFINITION. A sector of a circle isa figure / .

bounded by two radii and the are intercepted | '
between them. ‘ <Yy
166 GEOMETRY.

THEOREM 42. (Euclid III. 26.]

In equal circles, arcs which subtend equal angles, either at the
centres or at the cireumsfer ences, are equal.

B Cc
K

Let ABC, DEF be equal circles, and let the 2 BGC = the LEHF
at the centres ; and consequently
the 2BAC=the J EDF at the O*. Theor, 38.
It is required to prove that the are BKC= the are ELF.

Proof. Apply the ©ABC to the © DEF, so that the centre G

falls on the centre H, and GB falls along HE.
Then because the 2 BGC = the 2 EHF,
GC will fall along HF.
And because the circles have equal radii, B will fall on E,
and C on F, and the circumferences of the circles will coincide
entirely.
*. the are BKC must coincide with the are ELF;
*. the are BKC =the are ELF. Q.E.D.

CoroLiaRy. In equal circles sectors which have equal angles

are equal.

Obs. It is clear that any theorem relating to arcs, angles,

and chords in equal circles must also be true in the same circle.
 ARCS AND ANGLES. 167

THEorEM 43. [Euclid II. 27.]

In equal circles angles, either at the centres or at the circum-
ferences, which stand on equal ares are equal.

A D

K tC

Let ABC, DEF be equal circles;

and let the are BKC = the are ELF.
It is required to prove that
the L BGC =the 2 EHF at the centres ;
also the L BAC=the L EDF at the O™™.

Proof. Apply the © ABC to the ©DEF, so that the centre G

falls on the centre H, and GB falls along HE.
Then because the circles have equal radii,
*. B falls on E, and the two O*™ coincide entirely.
And, by hypothesis, the are BKC =the arc ELF.
.. C falls on F, and consequently GC on HF;
: .. the L BGC=the 2 EHF.
And since the 2 BAC at the O*=half the 2 BGC at the centre ;
and likewise the 2 EDF = half the 2 EHF;
“. the 2 BAC=the 2 EDF. Q.E.D.
168 GEOMETRY.

THEOREM 44, [Euclid III. 28.]

In equal circles, arcs which are cut off by equal chords are equal,
the major arc equal to the major arc, and the minor to the minor.

K L

Let ABC, DEF be equal circles whose centres are G and H;

and let the chord BC =the chord EF.
It is required to prove that
the major arc BAC = the major arc EDF,
and the minor arc BKC = the minor are ELF.
Join BG, GC, EH, HF.

Proof. In the A* BGC, EHF,

BG =EH, being radii of equal circles,
because GC =HF, for the same reason,
ih BC=EF, by hypothesis ;
.. the 2 BGC=the 2 EHF; Theor. 7.
.. the are BKC=the arc ELF; Theor. 42.
and these are the minor arcs.
+
But the whole O° ABKC =the whole O° DELF ;
.. the remaining are BAC = the remaining are EDF :
and these are the major arcs. Q.E.D.
 ARCS AND CHORDS. 169

THEOREM 45. [Euclid HI. 29.]

In equal circles chords which cut off equal arcs are equal.

Let ABC, DEF be equal circles whose centres are G and H;

and let the are BKC=the arc ELF.
It is required to prove that the chord BC =the chord EF.
Join BG, EH.

Proof. Apply the ©ABC to the ©DEF, so that G falls on H

and GB along HE.
Then because the circles have equal radii,
‘, B falls on E, and the O™ coincide entirely.
And because the are BKC= the arc ELF,
77 (C datisvonuk:
*. the chord BC coincides with the chord EF;
... the chord BC=the chord EF. Q.E.D.
170 GEOMETRY.

EXERCISES ON ANGLES IN A CIRCLE.

1. P is any point on the are of a segment of which AB is the chord.

Shew that the sum of the angles PAB, PBA is constant.
2. PQ and RS are two chords of a circle intersecting at X: prove
that the triangles PXS, RXQ are equiangular to one another.

3. Two circles intersect at A and B; and through A any cag

line PAQ is drawn terminated by the circumferences: shew that PQ
subtends a constant angle at B.

4. Two circles intersect at A and B; and through A any two

straight lines PAQ, XAY are drawn terminated by the circumferences;
shew that the ares PX, QY subtend equal angles at B.

5. P is any point on the are of a segment whose chord is AB: and

the angles PAB, PBA are bisected by straight lines which intersect at O.
Find the locus of the point O.

6. If two chords intersect within a circle, they form an angle equal to

that at the centre, subtended by half the sum of the arcs they cut off.

7. If two chords intersect without a circle, they form an angle equal

to that at the centre subtended by half the difference of the arcs they cut off.

8. The sum of the ares cut off by two chords of a circle at right
angles to one another is equal to the semi-cireumference.

9. If AB is a fixed chord of a circle and P any point on one of the

arcs cut offby it, then the bisector of the angle APB cuts the conjugate arc
in the same point for all positions of P.

10. .AB, AC are any two chords of a circle ; and P, Q are the middle
points of the minor ares cut off by them; if PQ is joined, cutting AB
in X and AC in Y, shew that AX=AY.

ll. A triangle ABC is inscribed in a circle, and the bisectors of the

angles meet the circumference at X, Y, Z. Show that the angles of
the triangle XYZ are respectively
A n 5,
OB 90°-5.
90°- ced

12. Two circles intersect at A and B; and through these points

lines are drawn from any point P onthe circumference of one of the
circles: shew that when produced they intercept on the other circum-
ference an are which is constant for all positions of P.
 EXERCISES ON ANGLES IN A CIRCLE. Vi

13. The straight lines which join the extremities of parallel chords
in a circle (i) towards the same parts, (ii) towards opposite parts, are
equal.

14. Through A, a point of intersection of two equal circles, two

straight lines PAQ, XAY are drawn: shew that the chord PX is equal
to the chord QY.
15. Through the points of intersection of two circles two parallel
straight lines are drawn terminated by the circumferences: shew that
the straight lines which join their extremities towards the same parts
are equal.

16. Two equal circles intersect at A and B; and through A any

straight line PAQ is drawn terminated by the circumferences: shew
that BP=BQ.

17. ABC is an isosceles triangle inscribed in a circle. and the

bisectors of the base angles meet the circumference at X and Y. Shew
that the figure BXAYC must have four of its sides equal.
What relation must subsist among the angles of the triangle ABC, in
order that the figure BXKAYC may be equilateral?

18. ABCD is a cyclic quadrilateral, and the opposite sides AB, DC

are produced to meet at P, and CB, DA to meet at Q: if the circles
circumscribed about the triangles PBC, QAB intersect at R, shew that
the points P, R, Q are collinear.

19. P, Q, R ave the middle points of the sides of a triangle, and X is

the foot of the perpendicular let fall from one vertex on the opposite side :
shew that the four points P, Q, R, X are concyclic.
[See page 64, Ex. 2: also Prob. 10, p. 83.]

20. Use the preceding exercise to shew that the middle points of the
sides of a triangle and the feet of the perpendiculars let fall from the
vertices on the opposite sides, are concyclic.

21. If a series of triangles are drawn standing on a fixed base, and

having a given vertical augle, shew that the bisectors of the vertical
angles all pass through a-fixed point.

22. ABC is a triangle inscribed in a circle, and E the middle point

of the arc subtended by BC on the side remote from A: if through E
a diameter ED is drawn, shew that the angle DEA is half the difference
of the angles at B and C.
Le GEOMETRY.

TANGENCY.

DEFINITIONS AND First PRINCIPLES.

1. A secant of a circle is a straight line of indefinite

length which cuts the circumference at two points,
2. If a secant moves in such a way that the two points
in which it cuts the circle continually approach one another,
then in the ultimate position when these two points become
one, the secant becomes a tangent to the circle, and is said to
touch it at the point at which the two intersections coincide.
This point is called the point of contact.
For instance :
(i) Let a secant cut the circle at the points oe
P and Q, and suppose it to recede from the P.
centre, moving always parallel to its original Q
position; then the two points P and Q will
clearly approach one another and finally coin- Q
cide.
In the ultimate position when P and Q Q
become one point, the straight line becomes a
tangent to the circle at that point.

(ii) Let a secant cut the circle at the points

P and Q, and suppose it to be turned about
the point P so that while P remains fixed, Q
moves on the circumference nearer and nearer
to P. Then the line PQ in its ultimate
position, when Q coincides with P, is a
tangent at the point P.

Since a secant can cut a circle at two points only, it is clear

that a tangent can have only one point in common with the
circumference, namely the point of contact, at which two
oe of section coincide. Hence we may define a tangent as
follows:
3. A tangent to a circle is a straight line which meets
the circumference at one point only; and though produced
indefinitely does not cut the circumference.
 TANGENCY. 173

Fig. Bign2s =) Fig. 3.

4. Let two circles intersect (as in Fig. 1) in the points

P and Q, and let one of the circles turn about the point P,
which remains fixed, in such a way that Q continually
approaches P. Then in the ultimate position, when Q coin-
cides with P (as in Figs. 2 and 3), the circles are said to
touch one another at P.
Since tio circles cannot intersect in more than two points,
two cucles which touch one another cannot have more than
one point in common, namely the point of contact at which the
two points of section coincide. Hence circles are said to touch
one another when they meet, but do not cut one another.
Notre. When each of the circles which meet is outside the other, as
in Fig. 2, they are said to touch one another externally, or to have
external contact: when one of the circles is within the other, as in
Fig. 3, the first is said to touch the other internally, or to have
internal contact with it.

INFERENCE FROM DEFINITIONS 2 AND 4.

If in Fig. 1, TQP is a common chord of two circles one
of which is made to turn about P, then when Q is brought
into coincidence with P, the line TP passes through two coinci-
dent points on each circle, as in Figs. 2 and 3, and therefore
becomes a tangent to each circle. Hence
Two circles which touch one another have a common tangent at
their point of contact.
174 GEOMETRY.

THEOREM 46.
The tangent at any point of a circle is perpendicular to the
radius drawn to the point of contact.

P a FT

Let PT be a tangent at the point P to a circle whose centre

is O.
It is vequired to prove that PT is perpendicular to the radius-OP.

Proof. Take any point Q in PT, and join OQ.

Then since PT is a tangent, every point in it except P is
outside the circle.
.. OQ is greater than the radius OP.
And this is true for every point Q in PT;
.. OP is the shortest distance from O to PT.
Hence OP is perp. to PT. Theor. 12, Cor. 1.
Q.E.D.

Corouiary 1. Since there can be only one perpendicular

to OP at the point P, it follows that one and only one tangent
can be drawn to a circle at a given point on the circumference.
CoROLLarRy 2. Since there can be only one perpendicular
to PT at the point P, it follows that the perpendicular to a
tangent at its point of contact passes through the centre.
COROLLARY 3. Since there can be only one perpendicular
from O to the line PT, it follows that the radius drawn perpen-
dicular to the tangent passes through the point of contact.
 TANGENCY. PTD

THEOREM 46. [By the Method of Limits.]

The tangent at any point of a circle is perpendicular to the
radius drawn to the point of contact.

(Q)

Fig.1. ; Fig. 2.

Let P be a point on a circle whose centre is O.

It is required to prove that the tangent at P is perpendicular to
the radius OP.
Let RQPT (Fig: 1) be a secant cutting the circle at Q and P.
Join OQ, OP.

Proof. Because OP= OQ,

., the 2OQP=the Zz OPQ;
.. the supplements of these angles are equal;
that is, the 2 OQR= the 2 OPT,
and this is true however near Q is to P.
Now let the secant QP be turned about the point P so that
Q continually approaches and finally coincides with P; then
in the ultimate position,
(i) the secant RT becomes the tangent at ‘aFie. 2
(ji) OQ coincides with OP; one
and therefore the equal L*OQR, OPT become adjacent,
.-. OP is perp. to RT. Q.E.D.
Nort. The method of proof employed here is known as the Method
ef Limits.
176 GEOMETRY.
~

THEOREM 47.
Two tungents can be drawn to a circle from an external point.

Let PQR be a circle whose centre is O, and let T be an

external point.
It is required to prove that there can be two tangents drawn to
the circle from T.
Join OT, and let TSO be the circle on OT as diameter.
This circle will cut the ©PQR in two points, since T is
without, and O is within, the ©PQR. Let P and Q be these
points.
Join TP, TQ; OP, OQ.

Proof. Now each of the z*TPO, TQO, being in a semi-

circle, is a rt. angle;
. TP, TQ are perp. to the radii OP, OQ respectively.
.. TP, TQ are tangents at Pand Q. Theor. 46.
Q.E.D.

CoroLLaRY. The two tangents to a circle from an external

point are equal, and subtend equal angles at the centre.
For in the A* TPO, TQO,
the 2* TPO, TQO are right angles,
because {the hypotenuse TO is common,
and OP=0Q, being radii ;
“. TP=TQ,
and the 2 TOP=the 2 TOQ. Theor. 18.
 TANGENCY. Wz

EXERCISES ON THE TANGENT.

(Numerical and Graphical.)

1. Draw two concentric circles with radii 5°0 cm. and 3:0 cm.
Draw a series of chords of the former to touch the latter. Calculate
and measure their lengths, and account for their being equal.
2. In a circle of radius 1-0” draw a number of chords each 1°6” in
length. Shew that they all touch a concentric circle, and find its
radius.
3. The chanieters of two concentric circles are respectively 10:0 cm.
and 5°0 cm.: find to the nearest millimetre the length of any chord of
the outer circle which touches the inner, and check your work by
measurement.
4. In the figure of Theorem 47, if OP=5’, TO=13", find the length
of the tangents from T. Draw the figure (scale 2 cm. to 5”), and
measure to the nearest degree the angles subtended at O by the
tangents.
5. The tangents from T to a circle whose radius is 0°7” are each 2-4”
in length. Find the distance of T from the centre of the circle. Draw
the figure and check your result graphically.

(Theoretical.)
6. The centre of any circle which touches two intersecting straight
lines must lie on the bisector of the angle Letween them.
7. AB and AC are two tangents to a circle whose centre isO; shew
that AO bisects the chord of contact BC at right angles.
8. If PQ is joined in the figure of Theorem 47, shew that the angle
PTQ is double the angle OPQ.
9. Two parallel tangents to a circle intercept on any third tangent
a segment which subtends a right angle at the centre.
10. The diameter of a circle bisects all chords which are parental to
the tangent at either extremity.
1l. Find the locus of the centres of all circles which touch a given
straryht line at a given point.
12. Find the locus of the centres of all circles which touch each of
two parallel straight lines.
13. Find the locus of the centres of all circles which touch each of two
intersecting straight lines of unlimited length.
14. In any quadrilateral circumscribed about a circle, the sum of one
pair of opposite sides 13 equal to the sum of the other pair.
State and prove the converse. theorem.
15. If a quadrilateral is described about a circle, the angles sub-
tended at the centre by any two opposite sides are supplementary.
H.S.G. M
178 GEOMETRY.

THEOREM 48.
If two circles touch one another, the centres and the point of
contact are in one straight line.

Af 0

. :

Let two circles whose centres are O and Q touch at the

point P.
It is required to prove that O, P, and Q are in one straight line.
Join OP, QP.

Proof. Since the given circles touch at P, they have a

common tangent at that point. Page 173.
Suppose PT to touch both circles at P.
Then since OP and QP are radii drawn to the point of
contact,
‘*, OP and QP are both perp. to PT ;
.. OP and QP are in one st. line. Theor, 2.
That is, the points O, P, and Q are in one st. line. Q.E.D.

CoroLuaRigs. (i) Jf two circles touch externally the distance

between their centres is equal to the swm of their radii.
(ii) If two circles touch internally the distance between their
centres is equal to the difference of their radii.
 THE CONTACT OF CIRCLES. 179

EXERCISES ON THE CONTACT OF CIRCLES.

(Numerical and Graphical.)

1. From centres 2°6” apart draw two circles with radii 1°7’ and 0°9”
respectively. Why and where do these circles touch one another?
If circles of the above radii are drawn front centres 08” apart, prove
that they touch. How and why does the contact differ from that in
the former case?
2. Draw a triangle ABC in which a=8 em., b=7 em., and c=6 cm.
From A, B, and C as centres draw circles of radii 2°5 em., 3°5 cm., and
4°5 cm. respectively ; and shew that these circles touch in pairs.
3. In the triangle ABC, right-angled at C, a=8 cm. and b=6 cm.;
and from centre A with radius 7 em. a circle is drawn. What must be
the radius of a circle drawn from centre B to touch the first circle?

4. A and B are the centres of two fixed circles which touch in-
ternally. If P is the centre of any circle which touches the larger circle
internally and the smaller externally, prove that AP + BP is constant.
If the fixed circles have radii 5:0 cm. and 3:0 cm. respectively, verify
the general result by taking different positions for P.
5. AB is a line 4” in length, and C is its middle point. On AB,
AC, CB semicircles are described. Shew that if a circle is inscribed in
the space enclosed by the three semicircles its radius must be 3”.

(Theoretical.)
6. A straight line is drawn through the point of contact of two circles
whose centres are A and B, cutting the circumferences at P and Q re-
spectively ; shew that the radu AP and BQ are parallel.
7. Two circles touch externally, and through the point of contact a
straight line is drawn terminated by the circumferences ; shew that the
tangents at its extremities are parallel.
8. Find the locus of the centres of all circles
(i) which touch a given circle at a given point ;
(ii) which are of given radius and touch a given circle.
9. From a given point as centre describe a circle to touch a given
circle. How many solutions will there be?
10. Describe a circle of radius a to touch a given circle of radius b
at a given point. How many solutions will there be?
180 GEOMETRY.

THEOREM 49. [Euclid III. 32.]

The angles made by a tangent to a circle with a chord drawn
from the point of contact are respectively equal to the angles in
the alternate seyments of the circle.
A

E B tr
Let EF touch the ©ABC at B, and let BD be a chord drawn
from B, the point of contact.
It is required to prove that
(i) the LFBD = the angle in the alternate segment BAD ;
(ii) the LEBD =the angle in the alternate segment BCD.
Let BA be the diameter through B, and C any point in the
are of the segment which does not contain A.
Join AD, DC, CB.
Proof. Because the 2 ADB in asemi-circle is a rt. angle,
‘. the 2*DBA, BAD together =a rt. angle.
But since EBF is a tangent, and BA a diameter,
‘, the 2 FBA is a rt. angle.
. the 2 FBA=the Z*DBA, BAD together.
Take away the common Z DBA,
then the 2 FBD = the 2 BAD, which is in the alternate segment.
Again because ABCD is a cyclic quadrilateral,
*, the 2BCD=the supplement of the 2 BAD
=the supplement of the 2 FBD
=the .EBD;
‘, the 2EBD=the 2 BCD, which is in the alternate segment.
Q.E.D.
 ALTERNATE SEGMENT. 181

EXERCISES ON THEOREM 49.

1. In the figure of Theorem 49, if the ©FBD=72°, write down the
values of the 47 BAD, BCD, EBD.
2. Use this theorem to shew that tangents to a circle from an
external point are equal.
3. Through A, the point of contact of two circles, chords APQ,
AXY are drawn: shew that PX and QY are parallel.
Prove this (i) for internal, (11) for external contact.
4. AB is the common chord of two circles, one of which passes
through O, the centre of the other: prove that OA bisects the angle
between the common chord and the tangent to the first circle at A.
5. Two circles intersect at A and B; and through P, any point on
one of them, straight lines PAC, PBD are drawn to cut the other at C
and D: shew that CD is parallel to the tangent at P.
6. If from the point of contact of a tangent to a circle a chord is
drawn, the perpendiculars dropped on the tangent and chord from the
middle point of either are cut off by the chord are equal.

EXERCISES ON THE METHOD OF LIMITS.

1. Prove Theorem 49 by the Method of Limits.

[Let ACB be a segment of a circle of CG
which AB is the chord ; and let PAT’ be any
secant through A. Join PB. S
Then the 2 BCA=the L BPA;
Theor. 39. B
and this is true however near P approaches
to A.
If P moves up to coincidence with A,
then the secant PAT’ becomes the tangent
AT, and the 2 BPA becomes the Z BAT. JS
“., ultimately, the 2 BAT =the ZBCA, in liso T
the alt. segment. ] “d
2. From Theorem 31, prove by the ao th
Method of Limits that
The straight line drawn perpendicular to the diameter of a circle at its
extremity 1s a tangent.
3. Deduce Theorem 48 from the property that the line of centres
bisects a common chord at right angles.
4. Deduce Theorem 49 from Ex. 5, page 163.
5. Deduce Theorem 46 from Theorem 41.
182. GEOMETRY.

PROBLEMS.

GEOMETRICAL ANALYSIS.

-Hitherto the Propositions of this text-book have been

arranged Synthetically, that is to say, by building up known
results in order to obtain a new result.
But this arrangement, though convincing as an argument,
in most cases affords little clue as to the way in which the
construction or proof was discovered. We therefore draw the
student's attention to the following hints.
In attempting to solve a problem begin by asswming the
required result; then by working backwards, trace the conse-
quences of the assumption, and try to ascertain its dependence
on some condition or known theorem which suggests the
necessary construction. If this attempt is successful, the
steps of the argument may in general be re-arranged in
reverse order, and the construction and proof presented in a
synthetic form.
This unravelling of the conditions of a proposition in order
to trace it back to some earlier principle on which it depends,
is called geometrical analysis: it is the natural way of attack-
ing the harder types of exercises, and it is especially useful in
solving problems.
Although the above directions do not amount to a method,
they often furnish a very effective mode of searching for a
suggestion. The approach by analysis will be illustrated in
some of the following problems. [See Problems 23, 28, 29.]
 PROBLEMS ON CIRCLES. 183

PROBLEM 20.
Given a circle, or an are of a circle, to find its centre.
Let ABC be an are of a circle
whose centre is to be found.

Construction, Take two chords

AB, BC, and bisect them at right
angles by the lines DE, FG, meeting
at O. Prob, 2.
Then O is the required centre.
Proof. Every point in DE is equi-
distant from A and B. Prob. 14. ~
And every point in FG is equidistant from B and C.
*, O is equidistant from A, B, and C.
.. O is the centre of the circle ABC. Theor. 33.

PROBLEM 21.

To bisect a given are.

Let ADB be the given arc to be bisected. 2
Construction. Join AB, and bisect it at
right angles by CD meeting the arc at D.
Prob. 2.
Then the arc is bisected at D. A Cc B
Proof. Join DA, DB. x
Then every point on CD is equidistant from A and B;
Prob. 14.
*, DA=DB;
.. the .DBA=the 2 DAB; Theorem 6.
.. the arcs, which subtend these angles at the O*, are equal;
that is, the are DA=the arc DB.
4
 184 GEOMETRY,

PROBLEM 22.
To draw a tangent to a circle from a given external point.

----

Let PQR be the given circle, with its centre at O; and let T
be the point from which a tangent is to be drawn.

Construction. Join TO, and on it describe a semi-circle TPO

to cut the circle at P.
Join TP.
Then TP is the required tangent.

Proof. Join OP.

Then since the 2 TPO, being in a semi-circle, is a rt, angle,
.. TP is at right angles to the radius OP.
*, TP is a tangent at P. Theor. 46.
Since the semi-circle may be described on either side of TO,
a second tangent TQ can be drawn from T, as shewn in the
figure.
Notr._ Suppose the point T to approach the given circle, then the
angle PTQ gradually increases. When T reaches the circumference,
the angle PTQ becomes a straight angle, and the two tangents coincide.
When T enters the circle, no tangent can be drawn. [See Obs. p. 94.]
 COMMON TANGENTS. 185

PROBLEM 23.
To draw a common tangent to two circles.

Let A be the centre of the greater circle, and a its radius;

and let B be the centre of the smaller circle, and 6 its radius.
Analysis. Suppose DE to touch the circles at D and E.
Then the radii AD, BE are both perp. to DE, and therefore
par’ to one another.
Now if BC were drawn par’ to DE, then the fig. DB would
be a rectangle, so that CD=BE=0.
And if AD, BE are on the same side of AB,
then AC=a-—), and the Z ACB is a rt. angle. -
These hints enable us to draw BC first, and thus lead to the
following construction.
Construction. With centre A, and radius equal to the
difference of the radu of the given circles, describe a circle,
and draw BC to touch it.
Join AC, and produce it to meet the circle (A) at D.
Through B draw the radius BE par’ to AD and in the same
sense. Join DE.
Then DE is a common tangent to the given circles.

Obs. Since two tangents, such as BC, can in general be

drawn from B to the circle of construction, this method will
furnish two common tangents to the given circles. These are
called the direct common tangents.
186 GEOMETRY.

PROBLEM 23. (Continued.)

C
Li
LN mA

Again, if the circles are external to one another two more

common tangents may be drawn.

Analysis. In this case we may suppose DE to touch the

circles at D and E so that the radii AD, BE fall on opposite sides
of AB.
Then BC, drawn par' to the supposed common tangent DE,
would meet AD produced at C ; and we should now have
AC=AD+DC=a+0; and, as before, the 2 ACB is a rt. angle.
Hence the following construction.

Construction. With centre A, and radius equal to the swm

of the radii of the given circles, describe a circle, and draw
BC to touch it.
Then proceed as in the first case, but draw BE in the sense
opposite to AD.

Obs. As before, two tangents may be drawn from B to the

circle of construction ; hence two common tangents may be
thus drawn to the given circles. These are called the
transverse common tangents.

[We leave as an exercise to the student the arrangement of the proof

in synthetic form.]
 COMMON TANGENTS. 187

EXERCISES ON COMMON TANGENTS.

(Numerical and Graphical.)

1. How many common tangents can be drawn in each of the
following cases ?
(i) when the given circles intersect;
(ii) when they have external contact ;
(iii) when they have internal contact.
Tilustrate your answer by drawing two circles of radii 1:4” and 1:0”
respectively,
(i) with 1:0” between the centres ;
(ii) with 2:4” between the centres ;
(iii) with 0°4” between the centres ;
(iv) with 3:0” between the centres.
Draw the common tangents in each case, and note where the general
construction fails, or is modified.
2. Draw two circles with radii 2:0’ and 0°8”, placing their centres
2-0" apart. Draw the common tangents, and find their lengths between
the points of contact, both by calculation and by measurement.
3. Draw all the common tangents to two circles whose centres are
1‘8” apart and whose radii are 0°6’ and 1:2” respectively. Calculate and
measure the length of the direct common tangents.
4. Two circles of radii 1°7” and 1:0” have their centres 21” apart.
Draw their common tangents and find their lengths. Also find the
length of the common chord. Produce the common chord and shew
by measurement that it bisects the common tangents.
5. Draw two circles with radii 1°6” and 0°8” and with their centres
3:0’ apart. Draw all their common tangents.
6. Draw the direct common tangents to two equal circles.

(Theoretical.)
7. If the two direct, or the two transverse, common tangents are
drawn to two circles, the parts of the tangents intercepted between the
points of contact are equal.
8., If four common tangents are drawn to two circles external to
one another, shew that the two direct, and also the two transverse,
tangents intersect on the line of centres.
9, Two given circles have external contact at A, and a direct common
tangent is drawn to touch them at P and Q: shew that PQ subtends a
right angle at the point A.
188 GEOMETRY.

ON THE CONSTRUCTION OF CIRCLES.

In order to draw a circle we must know (i) the position of
the centre, (ii) the length of the radius.
(i) To find the position of the centre, two conditions are
needed, each giving a locus on which the centre must lie; so
that the one or more points in which the two loci intersect
are possible positions of the required centre, as explained on
page 93.
(ii) The position of the centre being thus fixed, the radius
is determined if we know (or can find) any point on the
circumference.
Hence in order to draw a circle three independent data are
required,
Tor example, we may draw a circle if we are given
(i) three points on the circumference ;
or (ii) three tangent lines; "
or (iii) one point on the circumference, one tangent, and its point
of contact.
It will however often happen that more than one circle can be drawn
satisfying three given conditions.

Before attempting the constructions of the next Exercise the

student should make himself familiar with the following loci.
(i) The locus of the centres of circles which pass through two
given points.
(ii) The locus of the centres of circles which touch a given straight
line at a given point.
(iii) The locus of the centres of circles which touch a given circle
at a given point.
(iv) The locus of the centres of circles which touch a given straight
line, and have a given radius.
(v) The locus of the centres of circles which touch a given circle,
and have a given radius.
(vi) The locus of the centres of circles which touch two given
straight lines,
 THE CONSTRUCTION OF CIRCLES. 189

EXERCISES.

1. Draw a circle to pass through three given points.

2. Ifa circle touches a given line PQ at a point A, on what line
must its centre lie?
If a circle passes through two given points A and B, on what line
must its centre lie?
Hence draw a circle to touch a straight line PQ at the point A, and
to pass through another given point B.
3. Ifa circle touches a given circle whose centre is C at the point A,
on what line must its centre lie?
Draw a circle to touch the given circle (C) at the point A, and to pass
through a given point B.
4. A point P is 4°5 cm. distant from a straight line AB.. Draw two
circles of radius 32 cm. to pass through P and to touch AB.
~
5. Given two circles of radius 3‘0 cm. and 2:0 cm. respectively,
their centres being 60 cm. apart ; draw a circle of radius 3°5 cm. to
touch each of the given circles externally.
How many solutions will there be? What is the radius of the
smallest circle that touches each of the given circles externally?
6., If a circle touches two straight lines OA, OB, on what line must
its centre he?
Draw OA, OB making an angle of 76°, and describe a circle of radius
1:2” to touch both lines.

7. Given a circle of radius 3°5 cm., with its centre 5:0 cm. from a
given straight line AB; draw two circles of radius 2°5 cm. to touch the
given circle and the line AB.
8. Devise a construction for drawing a circle to touch each of two
parallel straight lines and a transversal.
Shew that two such circles can be drawn, and that they are equal.
9. Describe a circle to touch a given circle, and also to touch a
given straight line at a given point. _[Sce page 311.]
10. Describe a circle to touch a given straight line, and to touch a
given circle at a given point.
11. Shew how to draw a circle to touch each of three given straight
lines of which no two are parallel.
How many such circles can be drawn?

[Further Examples on the Construction of Circles will be found on

pp. 246, 311.]
190 GEOMETRY.

PROBLEM 24,
On a given straight line to describe a segment of a circle which
shall contain an angle equal to a given angle.

Let AB be the given st. line, and C the given angle.

It is required to describe on AB a segment of a circle containing
an angle equal to C.

Construction. At A in BA, make the 2 BAD equal to the LC.

From A draw AG perp. to AD.
Bisect AB at rt. angles by FG, meeting AGin G. Prob. 2.

Proof. Join GB.

Now every point in FG is equidistant from A and B;
Prob, 14.
so GA aD,
With centre G, and radius GA, draw a circle, which must
pass through B, and touch AD-at A. Theor. 46.
Then the segment AHB, alternate to the 2 BAD, contains an
angle equal to C. : Theor. 49.

Notr. In the particular case when the given angle isa rt. angle, the
segment required will be the semi-circle on AB asdiameter. [Theorem41.]
 PROBLEMS. 191

CoroLuaRy. To cut off from a given circle a segment containing

a given angle, it is enough to draw a tangent to the circle, and from
the point of contact to draw a chord making with the tangent an
angle equal to the given angle.

It was proved on page 161 that

The locus of the vertices of triangles which stand on the same base
and have a given vertical angle, is the arc of the segment standing
on this base, and containing an angle equal to the given angle.

The following Problems are derived from this result by the

Method of Intersection of Loci [page 93].

EXERCISES.

1. Describe a triangle on a given base having a given vertical angle

and having its vertex on a given straight line.
2 Construct a triangle having given the base, the vertical angle, and
(1) one other side.
(11) the altitude.
(iii) the length of the median which bisects the base.
(iv) the foot of the perpendicular from the vertex to the base.
3. Construct a triangle having given the base, the vertical angle, and
the point at which the base is cut by the bisector of the vertical angle.
[Let AB be the base, X the given point in it, and K the given angle.
On AB describe a segment of a circle containing an angle equal to K;
complete the Oct by drawing the arc APB. Bisect the arc APB at P:
join PX, and produce it to meet the Ot at C. Then ABC is the
required triangle.]
4. Construct a triangle having given the base, the vertical angle, and
the sum of the remaining sides.
[Let AB be the given base, K the given angle, and H a line equal to
the sum of the sides. On AB describe a segment containing an angle
equal to K, also another segment containing an angle equal to half the
LK. With centre A, and radius H, describe a circle cutting the arc of
the latter segment at X and Y. Join AX (or AY) cutting the arc of the
first segment at C. Then ABC is the required triangle.]
5. Construct a triangle having given the base, the vertical angle, and
the difference of the remaining sides.
192 GEOMETRY.

CIRCLES IN RELATION TO RECTILINEAL FIGURES.

DEFINITIONS.

1. <A Polygon is a rectilineal figure bounded by more than

four sides.
A Polygon of five sides is called a Pentagon,
if sia sides ” Hexagon,
ms seven sides cs Heptagon,
ad ~ eight sides ” Octagon,
, ten sides ™ Decagon,
ie twelve sides ¥ Dodecagon,
. jifteen sides ir Quindecagon.

2. A Polygon is Regular when all its sides are equal, and

all its angles are equal.

3. <A rectilineal figure is said to be in-

scribed in a circle, when all its angular points
are on the circumference of the circle; and a
circle is said to be circumscribed about a recti-
lineal figure, when the circumference of the
circle passes through all the angular points of
the figure.

4. <A circle is said to be inscribed in a

rectilineal figure, when the circumference of
the circle is touched by each side of the figure;
and a rectilineal figure is said to be cireum-
scribed about a circle, when each side of the
figure is a tangent to the circle, =
 PROBLEMS ON TRIANGLES AND CIRCLES. 193

PROBLEM 25.
To circumscribe a circle about a given triangle.

A aie A

NY

BS Ce

Let ABC be the triangle, about which a circle is to be

drawn.

Construction. Bisect AB and AC at rt. angles by DS and

ES, meeting at S. Prob. 2.
Then § is the centre of the required circle.

Proof. Now every point in DS is equidistant from A

and B; Prob. 14.
and every point in ES is equidistant from A and C;
. S is equidistant from A, B, and C.
With centre S, and radius SA describe a circle; this will
pass through B and C, and is, therefore, the required circum-
circle.

Obs. It will be found that if the given triangle is acute-

angled, the centre of the circum-circle falls within it: if it
is a right-angled triangle, the centre falls on the hypotenuse :
if it is an obtuse-angled triangle, the centre falls without the
triangle.

Notre. From page 94 it is seen that if S is joined to the middle

point of BC, then the joining line is perpendicular to BC.
Hence the perpendiculars drawn to the sides of a triangle from their
middle points are concurrent, the point of intersection being the centre
of the circle circumscribed about the triangle.
H.S.G. N
194 GEOMETRY.

PROBLEM 26.
To inscribe a circle in a given triangle.

Let ABC be the triangle, in which a circle is to be inscribed.

Construction, Bisect the 2*ABC, ACB by the st. lines BI,
Cl, which intersect at |. Prob. 1.
Then | is the centre of the required circle.
Proof. From | draw ID, IE, IF perp. to BC, CA, AB.
Then every point in BI is equidistant from BC, BA; Prob. 15.
“. ID=IF.
And every point in Cl is equidistant from CB, CA;
*, ID=IE,
‘, ID, IE, IF are all equal.
With centre | and radius ID draw a circle;
this will pass through the points E and F.
Also the circle will touch the sides BC, CA, AB,
because the angles at D, E, F are right angles.
’, the © DEF is inseribed in the A ABC.
Note. From IL., p. 96 it is seen that if Al is joined, then Al bisects
the angle BAC : hence it follows that
The bisectors of the angles of a triangle are concurrent, the point of
intersection being the centre of the inscribed circle.

DEFINITION.
A cirele which touches one side of a triangle and the other
two sides produced is called an escribed circle of the triangle.
 PROBLEMS ON TRIANGLES AND CIRCLES. 195

PROBLEM 27.
To draw an escribed circle of a given triangle.

A ) aa D
Let ABC be the given triangle of which the sides AB, AC are
produced to D and E.
It is required to describe a circle touching BC, and AB, AC
produced,
Construction. Bisect the 2°CBD, BCE by the st. lines BI,,
Cl, which intersect at |,.
Then |, is the centre of the required circle.
Proof. From |, draw |,F, |,G, |,H perp. to AD, BC, AE.
Then every point in BI, is equidistant from BD, BC; Prob. 15.
‘ LF=LG
Similarly 1,G=1,H.
*. IF, 1,G, 1,H are all equal.
With centre |, and radius |,F describe a circle;
this will pass through the points G and H.
Also the circle will touch AD, BC, and AE,
because the angles at F, G, H are rt. angles.
*. the ©FGH is an escribed circle of the A ABC.
Nore 1. It is clear that every triangle has three escribed circles.
Their centres are known as the Ex-centres.
Note 2. It may be shewn, as in II., page 96, that if Al, is joined,
then Al, bisects the angle BAC: hence it follows that
The bisectors of two exterior angles ofa triangle and the bisector of the
third angle are concurrent, the point of intersection being the centre of an
escribed circle.
196 GEOMETRY.

PROBLEM 28.
In a given circle to inscribe a triangle equiangular to a given
triangle.
G

E F B Cc

Let ABC be the given circle, and DEF the given triangle.

Analysis. A AABC, equiangular to the A DEF, is inscribed

in the circle, if from any point A on the O% two chords AB, AC
can be so placed that, on joining BC, the ~B=the ZE, and
the C= the .F; for then the .A=the 2D. Theor. 16.
Now the 2B, in the segment ABC, suggests the equal angle
between the chord AC and the tangent at its extremity
(Theor. 49.) ; so that, if at A we draw the tangent GAH,
then the 2 HAC=the 2E;
and similarly, the 2 GAB= the ZF.
Reversing these steps, we have the following construction.

Construction. At any point A on the O” of the © ABC

draw the tangent GAH. Prob. 22.
At A make the 2 GAB equal to the ZF,
and make the L HAC equal to the 2 E.
Join BC.
Then ABC is the required triangle.

Note. In drawing the figure on a larger scale the student should

shew the construction lines for the tangent GAH and for the angles
GAB, HAC. A similar remark applies to the next Problem.
 PROBLEMS ON CIRCLES AND TRIANGLES. 197

PROBLEM 29.
About a given circle to cirewmscribe a triangle equiangular to
a given triangle.
L

M B N

Let ABC be the given circle, and DEF the given triangle.
Analysis. Suppose LMN to be a circumscribed triangle in
which the 2_M=the ZE, the ~N=the ZF, and consequently,
the =the 2D.
Let us consider the radii KA, KB, KC, drawn to the points of
contact of the sides ; for the tangents LM, MN, NL could be
drawn if we knew the relative positions of KA, KB, KC, that is,
if we knew the 2* BKA, BKC.
Now from the quad’ BKAM, since the 2° B and A are rt. 2’,
the £ BKA= 180° -M=180°-E;
similarly the 2BKC=180°-—N=180°-F.
Hence we have the following construction.
Construction. Produce EF both ways to G and H.
Find K the centre of the © ABC,
and draw any radius KB.
At K make the 2 BKA equal to the 2 DEG;
and make the 2 BKC equal to the 2 DFH.
Through A, B, C draw LM, MN, NL perp. to KA, KB, KC.
Then LMN is the required triangle.

[The student should now arrange the proof synthetically. }

198 GEOMETRY.

EXERCISES.

On CircLes AND TRIANGLES.

(Inscriptions and Circumscriptions.)
1. Ina circle of radius 5 em. inscribe an equilateral triangle ; and
about the same circle circumscribe a second equilateral triangle. In
each case state and justify your construction.
2. Draw an equilateral triangle on a side of 8 em., and find by
calculation and measurement (to the nearest millimetre) the radii of
the inseribed, circumscribed, and escribed circles.
Explain why the second and third radii are respectively double and
treble of the first.
3. Draw triangles from the following data:
(i) @=2'0", B=665 C=60"4
(ii) 2=2°5", B=72,, C=4e°s
Gii).a=2°3", B=4r, O=23.
Cireumscribe a circle about each triangle, and measure the radii
to the nearest hundredth of an inch. Account for the three results
being the same, by comparing the vertical angles.
4. In a circle of radius 4 em. inscribe an equilateral triangle.
Calculate the length of its side to the nearest millimetre ; and verify
by measurement.
Find the area of the inscribed equilateral triangle, and shew that it
is one quarter of the circumscribed equilateral triangle.
5. In the triangle ABC, if | is the centre, and r the length of the
radius of the in-circle, shew that
AIBC=}har; AICA=hbr; AIAB=ther.
Hence prove that ASABC=3(a+b+c)r.
Verify this formula by measurements for a triangle whose sides are
9 cm., 8 cm., and 7 em.
§. Ifr, is the radius of the ex-circle opposite to A, prove that
AABC=4(b+c-a)r,.
If a=5 em., b=4 em., c=3 em., verify this result by measurement.
7. Find by measurement the cireum-radius of the triangle ABC in
which a=6'3 em., b=3‘0 em., and c=5'1 em.
Draw and measure the perpendiculars from A, B, C to the opposite
sides. If their lengths are represented by p,, p., ps, verify the following
statement:
circum-radius = be ate ah
2p, 2p, 2p,
 PROBLEMS ON CIRCLES AND SQUARES. 199

EXERCISES.

On CrrcLes AND SQUARES.

(Inscriptions and Circumscriptions.)

1. Draw a circle of radius 1°5”, and find a construction for inscribing
a square in it.
Calculate the length of the side to the nearest hundredth of an inch,
and verify by measurement.
Find the area of the inscribed square.
2. Circumscribe a square about a circle of radius 1:5’, shewing all
lines of construction.
Prove that the area of the square circumscribed about a circle is
double that of the inscribed square.
3. Draw a square on a side of 7°5 cm., and state a construction for
inscribing a circle in it.
Justify your construction by considerations of symmetry.
4. Circumscribe a circle about a square whose side is 6 cm.
Measure the diameter to the nearest millimetre, and test your
drawing by calculation.
5. In a circle of radius 1°8” inscribe a rectangle of which one side
measures 3 0”. Find the approximate length of the other side.
Of all rectangles inscribed in the circle shew that the square has the
greatest area.
6. A square and an equilateral triangle are inscribed in a circle.
If a and 6 denote the lengths of their sides, shew that
30° =26".
7. ABCD is a square inscribed in a circle, and P is any point on the
arc AD: shew that the side AD subtends at P an angle three times as
great as that subtended at P by any one of the other sides.

(Problems. State your construction, and give a theoretical proof.)

8. Circumscribe a rhombus about a given circle.
9. Inscribe a square in a given square ABCD, so that one of its
angular points shall be at a given point X in AB.
10. Ina given square inscribe the square of minimum area.
11. Describe (i) a circle, (ii) a square about a given rectangle.
12. Inscribe (i) a circle, (ii) a square in a given quadrant.
200 GEOMETRY.

ON CIRCLES AND REGULAR POLYGONS.

PROBLEM 30.
To draw a regular polygon (i) in (ii) about a given circle.
Let AB, BC, CD, ... be consecutive
sides of a regular polygon inscribed in
a circle whose centre is O.
Then AOB, BOC, COD, ... are con-
gruent isosceles triangles. And _ if
the polygon has m sides, each of the
Z*AOB, BOC, COD, ...= at
(i) Thus to inseribe a polygon of n sides in a given circle,
draw an angle AOB at the centre equal to ie This gives
the length of a side AB; and chords equal to AB may now be
set off round the circumference. The resulting figure will
clearly be equilateral and equiangular.
(ii) To cireumscribe a polygon of n sides about the circle,
the points A, B, C, D, ... must be determined as before, and
tangents drawn to the circle at these points. The resulting
figure may readily be proved equilateral and equiangular.
Note. This method gives a strict geometrical construction only when
the angle aa can be drawn with ruler and compasses.

EXERCISES.
1. Give strict constructions for inscribing in a circle (radius 4 em.)
(i) a regular hexagon ; (ii) a regular octagon ; (iii) a regular dodecagon,
2. About a circle of radius 1:5” cireumscribe
(i) a regular hexagon; (ii) a regular octagon.
Test the constructions by measurement, and justify them by proof.
3. An equilateral triangle and a regular hexagon are inscribed in a
given circle, and a and b denote the lengths of their sides: prove that
(i) area of triangle=4(area of hexagon); (ii) a?=3b*.
4. By means of your protractor inscribe a regular heptagon in a
circle of radius 2”. Calculate and measure one of its angles; and
measure the length of a side.
 PROBLEMS ON CIRCLES AND POLYGONS, 201

PROBLEM 31.
To draw a cirele (1) in (ii) about a regular polygon.

Let AB, BC, CD, DE, ... be con-

secutive sides of a regular polygon
of n sides.
Bisect the 2*ABC, BCD by BO, A
CO meeting at O.
Then O is the centre both of the
inscribed and circumscribed circle. Ve
B = Omea(o

Outline of Proof. Join OD; and from the congruent

4*OCB, OCD, shew that OD bisects the LCDE. Hence we
conclude that
All the bisectors of the angles of the polygon meet at O.
(i) Prove that OB=OC=OD=...; from Theorem 6.
Hence O is the circum-centre.
(i) Draw OP, OQ, OR, ... perp. to AB, BC, CD,....
Prove that OP=OQ=OR=...; from the congruent A*OBP,
OBQ, .... Hence O is the in-centre.

EXERCISES.
1. Draw a regular hexagon on a side of 2:0’. Draw the inscribed
and circumscribed circles. Calculate and measure their diameters to
the nearest hundredth of an inch.

2. Shew that the area of a regular hexagon inscribed in a circle is

three-fourths of that of the circumscribed hexagon.
Find the area of a hexagon inscribed in a circle of radius 10 cm. to
the nearest tenth of a sq. cm.
3. If ABC is an isosceles triangle inscribed in a circle, having each
of the angles B and C double of the angle A; shew that BC is a side of
a regular pentagon inscribed in the circle.
4, Ona side of 4 cm. construct (without protractor)
(i) a regular hexagon; (ii) a regular octagon.
In each case find the approximate area of the figure.
202 GEOMETRY.

THE CIRCUMFERENCE OF A CIRCLE.

By experiment and measurement it is found that the length

of the circumference of a circle is roughly 3} times the length of
its diameter: that is to say
circumference
= 3} nearly;
diameter
and it can be proved that this is the same for all circles.
A more correct value of this ratio is found by theory to be
31416; while correct to 7 places of decimals it is 31415926.
Thus the value 3} (or 3°1428) is too great, and correct to 2
places only.
The ratio which the circumference of any circle bears to its
diameter is denoted by the Greek letter 7; so that
circumference = diameter x m.
Or, if 7 denotes the radius of the circle,
circumference = 2r x 7 = 2rr ;
where to 7 we are to give one of the values 3}, 3:1416, or
31415926, according to the degree of accuracy required in
the final result. .
Notr. The theoretical methods by which 7 is evaluated to any
required degree of accuracy cannot be explained at this stage, but its
value may be easily verified by experiment to two decimal places.
For example: round a eylinder wrap a strip of paper so that the
ends overlap. At any point in the overlapping area prick a pin
through both folds. Unwrap and straighten the strip, then measure
the distance between the pin holes: this gives the length of the cireum-
ference. Measure the diameter, and divide the first result by the second.

Ex. 1. From these data CIRCUMFERENCE, | DIAMETER. | VALUE OF mr,

find and record the value |————— . eas T°
of r. 16°0 em. 51 om.
Find the mean of the
three results.

Ex. 2. A fine thread is wound evenly round a cylinder, and it is

found that the length required for 20 complete turns is 75°4’. The
diameter of the cylinder is 1°2": find roughly the value of 7.
ix. 3. <A bieycle wheel, 28” in diameter, makes 400 revolutions
in travelling over 977 yards. From this result estimate the value of 1.
 CIRCUMFERENCE AND AREA OF A CIRCLE. 203

THE AREA OF A CIRCLE.

Let AB be a side of a polygon of

n sides circumscribed about a circle
whose centre is O and radius 7, Then
we have
Area of polygon
=n. AAOB
=n. 4AB x OD
=4.nABxr sfal INS
=4 (perimeter of polygon) xr ; ADB
and this is true however many sides the polygon may have.
Now if the number of sides is increased without limit, the
perimeter and area of the polygon may be made to differ from
the circumference and area of the circle by quantities smaller
than any that can be named ; hence ultimately
Area of circle =4. circumference x r
=4.257T xr
any
=T7T-.

ALTERNATIVE METHOD.

A B

D C

Suppose the circle divided into any even number of sectors having
equal central angles : denote the number of sectors by n.
Let the sectors be placed side by side as represented in the diagram ;
then the area of the circle=the area of the fig. ABCD ;
and this is true however great n may be.
Now as the number of sectors is increased, each arc is decreased;
s0 that (i) the outlines AB, CD tend to become straight, and
(ii) the angles at D and B tend to become rt. angles.
204 GEOMETRY.

Thus when n is increased without limit, the fig. ABCD ultimately

becomes a ectangle, whose length is the semi-circumference of the circle,
and whose br eagts is its radius.
. Area of circle=}. circumference x radius
=h. 2x =

THE AREA OF A SECTOR.

Cc

it two radii of a circle make an angle of 1°, they cut off

(i) an are whose length = 5,5 of the circumference;
and (ii) a sector whose area= 3}, of the circle;
‘. if the angle AOB ae D degrees, then
(i) the arc AB of the circumference;
“m0

(ii) the sector AOB= of the area of the circle

360

me 55
5 of (4 circumference x radius)
=}. arc AB x radius.

THE AREA OF A SEGMENT,

The area of a minor segment is found by subtracting from

the corresponding sector the area of the triangle formed by
the chord and the radii. Thus
Area of segment ABC = sector OACB = triangle AOB.
The area of a major segment is most simply found by
subtracting the area of the corresponding minor segment from
the area of the circle.
 CIRCUMFERENCE AND AREA OF A CIRCLE. 205

EXERCISES.

[In each case choose the value of aw so as to give a result of the assigned
degree of accuracy. |
1. Find to the nearest millimetre the circumferences of the circles
whose radii are (i) 4°55 em. (ii) 100 cm.
2. Find to the nearest hundredth of a square inch the areas of the
circles whose radii are (i) 2°3”. (ii) 10°6’.
3. Find to two places of decimals the circumference and area of a
circle inscribed in a square whose side is 3°6 cm.
4. Ina circle of radius 7-0 cm. a square is described: find to the
nearest square centimetre the difference between the areas of the circle
and the square.
5. Find to the nearest hundredth of a square inch the area of the
circular ring formed by two concentric circles whose radii are 5°7” and
4: Se

6. Shew that the area of a ring lying between the circumferences of

two concentric circles is equal to the area of a circle whose radius is
the length of a tangent to the inner circle from any point on the outer.
7. A rectangle whose sides are 8:0 cm. and 6:0 em. is inscribed in a
circle. Calculate to the nearest tenth of a square centimetre the total
area of the four segments outside the rectangle.
8. Find to the nearest tenth of an inch the side of a siuebe whose
area is equal to that of a circle of radius 5”.
-9. A circular ring is formed by the circumference of two concentric
circles. The area of the ring is 22 square inches, and its width is 1:0’;
taking 7 as 24, find approximately the radii of the two circles.
10. Find to the nearest hundredth of a square inch the difference
between the areas of the circumscribed and inscribed circles of an
equilateral triangle each of whose sides is 4”.

11. Draw on squared paper two circles whose centres are at the
points (1°5”, 0) and (0, ‘8”), and whose radii are respectively °7” and
1-0’. Prove that the circles touch one another, and find approximately
their circumferences and areas.
12. Draw a circle of radius 1:0” having the point (1°6”, 1:2’) as
centre. Also draw two circles with the origin as centre and of radii
1:0” and 3:0" respectively. Shew that each of the last two circles
touches the first.
206 GEOMETRY.

EXERCISES.

On tHE INSCRIBED, CIRCUMSCRIBED, AND EscrIBED CIRCLES OF A

TRIANGLE.
(Theoretical.)
1. Describe a circle to touch two parallel straight lines and a third
straight line which meets them. Shew that two such circles can be
drawn, and that they are equal.
2. Triangles which have equal bases and equal vertical angles have
equal circumscribed circles.
3. ABC is a triangle, and |, S are the centres of the inseribed and
circumscribed circles ; if A, |, S are collinear, shew that AB=AC.
4. The sum of the diameters of the inscribed and circumscribed
circles of a right-angled triangle is equal to the sum of the sides con-
taining the right angle.
5. If the circle inscribed in the triangle ABC touches the sides
at D, E, F; shew that the angles of the triangle DEF are respectively
A B C
90-3» 90-5 90- >:

6. If | is the centre of the circle inscribed in the triangle ABC,

and |, the centre of the escribed circle which touches BC; shew that
1, B, !,, C are concyelic.
7. In any triangle the difference of two sides is equal to the
difference of the segments into which the third side is divided at the
point of contact of the inscribed circle,
' §. Jn the triangle ABC, | and § are the centres of the inscribed and
circumscribed circles: shew that IS subtends at A an angle equal to
half the difference of the angles at the base of the triangle.
Hence shew that if AD is drawn perpendicular to BC, then Al is the
bisector of the angle DAS,
9. The diagonals of a quadrilateral ABCD intersect at O: shew
that the centres of the cireles circumscribed about the four triangles
AOB, BOC, COD, DOA are at the angular points of a parallelogram.
10. In any triangle ABC, if | is the centre of the inscribed circle,
and if Al is produced to meet the circumscribed circle at O; shew that
O is the centre of the circle cireumscribed about the triangle BIC.
1l. Given the base, altitude, and the radius of the circumscribed
circle ; construct the triangle.
12. Three circles whose centres are A, B, C touch one another
externally two by two at D, E, F: shew that the inscribed circle of
the triangle ABC is the circumscribed circle of the triangle DEF.
 THE ORTHOCENTRE OF A TRIANGLE. 207

THEOREMS AND EXAMPLES ON CIRCLES AND

TRIANGLES.

THE ORTHOCENTRE OF A TRIANGLE.

I. The perpendiculars drawn from the vertices of a triangle to the

opposite sides wre concurrent.
In the AABC, let AD, BE be the
perp* drawn from A and B to the opposite
sides ; and let them intersect at O. : A
Join CO; and produce it to meet AB at \\
F. E
It is required to shew that CF is perp.
to AB. iN
Join DE.
B
Then, because the LSOEC, ODC are rt.
angles,
. the points O, E, C, D are concyclic :
. the ~DEC=the ZL DOC, in the same segment ;
=the vert. opp. L FOA.
) Again, because the L*AEB, ADB are rt. angles,
.. the points A, E, D, B are concyelic :
| .. the ©_DEB=the 2 DAB, in the same segment.
, *, the sum of the 4s FOA, FAO=the sum of the 4: DEC, DEB
=a rt. angle:
. the remaining LAFO=art. angle: Theor. 16.
that is, CF is perp. to AB.
Hence the three perps AD, BE, CF meet at the point O.
Q.E.B,

DEFINITIONS.

(i) The intersection of the perpendiculars drawn from the

vertices of a triangle to the opposite sides is called its
orthocentre.
(ii) The triangle formed by joining the feet of the perpen-
diculars is called the pedal or orthocentric triangle.
208 GEOMETRY.

Il. Jn an acute-angled triangle the perpendiculars drawn from the

vertices to the opposite sides bisect the angles of the pedal triangle through
which they pass.
In the acute-angled A ABC, let AD, BE,
CF be the perp* drawn from the vertices to
the opposite sides, meeting at the ortho-
centre O; and let DEF be the pedal triangle. F,
It is required to prove that
AD, BE, CF bisect respectively
the L* FDE, DEF, EFD.
It may be shewn, as in the last theorem,
that the points O, D, C, E are concyclic ; B D C
. the ©ODE=the LOCE, in the same segment.
Similarly the points O, D, B, F are concyclic ;
*. the ©ODF =the LOBF, in the same segment.

But the 2OCE=the LOBF, each being the compt of the 2 BAC,
*. the ,ODE=the LODF.

Similarly it may be shewn that the £* DEF, EFD are bisected by

BE and CF. Q.E.D.

CoroLuarRy. (i) Lvery two sides of the pedal triangle are equally
inclined to that side of the original triangle in which they meet.
For the LEDC=the compt of the LODE
=the compt of the LOCE
=the 2 BAC.

Similarly it may be shewn that the .FDB=the ZLBAC,

.. the LEDC=the .FDB=the ZA.

In like manner it may be proved that

the 2.DEC=the LFEA=the 2B,
and the 2DFB=the LEFA=the LC,

Corotuary. (ii) The triangles DEC, AEF, DBF are equiangwar to

one another and to the triangle ABC.

Nots. If the angle BAC is obtuse, then the perpendiculars BE, CF

bisect eavernal/y the corresponding angles of the pedal triangle.
 YHE ORTHOCENTRE OF A TRIANGLE, 209

EXERCISES.

te ‘If O is the orthocentre of the triangle ABC and vf the perpendicular

AD -s produced to meet the circum-circle in G, prove that OD=DG.

2. Inanacute-angled triangle the three sides are the external bisectors

of the angles of the pedal triangle: and in an obtuse-angled triangle the
sides containing the obtuse angle are the internal bisectors of the correspond-
sng angles of the pedal triangle.

3. If O is the orthocentre of the triangle ABC, shew that the angles

BOC, BAC are supplementary.

4. If O 1s the orthocentre of the triangle ABC, then any one of the

four points O, A, B, C ts the orthocentre of the triangle whose vertices are
the other three.

5. The three circles which pass through two vertices of a triangle and
sts. orthocentre are each equal to the circum-circle of the triangle.

6. D, E are taken on the circumference of a’ semi-circle described

on a given straight line AB: the chords AD, BE and AE, BD intersect
(produced if necessary) at F and G: shew that FG is perpendicular
to AB.

7. ABC isa triangle, O is its orthocentre, and AK a diameter of the

circum-circle: shew that BOCK is a parallelogram.

8. The orthocentre of a triangle is joined to the middle point of the

base, and the joining line is produced to mect the cireum-circle: prove
that it will meet it at the same point as the diameter which passes
through the vertex.

9. The perpendicular from the vertex of a triangle on the base, and

the straight line joining the orthocentre to the middle point of the
base, are produced to meet the circum-circle at P and Q: shew that
PQ is parallel to the base.
10. The distance of each vertex of a triangle from the orthocentre is
double of the perpendicular drawn from the centre of the circum-circle to
the opposite side.

1l. Three circles are described each passing through the orthocentre
of a triangle and two of its vertices: shew that the triangle formed by
joining their centres is equal in all respects to the original triangle.

12. Construct a triangle, having given a vertex, the orthocentre,

and the centre of the circum-circle.
H.S.G. Oo
210 GEOMETRY

LOCI.

III. Given the base and vertical angle of a triangle, find the locus of
tts orthocentre.
Let BC be the given base, and X the
given angle ; and let BAC be any triangle x A
en the base BC, having its vertical LA
equal to the 2X.
Draw the perp* BE, CF, intersecting at E
the orthocentre O.
It is required to find the locus of O.
Proof. Since the 2*OFA, OEA are rt.
angles, B Cc
., the points O, F, A, E are concyclic ;
“. the FOE is the supplement of the LA:
. the vert. opp. 2 BOC is the supplement of the LA.
But the ZA is constant, being always equal to the LX;
*, its supplement is constant ;
that is, the A BOC has a fixed base, and constant vertical angle ;
hence the locus of its vertex O is the are of a segment of which BC is
the chord.

IV. Given the base and vertical angle of a triangle, find the locus of
the in-centre.
Let BAC be any triangle on the given
base BC, having its vertical angle equal to A
the given LX; and let Al, BI, Cl be the x
bisectors of its angles. Then | is the in- As
centre.
It is required to find the locus of |.
Proof. Denote the angles of the AABC
by A, B, C; and let the L BIC be denoted
by |. B Cc
Then from the A BIC,
(i) 1+43B+4C=two rt. angles ; Theor, 16
and from the A ABC,
A+B+C=two rt. angles ;
(ii) so that 4A+4B+4C=one rt. angle.
., taking the differences of the equals in (i) and (ii),
|-4A=one rt. angle :
or, |=one rt. angle + 4A.
But A is constant, being always equal to the LX ;
| is constant :
*. the locus of | is the are of a segment on the fixed chord BC
 EXERCISES ON LOCI. 211

EXERCISES ON LOCI.

1, Given the base BC and the vertical angle A of a triangle; find

the locus of the ex-centre opposite A.

2. ‘Through the extremities of a given straight line AB any two

paraliel straight lines AP, BQ are drawn; find the locus of the inter
section of the bisectors of the angles PAB, QBA.

3. Find the locus of the middle points of chords of a circle drawn

through a fixed point.
Distinguish between the cases when the given point is within, on,
or without the circumference.

4. Find the locus of the points of contact of tangents drawn from

s fixed point to a system of concentric circles.

5. Find the locus of the intersection of straight lines which pass

through two fixed points on a circle and intercept on its circumference
an are of constant length.

6. A and B are two fixed points on the circumference of a circle,

and PQ is any diameter: find the locus of the intersection of PA
and QB.
7. BAC is any triangle described on the fixed base BC and having
% constant vertical angle; and BA is produced to P, so that BP is
equal to the sum of the sides containing the vertical angle: find the
locus of P.
8. AB is a fixed chord of a circle, and AC is a moveable chord
passing through A: if the parallelogram CB is completed, find the
locus of the intersection of its diagonals.

9. A straight rod PQ slides between two rulers placed at right

angles to one another, and from its extremities PX, QX are drawn
perpendicular to the rulers: find the locus of X.

10. Two circles intersect at A and B, and through P, any point on

the circumference of one of them, two straight lines PA, PB are
drawn, and produced if necessary, to cut the other circle at X and Y:
find the locus of the intersection of AY and BX.

11. Two circles intersect at A and B; HAK is a fixed straight line

drawn through A and terminated by the circumferences, and PAQ ia
any other straight line similarly drawn: find the locus of the inter:
section of HP and QK.
212 GEOMETRY.

SIMSON’S LINE.
V. The feet of the perpendiculars drawn to the three sides of @
triangle from any point on its circum-circle are collinear.

Let P be any point on the circum-circle of

the A ABC; and let PD, PE, PF be the perps. -
drawn from P to the sides.
It is required to prove that the points D, E, F
are collinear.
Join FE and ED:
then FE and ED will be shewn to be in the
game straight line.
Join PA, PC.

Proof. Because the Lt PEA, PFA are rt. angles,

*. the points P, E, A, F are concyelic:
”. the LPEF=the PAF, in the same segment
=the yeu of the 2 PAB
=the £ PCD,
since the points A, P, C, B are concyelic.
Again because the Ls PEC, PDC are rt. angles,
. the points P, E, D, C are concyclic.
*. the £4PED=the suppt of the 2 PCD
=the supp* of the 2 PEF.
‘. FE and ED are in one st. line.
Obs. The line FED is known as the Pedal or Simson’s Line of the
triangle ABC for the point P.

EXERCISES.
1. From any point P on the circum-circle of the triangle ABC,
perpendiculars PD, PF are drawn to BC and AB: if FD, or FD
produced, cuts AC at E, shew that PE is perpendicular to AC.
2. Find the locus of a point which moves so that if perpendiculars
are drawn from it to the sides of a given triangle, their feet are
collinear.
3. ABC and AB’C’ are two triangles with a common angle, and
their circum-circles meet again at P; shew that the feet of perpen-
diculars drawn from P to the lines AB, AC, BC, B’C’ are collinear.
4. A triangle is inseribed in a circle, and any point P on the circum-
ference is joined to the orthocentre of the triangle: shew that this
joining line is bisected by the pedal of the point P.
 THE TRIANGLE AND ITS CIRCLES. 213

THE TRIANGLE AND ITS CIRCLES.

VI. OD, E, F ave the points of contact of the inscribed circle of the
wsangle ABC, and D,, E,, F, the points of contact of the escribed circle,
which touches BC and the other sides produced : a, b, ce denote the length
of the sides BC, CA, AB; s the semi-perimeter of the triangle, and r, ry
he radu of the inscribed and escribed circles.

Prove the foliowing equalities :

(i) AE =AF =s-a,
BD =BF =s--6,
C)9) =(ClS =ooe,
(i) Ae, Alyss
(mi) CD, =CE,=s—b;
BD} =BF,=s=c.
(iv) CD =BD,, and BD=C0,.
W)EE Fria.
(vi) The area of the A ABC=rs
=r,(3—-a),

(vii) Draw the above figure in the case when C is a right angle. and
prove that r=8—-c3 r,=s—b.
 214 GEOMETRY,

VII. In the triangle ABC, | is the centre of the inscribed circle, and
i, lo, Is the centres of the escribed circles touching respectively the sides
8c, CA, AB and the other sides produced.

Prove the following properties

(i) The points A, |, |, are collinear: so are B, |, 1,3 and C, |, hy.

(ti) The points |,, A, |, are collinear ; 80 are |,, B, |, ; and |,, ©, le.

(ili) The triangles BI,C, CIA, Al,B are equiangular to one another.

(iv) The triangle Ill, is equiangular to the triangle formed by

joining the points of contact of the inscribed circle.
(v) Of the four points |, |, |g, ly, each is the orthocentre of the
triangle whose vertices are the other three.
(vi) The four circles, each of which passes through three of the
points |, |,, |, ly, are all equal.
 THE TRIANGLE AND ITS CIRCLES. 215

EXERCISES.
1. With the figure given on page 214 shew that if the circles whose
sentres are |, |,, 1,, 1; touch BC at D, D,, D,, Dg, then
(i) DD,=D,D,=2. (ii) DD,=D,D,=c.
(iii) D,D,=b +c. (iv) DD, =b~e.
2. Shew that the orthocentre and vertices of a triangle are the centres
of the inscribed and escribed circles of the pedal triangle.
3. Given the base and vertical angle of a triangle, find the locus of the
centre of the escribed circle which touches the base.
4. Given the base and vertical angle of a triangle, shew that the centre
of the circum-circle ts fixed.
5. Given the base BC, and the vertical amgle A of the triangle, find
the locus of the centre of the escribed circle which touches AC.
6. Given the base, the vertical angle, and the point of contact with
the base of the in-circle;
; construct the triangle.
7. Given the base, the vertical angle, and the point of contact with
the base, or base produced, of an escribed circle ; construct the triangle.
8. | is the centre of the circle scribed in a triangle, and |,, ly, |s the
centres of the escribed circles; shew that \l,, ll,, Il, ave bisected by the
circumference of the circum-circle.
9. ABC is a triangle, and |,, |, the centres of the escribed circles
which touch AC, and "AB respectiv ely : shew that the points B, C, Ip, |,
lie upon a circle whose centre is on the circumference of the circum-
circle of the triangle ABC.
10. With three given points as centres describe three circles touch-
ing one another two by two. How many solutions will there be?
11. Given the centres of the three escribed circles; construct the
triangle.
12. Given the centre of the inscribed circle, and the centres of two
escribed circles ; construct the triangle.
13. Given the vertical angle, perimeter, and radius of the inscribed
circle ; construct the triangle.
14. Given the vertical angle, the radius of the inscribed circle, and
the length of the perpendicular from the vertex to the base ; construct
the triangle.
15. Ina triangle ABC, | is the centre of the inscribed circle; shew
that the centres of the circles circumscribed about the triangles BIC,
CIA, AIB lie on the circumference of the circle circumscribed about the
given triangle.
216 GEOMETRY

THE NINE-POINTS CIRCLE.

VIII. Jn any triangle the middle points of the sides, the feet of the
perpendiculars from the vertices to the opposite sides, and the middie
points of the lines joining the orthocentre to the vertices are concyclic.
In the A ABC, let X, Y, Z be the
middle points of the sides BC, CA,
AB; let D, E, F be the feet of the
yerp* to these sides from A, B, C; let
5 be the orthocentre, and a, B, y the
middle points of OA, OB, OC
It is required to prove that
the nine points X, Y, Z, D, E, F, a, 8, y
mre concyclic.
Join XY, XZ, Xa, Ya, Za.
Now from the A ABO,
since AZ=ZB, and Aa=aO,
“. Zais par'to BO. Ex. 2, p. 64.
And from the A ABC, since BZ=ZA, and BX=X6,
“. ZX is par! to AC.
But BO produced makes a rt, angle with AC;
. the LXZa is art. angle.
Similarly, the C XYa is a rt. angle.
*. the points X, Z, a, Y are concyelic :
that is, a lies on the O* of the circle which passes through X, ¥, Z5
and Xa is a diameter of this circle.
Similarly it may be shewn that 8 and ¥ lie on the O* of this circle.
Again, since aDX is a rt. angle,
*. the circle on Xa as diameter passes through D.
Similarly it may be shewn that E and F lie on the O* of this cirole>
. the points X, Y, Z, D, E, F, a, 8, y are concyclic. Q.8.D.

Obs. From this property the circle which passes through the middie
points of the sides of a triangle is called the Nine-Points Circle ; many
of its properties may be derived from the fact of its being the circum
sircle of the pedal triangle.
 THE NINE-POINTS CIRCLE. 21 oj

To prove that
_ (i) the centre of the nine-points circle is the middle point of the
sirarght line which joins the orthocentre to the circum-centre.
: (ii) the radius of the wine-points circle is half the radius of the
crrcum-circle,
(ili) the centrozd is collinear with the circum-centre, the nine-pointa
centre, and the orthocentre.

In the AABC, let X, Y, Z be the

middie points of the sides; D, E, F
the feet of the perp*; O the ortho-
centre; S and N the centres of the
circumscribed and nine-points circles
respectively.
(i) Vo prove that N is the middle
point of SO.
It may be shewn that the perp. to
XD from its middle point bisects SO ;
Theor. 22.
Similarly the perp. to EY at its B
middle point bisects SO:
that is, these perp* intersect at the middle point of SO:
And since XD and EY are chords of the nine-points circle,
«. the intersection of the lines which bisect XD and EY at rt. angles is
its centre: Theor. 31, Cor. i.
.. the centre N is the middle point of SO. Q.E.D.

‘:) To prove that the radius of the nine-points circle is half the
radwus of the circum-circle.
By the last Proposition, Xa is a diameter of the nine-points circle.
.. the middle point of Xa is its centre :
ibut the middle point of SO is also the centre of the nine-points circle.
(Proved. }
Hence Xa and SO bisect one another at N,
Then from the A* SNX, ONa,
f SN=ON,
because and NX=Na,
and the 2 SNX=the LONa:
EO Og,
aCe
And SX is also par! to Aa,
en A=
But SA is a radius of the circum-circle ;
and Xa is a diameter of the nine-points circle ;
§. the radius of the nine-points circle is half the radius of the circum
gircle. [See also p. 267, Examples 2 and 3.] Q.E. D.
218 GEOMETRY.

(iii) To prove that the centroid is collinear with points 8, N, O.

Join AX and draw ag par! to SO.

Let AX meet SO at G.
Then from the A AGO, since Aa=a0O,
and ay is par! to OG,
. Ag=9G. Ex, I, p. 64,
And from the A Xag, since aN=NX,
and NG is par’ to ag,
*. ga=GX.
. AG=8 of AX;
.. Gis the centroid of the triangle ABC.
Theor. III., Cor., p. 97. B Xx Dp G
That is, the centroid is collinear with
the points S,N, O. Q.5.p.

EXERCISES.

1. Given the base and vertical angle of a triangle, find the locus of the
centre of the nine-points circle.
2. The nine-points circle of any triangle ABC, whose orthocentre &
O, is also the nine-points circle of each of the triangles AOB, BOC,
COA.
3. If I, |, bl, |, are the centres of the inscribed and escribed circles
of a triangle ABC, then the circle circumscribed about ABC is the
nine-points circle of each of the four triangles formed by joining three
of the points |, 1), le, ly.
4. All triangles which have the same orthocentre and the same
circumscribed circle, have also the same nine-points circle.
5. Given the base and vertical angle of a triangle, shew that one
angle and one side of the pedal triangle are constant.
6. Given the base and vertical angle of a triangle, find the locus of
the centre of the circle which passes through the three escribed
centres.

Norg. For some other important properties of the Nine-pointe

Cirele see Ex. 54, page 210. :
 PART sky,

ON SQUARES AND RECTANGLES IN CONNECTION

WITH THE SEGMENTS OF A STRAIGHT LINE.

THE GEOMETRICAL EQUIVALENTS OF CERTAIN

ALGEBRAICAL FORMULAi.

DEFINITIONS.
A B
i. A rectangle ABCD is said to be
contained by two adjacent sides AB, AD ;
for these sides fix its size and shape.
D C
A rectangle whose adjacent sides are AB, AD is denoted by
the rect. AB, AD; this is equivalent to the product AB. AD.
Similarly a square drawn on the side AB is denoted by
the sg. on AB, or AB?.
2. Jf a point X is taken ina A =B
straight line AB, or in AB produced, re
then X is said to divide AB into the Fig. I.
two segments AX, XB; the segments
being in either case the distances of
the dividing point X from the extremities / B XxX
of the given line AB. Fig. 2.
In Fig. 1, AB is said to be divided internally at x.
Hoagie 25 AB: <2.ost easn'ens divided externally at X.

Obs. In internal division the given line AB is the sum of

the segments AX, XB.
In external division the given line AB is the difference of the
segments AX, XB.
220 GEOMETRY.

THEOREM 50. [Euclid II. 1.]

Tf of two straight lines, one is divided into any number of parts,
the rectangle contained by the two lines is equal to the sum of the
rectangles contained by the undivided line and the several parts of
the diwided line.

D E FG

Let AB and K be the two given st. lines, and let AB be

divided into any number of parts AX, XY, YB, which contain
respectively a, 6, and ¢ units of length ;3 so that AB contains
a+b+c units.
Let the line K contain / units of length.
It is required to prove that
the rect. AB, K=rect. AX, K+ rect. XY, K+rect. YB, K;
namely that
(a+ b+c)k= ak + bk + ck.
Construction. Draw AD perp. to AB and equal to K.
Through D draw DC par’ to AB.
Through X, Y, B draw XE, YF, BC par’ to AD.

Proof. The fig. AC=the fig. AE+the fig. XF + the fig. YC;
and of these, by construction,
fig. AC=rect. AB, K; and contains (a+b+c)k units of area;
fig. AE =rect. AX, K; and contains ak units of area;
feRE mePCI, Rig oss ce yvespee UM. nikebie unex saan’ ; —

fig, YO mreot, YB, Kj. cvsresccssvacs BE sbikadsicusheee ‘

Hence
the rect. AB, K=rect. AX, K+rect. XY, K+rect. YB, K;
or, (a+b+c)k= ak + Lk + ck.
Q.E.D.
 SQUARES AND RECTANGLES. 22%

*CoROLLARIES. [Euclid II. 2 and 3.]

Two special cases of this Theorem deserve attention.
(i) When AB is divided only at one point X, and when the
undivided line AD is equal to AB.
A 6B

D E. €
Then the sq. on AB=the rect. AB, AX + the rect. AB, XB.
Phat 1s,
The square on the given line is equal to the sum of the rectangles
contained by the whole line and each of the segments.
Or thus:
AB?=AB.AB
= AB(AX + XB)
=AB.AX+AB.
XB.
(ii) When AB is divided at one point X, and when ths
undivided line AD is equal to one segment AX.
A X B

D E Cc
Then the rect. AB, AX =the sq. on AX +the rect. AX, XB.
That is,
The rectangle contained by the whole line and\one segment 1:
equal to the square on that segment with the rectangle contained by
the two segments.
Or thus :
AB. AX = (AX + XB) AX
=AX?+AX.XB
222 GEOMETRY.

THEOREM 51. [Euclid II. 4.]

If a strarght line is divided internally at any point, the square
on the given line is equal to the sum of the squares on the two seg-
ments together with twice the rectangle contained by the segments.
a X..6.'B

(a) |(ab)

a F60O
Let AB be the a st. line divided internally at X ; and let
the segments AX, XB contain a and 0 units of length respectively.
Then AB is the swm of the segments AX, XB, and therefore
contains a+ units.
It is required to prove that
AB? = AX? + XB2 + 2AX. XB;
namely that
(a+bP?= a2 + B + ~— 2ab,
Construction. On AB describe a square ABCD. From AD
cut off AE equal to AX, ora. Then ED=XB=+. Through E
and X draw EH, XF par’ respectively to AB, AD and meeting
at G.
Proof. ‘Then the fig. AC=the figs. AG, GC + the figs. EF, XH
And of these, by construction,
fig. AC is the sq. on AB, and contains (a+)? units of area;
fig. AG=sq. on AX, and contains a? units of area ;
feGO=aq. On XB; oes cewevieee OP sien daiay teases ;
fig. EF =rect. EG, ED ;
mm OOt, AX; RBs iicvccssvens Ee pee ;
liXH= rect. GX, XB
me TCU. Ary HE cise vsrcscees Ss
Hence AB? = AX? + XB? + 2AX. XB;
that is, (a+b)?= a? + 0? + = ab.
QED
 SQUARES AND RECTANGLES, 223

THEOREM 52. [Euclid II. 7.]

If a straight line is divided externally at any point, the square
on the given line is equal to the swum of the squares on the two
segments diminished by twice the rectangle contained by the
segments,

a-b

(a-b) ;

E rece
Let AB be the given st. line divided eaternally at X ; and let
the segments AX, XB contain a and / units of length re-
spectively.
Then AB is the difference of the segments AX, XB, and
therefore contains a —0 units.
It 1s required to prove thut
AB?= AX? + XB? — 2AX. XB;
namely that (a—b)?= a? + 0? — 2ab.
Construction. On AX describe a square AXGE. From AE
cut off AD equal to AB, ora—). Then ED=XB=). Through
D and B draw DF, BH par’ respectively to AX, AE, meeting
at OC.
Proof. Then the fig. AC =the figs. AG, CG — the figs. EF, XH.
And of these, by construction,
fig. AC is the sq. on AB, and contains (a — 0)? units of area ;
9
fig. AG=sq. on AX, and contains a units of area;
fig. GCG=sdn oi XB, aie ncaame Peg SR aigSeta, :
fig. EF—rect. EG; ED
—rect. AX, XB ......... Pe ye
eee ee
fg. XH
= rect. GX, XB
=rect. AX, XBu... ccscesnee CUM cisaberatedeg
Hence AB? = AX? + XB? — 2AX. XB;
that is, (a—b)*= a? + 0 — 2ab.
Q.E.D.
224 GEOMETRY.

THEOREM 53. [Euclid II. 5 and 6€.]

The difference of the squares on two straight lines is equal to the
rectangle contained by their sum and difference.

m< H D
Let the given lines AB, AC be placed in the same st. line,
and let them contain a and 0) units of length respectively.
It is required to prove that
AB? — AC? = (AB + AC) (AB — AC) ;
namely that a* — b? =(a+b)(a—-5).

Construction. On AB and AC draw the squares ABDE,

ACFG ; and produce CF to meet ED at H.
Then GE =CB=a-bd units.

Proof. Now AB?—AC?=the sq. AD —the sq. AF

=the rect. CD + the rect. GH
=DB.BC +GF.GE
=AB.CB +AC.CB
=(AB+AC)CB
== (AB + AC) (AB ~ AC)
That is,
a® — 0% =(a+b\(a- b).
0.2.D
 SQUARES AND RECTANGLES. 225

Corottary. Jf a straight line is bisected, and also divided

(internally or externally) into two unequal segments, the rectangle
contained by these segments is equal to the difference of the squares on
half the line ard on the line between the points of section.
A-. NE NB) ASS
X By NG

Fig. 1. Fig. 2.

That is, if AB is bisected at X and also divided at Y, inter-

nally in Fig. 1, and externally in Fig. 2, then
in Fig. 1, AY . YB=AX?— XY?;
in Fig. 2, AY . YB= XY2— AX2,
For in the first case, AY. YB=(AX + XY) (XB — XY)
= (AX + XY) (AX — XY)
= AX? — XY”,
The second case may be similarly proved.

EXERCISES.

1, Draw diagrams on squared paper to shew that the square on a

straight line is
(i) four-times the square on ha/f the line ;
(ii) n¢ne-times the square on one-third of the line.
2. Draw diagrams on squared paper to illustrate the following
algebraical formule :
(i) (e+ 7)?=a2+ 14449.
(ii) (@+b4+c)?=a?+b? +0? +42be+ 2ca+ 2ab.
(iii) (a+6)(c+d) =ac+ad
+ be+bd.
(iv) (2 +7)(a+9)=a2 +16x +63.
3. In Theor. 50, Cor. (i), if AB=4 cm., and the fig, AE=9°6 sq. cm.,
find the area of the fig. XC.
4, In Theor. 50, Cor. (ii), if AX=2°1”, and the fig. XC=3°36 sq. in.,
find AB. :
5. In Theor. 51, if the fig. AG=36 sq. em., and the rect. AX, XB
=24 sq. cm., find AB.
6. In Theorem 52, if the fig. AG=9°61 sq. in., and the fig. D@=6°51
sq. in., find AB.
[For further Examples on Theorems 50-53 see p. 230.]
H.S.G. P
226 GEOMETRY.

THeorEM 54. (Euclid I. 12.]

In an obtuse-angled triangle, the square on the side subtending
the obtuse angle is equal to the sum of the squares on the sides
containing the obtuse angle together with twice the rectangle con-
tained by one of those sides and the projection of the other side
upon tt.
A

B Cc D
is
Let ABC be a triangle obtuse-angled at C; and let AD be
drawn perp. to BC produced, so that CD is the projection of
the side CA on BC. [See Def. p. 63.]
It is required to prove. that
AB? = BC? + CA? + 2BC. CD.

Proof. Because BD is the sum of the lines BC, CD,

*, BD?=BC?+CD?+2BC. CD. Theor. 51
To each of these equals add DA?
Then BD? + DA? = BC? + (CD? + DA?) + 2BC. CD.

But BD? + DA? = AB?

for the .Disart. 2.
and CD2+ ones:

Hence AB? = BC? + CA? + 2BC. CD.

Q.E.D.
 SQUARES AND RECTANGLES. 29

TuEoREM 55. [Euclid Il. 13.]

In every triangle the square on the side subtending an acute angle
is equal to the sum of the squares on the sides containing that angle
diminished by twice the rectangle contained by one of those sides
and the projection of the other side wpon it.

is. soe
B D C D B (}

Big. ot. 18h fear

Let ABC be a triangle in which the LC is acute; and let

AD be drawn perp. to BC, or BC produced ; so that CD is the
projection of the side CA on BC.
It is required to prove that
AB?= BC?+ CA? — 2BC. CD.

Proof. Since in both figures BD is the difference of the lines

BC, CD,
‘, BD?=BC?+ CD? - 2BC . CD. Theor. 52.
To each of these equals add DA?.
Then BD?+DA? =BC? + (CD? + DA?) - 2BC . CD. ......(i)
2
But BD?+DA 2—eels
AB?
for the LD is art. Z.
and CD?+ DA?=CA?|
Hence AB? = BC? + CA? - 2BC . CD.
228 GEOMETRY.

SuMMARY OF THEOREMS 29, 54 and 55.

Pres regia
B (oe 1D) B C(D) B D C

(i) If the 2 ACB is obtuse,

AB? = BC? + CA? + 2BC ..CD. Theor. 54.
(ii) If the 2 ACB is a right angle,
AB? = BC? + CA? Theor. 29.
(iii) If the 2 ACB is acute,
AB? = BC? + CA? — 2BC . CD. Theor, 55.
Observe that in (ii), when the 2 ACB is right, AD coincides
with AC, so that CD (the projection of CA) vanishes;
hence, in this case, 2BC.CD=0.

Thus the three results may be collected in a single

enunciation :
The square on a side of a triangle is greater than, equal to, or
less than the sum of the squares on the other sides, according as the
angle contained by those sides is obtuse, a right angle, or acute ; the
difference in cases of inequality being twice the rectangle contained
by one of the two sides and the projection on it of the other.

EXERCISES.

1. In a triangle ABC, a=21 cm., b=17 em., c=10 cm. By how

many square centimetres does c? fall short of a?4+6?? Hence or other-
wise calculate the projection of AC on BC.
2. ABC is an isosceles triangle in which AB=AC; and BE is drawn
perpendicular to AC. Shew that BC?=2AC. CE.
3. Inthe A ABC, shew that
(i) if the 2C=60°, then c?=a?+l?-ab;
(ji) if the 2C=120°, then c?=a*?+l?+
ab.
 SQUARES AND RECTANGLES. boboie}

THEOREM 56.

In any triangle the sum of the squares on iwo sides is equal to

twice the square on half the third side together with twice the square
on the median which bisects the third side.
A

B X D C
Let ABC be a triangle, and AX the median which bisects the
base BC.
it 1s required to prove that
AB? + AC? = 2BX? + 2AX?2.
Draw AD perp. to BC; and consider the case in which 4B
and AC are unequal, and AD falls within the triangle.
Then of the 2*AXB, AXC, one is obtuse, and the otl:or acute
Let the ~ AXB be obtuse.
Then from the A AXB,
AB? = BX? + AX? + 2BX.XD. Theor. 54.
And from the A AXC,
AC? = XC? + AX? — 2XC.. XD. Theor, 55
Adding these results, and remembering that XC=BX,
we have
AB? + AC? = 2BX?2 + 2AX?,
Q.E.D.
Norr.—The proof may easily be adapted to the case in which the
perpendicular AD falls outside the triangle.

EXERCISE.

In any triangle the difference of the squares on two sides is equal te

twice the rectangle contained by the base and the intercept between the
middle point of the base and the foot of the perpendicular drawn from
the vertical angle to the base.
230 GEOMETRY.

EXERCISES ON THEOREMS 50-53.

1. Use the Corollaries of Theorem 50 to shew that if a straight line
AB is divided internally at X, then
AB?=AX?+ XB?+2AX. XB.
2. Ifa straight line AB is bisected at X and produced to Y, and if
AY . YB=SAX?, shew that AY =2AB.
3. The sum of the squares on two straight lines is never less than
twice the rectangle contained by the straight lines.
Explain this statement by reference to the diagram of Theorem 52.
Also deduce it from the formula (a — b)?=a?+b* —2ab.
4. Inthe formula (a+b) (a —b)=a* - b*, substitute a= nS, b= 74,
and enunciate verbally the resulting theorem,
5. If a straight line is divided internally at Y, shew that the
rectangle AY, YB continually diminishes as Y moves from X, the mid-
point of AB.
Deduce this (i) from the Corollary of Theorem 53 ;
2 _h\2
(ii) from the formula ab= (>) - (3°) F

6. Ifa straight line AB is bisected at X, and also divided (i) inter-

nally, (ii) externally into two unequal segments at Y, shew that in either
case AY? + YB?=2(AX?+ XY*), {Euclid II. 9, 10.]
[Proof of case (i).
AY?+ YB?=AB?-2AY.YB Theor. 51.
=4AX?—-2(AX + XY) (AX — XY)
=4AX? —- 2(AX2— XY?) Theor. 53.
=2AX?+2XY?.
Case (ii) may be derived from Theorem 52 in a similar way.)
7. If AB is divided internally at Y, use the result of the last
example to trace the changes in the value of AY?+YB*, as Y moves
from A to B.
8. Ina right-angled triangle, if a perpendicular is drawn from the
right angle to the hypotenuse, the square on this perpendicular is equal to
the rectangle contained by the segments of the hypotenuse.
9. ABC is an isosceles triangle, and AY is drawn to cut the base BC
internally or externally at Y. Prove that
AY?=AC?-—BY.YC, for internal section ;
AY?=AC?+BY.YC, for external section.
 EXERCISES. ool

EXERCISES ON THEOREMS 54-56.

1. AB is a straight line 8 cm. in length, and from its middle point
O as centre with radius 5 em. a circle is drawn; if P is any point on
the circumference, shew that
AP?+ BP?=82 sq. cm.
2. In a triangle ABC, the base BC is bisected at X. If a=17cm.,
b=15 em., and c=8 em., calculate the length of the median AX, and
deduce the LA.
3. The base of a triangle=10 cm., and the sum of the squares on
the other sides= 122 sq. cm. ; find the locus of the vertex.
4. Prove that the sum of the squares on the sides of a parallelogram
is equal to the sum of the squares on its diagonals.
The sides of a rnombus and its shorter diagonal each measure 3” ;
find the longer diagonal to within ‘01’.
5. In any quadrilateral the squares on the diagonals are together
equal to twice the sum of the squares on the straight lines joining the
middle points of opposite sides. [See Ex. 7, p. 64.]
6. ABCD isa rectangle, and O any point within it: shew that
OA?+ OC?= OB?+ OD?.
If AB=6:0’, BC=2°5”, and OA?+ OC?=2]1} sq. in., find the distance
of O from the intersection of the diagonals.
7. The sum of the squares on the sides of a quadrilateral is greater
than the sum of the squares on its diagonals by four times the square
on the straight line which joins the middle points of the diagonals.
8. In a triangle ABC, the angles at B and C are acute; if BE, CF
are drawn perpendicular to AC, AB respectively, prove that
BC?=AB.BF+AC.CE.
9. Three times the sum of the squares on the sides of a triangle is
equal to four times the sum of the squares on the medians.
10. ABC is a triangle, and O the point of intersection of its
medians: shew that
AB?+ BC?+ CA?=3(OA?+ OB?+ OC?).
1]. If a straight line AB is bisected at X, and also divided (inter-
nally or externally) at Y, then .
AY?+ YB?=2(AX?+ XY?). [See p. 230 Ex. 6.]
Prove this from Theorem 56, by considering a triangle CAB in the
limiting position when the vertex C falls at Y in the base AB.
12. In a triangle ABC, if the base BC is divided at X so that
mBX=nXC, shew that
mAB? + nAC?=mBX? + nXC? + (m +n) AX?.
252 GEOMETRY.

RECTANGLES IN CONNECTION WITH CIRCLES.

THEOREM 57. [Euclid III. 35.]

If two chords of a vircle cut at a point within it, the rectangles
contained by their segments are equal.

In the © ABC, Iet AB, CD be chords cutting at the internal

point X.
It is required to prove that
the rect. AX, XB =the rect. CX, XD.
Let O be the centre, and 7 the radius, of the given circle.
Supposing OE drawn perp. to the chord AB, and therefore
bisecting it.
Join OA, OX.
Proof. The rect. AX, XB= (AE + EX)(EB — EX)
= (AE + EX) (AE — EX)
= AE? — EX? Theor. 53
= (AE? + OE?) - (EX? + OE*)
= 9 - OX? since
the 2" at E are rt. 2°.
Similarly it may be shewn that
une rect. CX, XD=77 — OX2.
*. the rect. AX, XB=the rect. CX, XD.
OE.D.

COROLLARY — ach rectangle is equal to tie square on half the

chord which is bisected at the given point X.
 RECTANGLES IN CONNECTION WITH CIRCLES. 263

THEOREM 58. (Euclid II. 36.]

If two chords of a circle, when produced, cut at a point outside it,
the rectangles contained by their segments are equal. And each
rectangle is equal to the square on the tangent from the point of
antersection.
r

Cz
NigES ees
pate

In the ©ABC, let AB, CD be chords cutting, when produced,

at the external point X; and let XT be a tangent drawn from
that point.
Lt is required to prove that
the rect. AX, XB=the rect. CX, XD = the sq. on XT.
Let O be the centre, and r the radius of the given circle.
Suppose OE drawn perp. to the chord AB, and therefore
bisecting it.
Jain OA,.0%, On.

Proof. The rect. AX, XB=(EX+AE)(EX — EB)

= (EX + AE)(EX — AE)
= EX? — AE? Theor. 53.
= (EX? + OE2) — (AE2 + OE?)
Ox? - 7, since
the 2* at E are rt. 2".
Similarly it may be shewn that
the rect. CX, XD = OX? — 72.
And since the radius OT is perp. to the tangent XT,
| seeelia — O e774. Theor. 29.
.. the rect. AX, XB=the rect. CX, XD=the sq. on XT.
Q.E.D.
234 GEOMETRY.

THEOREM 59. [Euclid HI. 37.]

If from a point outside a circle two straight lines are drawn, one
of which cuts the circle, and the other meets it ; and if the rectangle
contained by the whole line which cuts the circle and the part of it
outside the circle is equal to the square on the line which meets the
circle, then. the line which meets the circle is a tangent to it.

From X a point outside the © ABC, let two straight lines

XA, XC be drawn, of which XA cuts the circle at A and B, and
XC meets it at C;
and let the rect. XA, XB=the sq. on XC.
It is required to prove that XC touches the circle at C.

Proof. Suppose XC meets the circle again at D;

then XA. XB=XC. XD. Theor. 58.
But by hypothesis, XA.XB=XC?;
: . XC. XD=XC?;
XD= XC.
Hence XC cannot meet the circle again unless the points of
section coincide ;
that is, XC is a tangent to the circle.
Q.E.D.
Note on THrorems 57, 58.
Remembering that the segments into which the chord AB is divided
at X, internally in Theorem 57, and externally in Theorem 58, are
in each case AX, XB, we may include both Theorems in a single
enunciation.
If any number of chords of a circle are drawn through a given point
within or without a circle, the rectangles contained by the segmenis of the
chords are equal.

e
 RECTANGLES IN CONNECTION WITH CIRCLES. 200

EXERCISES ON THEOREMS 57-59.

(Numerical and Graphical.)
I. Draw a circle of radius 5 em., and within it take a point X
3 cm. from the centre O. Through X draw any two chords AB, CD.
(i) Measure the segments of AB and CD; hence find approximately
the areas of the rectangles AX.XB and CX.XD, and compare the
results.
(ii) Draw the chord MN which is bisected at X; and from the
right-angled triangle OXM calculate the value of XM?.
(iii) Find by how much per cent. your estimate of the rect. AX, XB
differs from its true value.
2. Draw a circle of radius 3 cm., and take an external point X
5cm. from the centreO. Through X draw any two secants XAB, XCD.
(i) Measure XA, XB and XC, XD; hence find approximately the
rectangles XA. XB and XC. XD, and compare the results.
(ii) Draw the tangent XT ; and from the right-angled triangle XTO
calculate the value of XT?.
(iii) Find by how much per cent. your estimate of the rect. AX, XB
differs from its true value.
3. AB, CD are two straight lines intersecting at X. AX=1°8",
XB=1:2’, and CX=2°7”. If A, C, B, D are concyclic, find the length
of XD.
Draw a circle through A, C, B, and check your result by measure-
ment.

4. A secant XAB and a tangent XT are drawn to a circle from an

external point X.
(i) If XA=0°6’, and XB=2'4”, find XT.
(ii) If XT=7'5 cm., and XA=4’5 em., find XB.
5. A semi-circle is drawn on a given line AB; and from X, any
point in AB, a perpendicular XM is drawn to AB cutting the circum-
ference at M: shew that
AX .XB=MX?.
(i) If AX=2°5’, and MX=2:0’, find XB; hence find the diameter
of the semi-circle.
(ii) If the radius of the semi-circle=3°'7 cm., and AX=4°9 em.,
find MX.
6. A point X moves within a circle of radius 4 cm., and PQ is any
chord passing through X; if in all positions PX. XQ=12 sq. em., find
the locus of X.
What will the locus be if X moves outside the same circle, so that
PX .XQ=20 sq. cm.?
236 GEOMETRY,

EXERCISES ON THEOREMS 57-59.

(Theoretical.)
1. ABC isa triangle right-angled at C; and from Ca perpendicular
CD is drawn to the hypotenuse: shew that
AD.DB=CD?.
2. If two circles intersect, and through any point X in their
common chord two chords AB, CD are drawn, one in each circle, shew
that AX.XB=CX.XD.
3. Deduce from Theorem 58 that the tangents drawn to a circle
from any external point are equal.
4. If two circles intersect, tangents drawn to them from any point
in. their common chord produced are equal.
5. Ifacommon tangent PQ is drawn to two circles which cut at A
and B, shew that AB produced bisects PQ.
6. If two straight lines AB, CD intersect at X so that AX.XB
=CX.XD, deduce from Theorem 57 (by reductio ad absurdum) that
the points A, B, C, D are concyclic.
7. In the triangle ABC, perpendiculars AP, BQ are drawn from A
and B to the opposite sides, and intersect at O: shew that
AO.OP=BO.0Q.
8. ABC isa triangle right-angled at C, and from C a perpendicular
CD is drawn to the hypotenuse ; shew that
AB.AD=AC?.
9. Through A, a point of intersection of two circles, two straight
lines CAE, DAF are drawn, each passing through a centre and ter-
minated by the circumferences: shew that
CA.AE=DA. AF.
10. If from any external point P two tangents are drawn to a
given circle whose centre is O and radius r; and if OP meets the chord
of contact at Q; shew that
OP.OQ=r".
tl. AB is a fixed diameter of a circle, and CD is perpendicular to
AB (or AB produced) ; if any straight line is drawn from A to cut CD
at P and the circle at Q, shew that
AP .AQ=constant.

12. A is a fixed point, and CD a fixed straight line; AP is an

straight line drawn from A to meet CD at P; if in AP a point &
is taken so that AP. AQ is constant, find the locus of Q.
 RECTANGLES IN CONNECTION WITH CIRCLES. et

EXERCISES ON THEOREMS 57-59.

(ALiscellaneous. )
1. The chord of an are of a circle=2c, the height of the are=h, the
radius=r. Shew by Theorem 57 that
h(2r -—h)=c?.
Hence find the diameter of a circle in which a chord 24” long cuts off
a segment 8” in height.
2. The radius of a circular arch is 25 feet, and its height is 18 feet :
find the span of the arch.
If the height is reduced by 8 feet, the radius remaining the same, by
how much will the span be reduced?
Cheek your calculated results graphically by a diagram in which 1’
represents 10 feet.
3. Employ the equation h(27—h)=c? to find the height of an are
whose chord is 16 ecm., and radius 17 cm.
Explain the double result geometrically.
4. If d denotes the shortest distance from an external point to a
circle, and ¢ the length of the tangent from the same point, shew by
Theorem 58 that d(d+2r)=t2.
Hence find the diameter of the circle when d=1'2”, and t=2°4”; and
verify your result graphically.
5. If the horizon visible to an observer on a cliff 330 feet above the
sea-level is 224 miles distant, find roughly the diameter of the carth.
Hence find the approximate distance at which a bright light raised
66 feet above the sea is visible at the sea level.
6. If his the height of an arc of radius 7, and b the chord of half the
arc, prove that §2= rh,

7. Asemi-circle is described on AB as diameter, and any two chords

AC, BD are drawn intersecting at P: shew that
AB?=AC.AP+BD. BP.
8. Two circles intersect at B and C, and the two direct common
tangents AE and DF are drawn: if the common chord is produced to
meet the tangents at G and H, shew that
Gh?=AE?+ BC,
9. If from an external point P a secant PCD is drawn to a circle,
and PM is perpendicular to a diameter AB, shew that
PM?=PC.PD+AM.MB.
238 GEOMETRY.

PROBLEMS.

PROBLEM 32.
To draw a square equal in area to a given rectangle.

Let ABCD be the given rectangle.

Construction. Produce AB to E, making BE equal to BC.
On AE draw a semi-circle; and produce CB to meet the
circumference at F.
Then BF is a side of the required square.
Proof. Let X be the mid-point of AE, and r the radius of
the semi-circle. Join XF.
Then the rect. AC=AB.BE
= (7 + XB) (7 — XB)
= 7? — XB?
= FB, from the rt. angled A FBX.

CoroLuaRy. To describe a square equal in area to any given

rectilineal figure.
Reduce the given figure to a triangle of equal area. Prob. 18.
Draw a rectangle equivalent to this triangle. Prob. 17.
Apply to the rectangle the construction given above,
 PROBLEMS ON CIRCLES AND RECTANGLES. 239

EXERCISES.

1. Draw a rectangle 8 cm. by 2 cm., and construct a square of equal

area. What is the length of each side?
2. Find graphically the side of a square equal in area to a rectangle
whose length and breadth are 3:0’ and 1°5”. Test your work by
measurement and calculation.
3. Draw any rectangle whose area is 3°75 sq. in.; and construct a
square of equal area. Find by measurement and calculation the length
cf each side.
4. Draw an equilateral triangle on a side of 3’, and construct a
rectangle of equal area [Problem 17]. Hence find by construction and
measurement the side of an equal square.
5. Draw a quadrilateral ABCD from the following data: A=65° ;
AB=AD=9cm.; BC=CD=5 cm. Reduce this figure to a triangle
[Problem 18], and hence to a rectangle of equal area. Construct an
equal square, and measure the length of its side.
6. Divide AB, a line 9 em. in length, internally at X, so that
AX .XB=the square on a side of 4 cm.
Hence give a graphical solution, correct to the first decimal place, of
the simultaneous equations :
e+ y=9, 07 = 16,
2 " 3 : :
7. Taking 75 as your unit of length, solve the following equations
by a graphical construction, correct to one decimal figure :
x+y=40, xcy=169:
8. The area of a rectangle is 25 sq. em., and the length of one side
is 7°2 cm.; find graphically the length of the other side to the nearest
millimetre, and test your drawing by calculation.
9. Divide AB, a line 8 cm. in length, externally at X, so that
AX.XB=the square on a side of 6cm. [See p. 245.]
Hence find a graphical solution, correct to the first decimal place, of
the equations : a—-y=8, ay =36.

10. Ona straight line AB draw a semi-circie, and from any point P
on the circumference draw PX perpendicular to AB. Join AP, PB, and
denote these lines by w and y.
Noticing that (i) 2?+7?=AB?; (ii) xy=2AAPB=AB.PX; devise a
graphical solution of the equations :
ey =100 3 my =2d.
240 GEOMETRY.

PROBLEM 33.
To divide a given straight line so that the rectangle contained by
the whole and one part may be equal to the square on the other part.

= >
2
C
Let AB be the st. line to be divided at a point X in such a
way that AB. BX = AX?.
Construction. Draw BC perp. to AB, and make BC equal to
half AB. Join AC.
From CA cut off CD equal to CB.
From AB cut off AX equal to AD.
Then AB is divided as required at X.
Proof. Let AB=a units of length, and let AX =a. *
Then BX=a—2; AD=2; BO=CD=5.
Now AB*= AC? — BC’, from the rt. angled A ABC ;
= (AC — BC) (AC + BC) ;
that is, a@=2(%+¢a)
= 27 + an.
From each of these equals take az ;
then a —ar=2?;
or a(a—«)=2
that is, fe; be— AX?, pee
x

EXERCISE. A|__x_X\a-x B
Let AB be divided as above at X. On AB, AX,
and on opposite sides of AB, draw the squares
ABEF, adn ; and produce GX to meet FE at K. @ a
In this diagram name rectangular figures equivalent
to a*, 2, a(a+a), av, and a(a-2). Hence illus-
trate the above proof graphically.
 MEDIAL SECTION. 24}

Nore. A straight line is said to be divided in Medial Section when

the rectangle contained by the given line and one segment is equal to
the square on the other segment.
This division may be internal or external; that is to say, AB may be
divided internally at X, and externally at X’, so that

(i) AB. BX =AX?,

(ii) AB. BX’=AX”.

Te cbtain X’, the construction of p. 240 must be modified thus :

CD is to be cut off from AC produced;
PORE ind csc aee aes from BA produced, in the negative sense.

ALGEBRAICAL ILLUSTRATION.

Tf a st. line AB is divided at X, internally or externally, so that

AB. BX=AX?,
and if AB=a, AX=x, and consequently BK=a-.a, then
a(a—x)=27,
or, x?+axn—-a?=0,
/5 fan!5
and the roots of this quadratic, namely, = -5 and — ean) are
the lengths of AX and AX’.

EXERCISES.

1. Divide a straight line 4” long internally in medial section.

Measure the greater segment, and find its length algebraically.
2. Divide AB, a line 2” long, externally in medial section at X’.
Measure AX’, and obtain its length algebraically, explaining the
geometrical meaning of the negative sign.
ans
3. In the figure of Problem 33, shew that LCs . [Theor. 29.]
i [5 Ge.
ONDE yx ,_ [and[rR +5):
a
Hence prove (i) SS ar 53 (il) AX'= (SS

4, If a straight line is divided internally in medial section, and

from the greater segment a part is taken equal to the less, shew that
the greater segment is also divided in medial section.
H.S.G. Q
'242 GEOMETRY.

PROBLEM 34. -
To draw an isosceles triangle having each of the angles at the base
double of the vertical angle.

Xx

B
Construction. Take any line AB, and divide it at X,
so that AB. BX = AX?. Prob. 33.
(This construction is shewn separately on the left.)
With centre A, and radius AB, draw the ©BCD ;
and in it place the chord BC equal to AX.
Join AC.
Then ABC is the triangle required.
Proof. Join XC, and suppose a circle drawn through A, X
and C.
Now, by construction, BA. BX = AX?
= BC?;
... BC touches the OAXC at C; Theor, 59.
.". the 2 BCX =the z XAG, in the alt. segment.
To each add the 2 XCA;
then the 2 BCA=the 2 XAC+ the 2 XCA
=the ext. 2 CXB.
And the 2 BCA=the 2 CBA, for AB =AC.
.. the CBX =the .CXB;
, OX=CB=AX,
.. the 2XAC=the 2 XCA;
.. the 2XAC+the 2 XCA=twice the LA.
But the 2ABC =the LACB=the 2 XAC +the 2XCA Proved.
= twice the LA.
 MEDIAL SECTION 243

EXERCISES.

1. How many degrees are there in the vertical angle of an isosceles

triangle in which each angle at the base is double of the vertical angle?

2. Shew how a right angle may be divided into five equal parts by
means of Problem 34.

3. In the figure of Problem 34 point out a triangle whose vertical

angle is three times either angle at the base.
Shew how such a triangle may be constructed.

_ 4. If in the triangle ABC, the _B=the LC=twice the LA, shew

that BC_ Rie L

$ ABS 2

5. In the figure of Problem 34, if the two circles intersect at F,

shew that
(i) BC=CF;
(ii) the cirele AXC=the circum-circle of the triangle ABC;
(iii) BC, CF are sides of a regular decagon inscribed in the
circle BCD ;
(iv) AX, XC, CF are sides of a regular pentagon inscribed in
the circle AXC.

6. In the figure of- Problem 34, shew that the centre of the circle
sircumscribed about the triangle CBX is the middle point of the
are XC.

7. In the figure of Problem 34, if | is the in-centre of the triangle

ABC, and |’, S’ the in-centre and circum-centre of the triangle CBX,
shew that S‘1=S'I’. .

8. lia straight line is divided in medial section, the rectangle con-

tained by the sum and difference of the segments is equal to the
rectangle contained by the segments.

9. Ifa straight line AB is divided internally in medial section at X,

shew that AB?+ BX?2=3AX2.
- Also verify this result by substituting the values given on page 241.
244 GEOMETRY.

THE GRAPHICAL SOLUTION OF QUADRATIC EQUATIONS.

From the following constructions, which depend on Problem 32, a

graphical solution of easy quadratic equations may be obtained.

I. To divide a straight line internally so that the rectangle contained

by the segments may be equal to a given square.

C. C_F

Rix PS xX B D E
Let AB be the st. line to be divided, and DE a side of the given
square.

Construction. On AB draw a semicircle; and from B draw BF

perp. to AB and equal to DE.
From F draw FCC’ par' to AB, cutting the Or at C and C’.
From C, C’ draw CX, C’X’ perp. to AB.
Then AB is divided as required at X, and also at X’.

Proof. / AX.XB=CX? Prob. 32.

= 5
=DE*,
Similarly AX’. X’B=DE?.

Application. The purpose of this construction is to find two straight

lines AX, XB, having given their swm, viz. AB, and their product, viz.
the square on DE,
Now to solve the equation 2?-13x+4+36=0, we have to find two
numbers whose swum is 13, and whose product is 36, or 6?
To do this graphically, perform the above construction, making AB
equal to 13 em., and DE equal to ¥36 or 6 em. The segments AX, XB
represent the roots of the equation, and their values may be obtained
by measurement.

Nors. If the last term of the equation is not a perfect square, as

in 2®-7x+11=0, V/11 must be first got by the arithmetical rule, or
graphically by means of Problem 32.
 GRAPHICAL SOLUTION OF QUADRATICS. 245

Il. To divide a straight line externally so that the rectangle contained

by the segments may be equal to a given square.

xX’ A O B X D E
Let AB be the st. line to be divided externally, and DE the side of
the given square. :

Construction. From B draw BF perp. to AB, and equal to DE.

Bisect AB at O.
With centre O, and radius OF draw a semi circle to cut AB pro-
duced at X and X’.
Then AB is divided externally as required at X, and also at X’.
Proof. AX.XB=X’B. BX, since AX=X’B,
=BF? Prob. 32.
= DE.

Application. Here we find two lines AX, XB, having given their
difference, viz. AB, and their product, viz. the square on DE.
Now to solve the equation x?-6x-16=0, we have to find two
numbers whose numerical difference is 6, and whose product is 16,
wr 4?
To do this graphically, perform the above construction, making AB
equal to 6 cm., and DE equal to 16 or 4cm. The segments AX, XB
represent the roots of the equation, and their values, as before, may be
obtained by measurement.

EXERCISES.

Obtain approximately the roots of the following quadratics by means

of graphical constructions ; and test your results algebraically.
1. 2?-10x+16=0. 2. «?-14a+49=0. 3. 2 -127+25=0.
4, w—52-36=0. . 5. x®-7x-49=0. 6. 2?-102+20=0.
246 GEOMETRY.

EXERCISES FOR SQUARED PAPER.

1. A circle passing through the points (0, 4), (0, 9) touches the
a-axis at P. Calculate and measure the length of OP.
2. With centre at the point (9, 6) a circle is drawn to touch the
y-axis. Find the rectangle of the segments of any chord through O.
Also find the rectangle of the segments of any chord through the
point (9, 12).
3. Draw a circle (shewing all lines of construction) through the
points (6, 0), (24, 0), (0, 9). Find the length of the other intercept on
the y-axis, and verify by measurement. Also find the length of a
tangent to the circle from the origin.
4. Draw a circle through the points. (10, 0), (0, 5), (0, 20); and
prove by Theorem 59 that it touches the a-axis.
Find (i) the coordinates of the centre, (ii) the length of the radius.
5. Ifa circle passes through the points (16, 0), (18, 0), (0, 12), shew
by Theorem 58 that it also passes through (0, 24).
Find (i) the coordinates of the centre, (ii) the length of the tangent
from the origin.
6. Plot the points A, B, C, D from the coordinates (12, 0), (-6, 0),
(0, 9), (0, —8); and prove by Theorem 57 that they are concyclic.
If r denotes the radius of the cirele, shew that
OA? + OB? + OC? + OD? =4r".
7. Draw a circle (shewing all lines of construction) to touch the
y-axis at the point (0, 9), and to eut the a-axis at (3, 0).
Prove that the circle must cut the x-axis again at the point (27, 0);
and find its radius. Verify your results by measurement.
8. Shew that two circles of radius 13 may be drawn through the
oint (0, 8) to touch the a-axis ;and by means of Theorem 58 find the
een of their common chord.
9. Given a cirele of radius 15, the centre being at the origin, shew
how to draw a second circle of the same radius touching the given circle
and also touching the 2-axis.
How many circles can be so drawn? Measure the coordinates of the
centre of that in the first quadrant.
10. A, B, C, D are four points on the a-axis at distances 6, 9, 15, 25
from the origin O. Draw two intersecting circles, one through A, B,
and the other through C, D, and hence determine a point P in the
x-axis such that
PA.PB=PC. PD.
Caleulate and measure OP.
If the distances of A, B, C, D from O are a, b, ¢, d respectively,
prove that “
OP =(ab —cd)/(a+b-c-d).
 PART V.
ON PROPORTION.
DEFINITIONS AND FIRST PRINCIPLES.

1. The ratio of one magnitude to another of the same kind

is the relation which the first bears to the second in regard to
quantity ; this is measured by the fraction which the first is
of the second.
Thus if two such magnitudes contain a and 0 units respectively, the
ratio of the first to the second is expressed by the fraction z

The ratio of a to d is generally denoted thus, a: 0; and a is

called the antecedent and 0 the consequent of the ratio.
The two magnitudes compared in a ratio must be of the same kind;
ior example, both must be lines, or both angles, or both areas. It is
slearly impossible to compare the /ength of a straight line with a
- magnitude of a different kind, such as the area of a triangle. Moreover,
@ratic is an abstract fraction. Thus the ratio which a line 6 em. long
bears to a line 8 em. long is § or 3, (not # em.),
Notes. It is not always possible to express two quantities of the same
xind in terms of a common unit. For instance, if the side of a square
is 1 inch, the diagonal is V2 inches. But since the numerical valus
of V2 cannot be exactly determined (though it can be found to any
uumber of decimal figures), the side and diagonal cannot be expressed
im terms of the same unit. Two such quantities are said to be incom-
mensurahle. But by choosing a sufficiently small quantity as unit, two
incommensurables, such as V2 inches and 1 inch, may be expressed to any
required degree of accuracy. Thus, remembering that J/2 =1:41421...,
it follows that V2 inches and 1 inch may be represented by
1414 and 1000, roughly, taking z)57” as unit,
14142 and 10000, more nearly, taking zp4;55” as unit;
and go on.

2. If a point X is taken in a
given line AB, or in AB produced, §*=———__—+——=
the ratio in which it divides AB is Bigei.
the ratio of the segments of AB,
namely AX:XB, whether the divi- a RX
sion is interna’ as in Fig. 1, or
external as in Fig. 2.
248 GEOMETRY.

3, Four magnitudes are in proportion, when the ratio of the

Jirst to the second is equal to the ratio of the third to the fourth.
When the ratio a to 6 is equal to that of z to y, the four
magnitudes are called proportionals.
This is expressed by saying “a is to b as x is to y”: and the
proportion is written
pe
b ¥
or a:b=n7y.
Here a and y are called the extremes, and 6 and z th
means ; and y is said to be a fourth proportionai to a, }, and x
In a proportion, terms which are both antecedents or both
consequents of the ratios are said to be corresponding terms.
Notr. Ina proportion such as a:b=a:y, the magnitudes compared
in each ratio must be of the same kind, though the magnitudes of ths
second ratio need not be of the same kind as those of the first. Foz
instance, a and b may denote areas, and w and y dines; in which case
the proportion asserts that the first area bears the same ratio to the
second area, as the first line bears to the second line.

4, Three magnitudes of the same kind are said to be pro

portionals, when the ratio of the first to the second is equal toe
that of the second to the third.
Thus a, 6, ¢ are proportiona!s if

er a:b=b:e,

Here 3 is called a mean proportional between a and ¢

and ¢c is called a third proportional to a and 2.

AXIOMS.

(i) Ratios which are equal to the same ratio are equal to one
another.
For instance, if a:b=a:y, and c:d=a:y,
then evidently a:b=c:&
 INTRODUCTORY THEOREMS, 249

iii) Magnitudes which bear the same ratio to the same magnitude
gre equal to one another. ;
¥or instance, if Che 00,3201
then evidently a=b,

INTRODUCTORY THEOREMS.

I. If four magnitudes are proportionals, they are also pro

vortionals when taken inversely.
That 1s, if OiLijemrtnnTD
then M0) Pe
: b
For, by hypothesis, ame hence -=%;
or Ot

ll. Lf four magnitudes of the sume kind are proportionais, they

wre also proportionals when taken alternately.
That is, if Dy,
then GLO
For, by hypothesis, =] :

multiplying both sides by -

ab «2 6b
we have Seppe gpg
Coys

that is, C0)

et)
or CeO si
Note. In this theorem the hypothesis requires that # and b shall be
of the same kind, also that x and y shall be of the same kind; while the
conclusion requires that a and = shall be of the same kind, and also é
snd y of the same kind.
250 GEOMETRY.

TM. Jf jour nwmbers are proportional. the product of the

extremes is eyual to the product of the means.
That is, if a:b=6:4,
then ad = be.
For, by hypothesis, += 53

multiplying each side of this equation by bd, we have

ad = be.

Coroiiary. If a, 2, c, d denote the lengths of four straight

ines in proportion, the above result states that the rectangle
contained by the extremes is equal to the rectangle contained by ths
mOCAns.
This is illustrated by the following diagram :

a—-—-

—
Cc a
c| (bc)

- a }

Similarly if three lines a, b, ¢ are proportionals,

that is, if a:b=b:c;
then . ac = b®,
Or, the rectangle contained by the extremes is equal to the square
wa the mean,

iV. If there are four magnitudes in proportion, the sum

(or difference) of the first and second is to the second as the
sum (or difference) of the third and fourth is to the fourth.
That is, if a:b=a2:y;
then (i) a+b:b=at+y:y;
(i) a—b:b=a-yry.
 INTRODUCTORY THEOREMS. 251]

For by hypothesis, ore

BRT : i ceeee
MB or a+b
a
ae aty,
ae:

that is, BEI URS Qin actif SCY sn Pitse demraaaedeohy aatoet (i)
This inference is sometimes referred to as componendo.

Similarly by subtracting 1 from the equal ratios %, 7, we

obtain by
Ga) ted.
ET
that is, = PVEYs Ye, poate aan en ee ereaet (ii)
This inference is sometimes referred to as dividéndo.

CoROLLARY. If @:b6=2:4,
then a+b:a-b=a+y:20-Y.
This is obtained by dividing the result of (i) by that of (ii),

V. Ina series of equal ratios (the magnitudes being ali of the

same kind), as any antecedent is to its consequent so is the sum of
the antecedents to the sum of the consequents.

That is, if Ud
BU
ore.
oe
ee a at+b+er+
eoUbyte+
Let each of the equal ratios * ; > ... be equal to &

Then b=ty OS hic Che, enc

°., by addition,
at+b+c+...=k(etyt+et+...)3
A+b+6+...— _4
“EEYRE +... Ok
or @22=A+b+e+...:24+Y4+2+...,
’
252 GEOMETRY.

VI. A given straight line can be divided internally in a given

ratio at one, and only one, point ; and externally at one, and only
one, pornt.

<---~--nz AED ----> ~<------- Mm

— 1 ------- >< - 2 -->(P)
| M=—— >yee-= => B A«- ----------- Mfa Ba >X

Fig:r. Fig.2.

Let AB be the given line, and m:n the given ratio, m being
greater than 7.

Internal Division. (i) Divide AB (Fig. 1) into m+n equal

parts [Prob. 7]; and of these parts make AX to contain m;
then XB must contain n.
Hence - AX: XB=m:n;
that is, AB is divided internally at X in the given ratio.
(ii) Again, since AX and AB contain respectively m and
m+n equal parts,
“. AX: AB=m:m+n.,
Similarly, if P divides AB in the given ratio m: ”,
AP: AB=m:m+n,
-, Ae te,
“AB AB’
AX
= AP.
Hence P and X coincide ; that is, X is the only point which
divides AB internally in the ratio m:n.

External Division. (i) Divide AB (Fig. 2) into m—n equal

parts ;and in AB produced make AX to contain m such parts ;
then XB must contain n.
Hence AX: XB=m:n;
that is AB is divided externally at X in the given ratio.
(ii) And it may be shewn, as above, that X is the only point
which divides AB externally in the ratio m:n.
:
 EXBROISES ON RATIO AND PROPORTION. 453

EXERCISES.

1. Insert the missing terms in the following proportions :

(i) 3:7=15:( )5
(ii) 2°5:( )=10:32;
(iii) ( ):ac?=be : bc%,
2, Oorrect the following statement :
£65 : 78 ft. = £25: 30 ft.

& If a straight line, 9°6” in length, is divided internally in the retic

5:7, oslculate the lengths of the segments.
4 If a straight line 4°5 cm. in Jength is divided externally in the
ratio 11:8, calculate the lengths of the segments.
5. AB is a straight line, 6°4 cm. in length, divided tniernally at &
and externoliy at ¥ in the ratio 5:3; calculate the lengths of the
segments, and shew that they satisfy the formula,
Zee a 1
AB AX" BY
€. If straight line, @ inches in length, is divided internaily in the
ratio m:n, shew that the lengths of the segments are respectively
7
.@ inches, _”_ a inches.
M+ m+n

7. If a straight line, a units in length, is divided externaily in the

ratio m:n, shew that the lengths of the segments are respectively
m : e
.@ units, .@ units.
m—-n m—2

8. If a:b=x:y, and 6:c=y:z, prove that a:c=x:2,

9, If a:b=x:y, shew that a+b:a=a+y:2.
10. If a, 5, © are three proportionals, shew that a:c=a?; 8%,
1). If two straight lines AB, CD are divided internally in the sams
ratio at X and Y respectively, shew that
(i) AB:XB=CD: YD;
(ii) AB: AX=CD: CY.

12, Ifa, 5, c, d are four straight lines such that the rectangle oon-
¢sined by a and d is equal to that contained by 6 and c, prove
a:b=c:d.
254 GEOMETRY.

PROPORTIONAL DIVISION OF STRAIGHT LINES.

THEOREM 60. [Euclid VI. 2.1
A straight line drawn parallel to one side of a triangle cuts thé
other two sides, or those sides produced, proportionally.
N¢

C
vé

A xB
Fig. 1. Pig. 2.

In the AABC, let XY, drawn par' to the side BO, cut AB, AQ
a¢ X and Y, internally in Fig. 1, externally in Fig. 2.
It is requered to prove in both cases that
AX :XB=AY: YC.

Proof. Suppose X divides AB in the ratio m:n; that is,

sf sini AX: XB=m:n;
so that, if AX is divided into m equal parts, then XB may be
divided into n such equal parts.
Through the points of division in AX, XB let parallels be
drawn to BC.
Then these parallels divide the segments AY, YC into parts
which are all equal ; Theor. 22,
and of these equal parts AY contains m,
and YC contains n;
hence AY: YC=m: n
v. AX: XB=AY:
YQ
VED,
 PROPORTIONAL DIVISION OF STRAIGHT LINES. 236

Conversely, if a line cuts two sides of a triangle proportionally,

# 1s parallel tu the inird side,

Bier. Fig.2.

Conversely, let XY cut the sides AB, AC proportionally, so that

AX: XB = AY + YC.
It is required to prove that XY is parallel to BC.

Let XP be drawn through X par' to BC, to meet AC in F,

Then AP: PC=AX: XB;
but, by hypothesis, AY: YC=AX: XB.
Thus AC is cut, internally in Fig. 1, and externally in Fig. 2
in the same ratio at P and Y.
Hence P coincides with Y, and consequentiy XP with XY.
Theor. Vi. p. 252
That is, XY is par’ to BC,
Q.E.D.

CoRCLLARY. If XY is parallel to BC, then

AX : AB=AY
: AC.
For, taking Fig. 1, it may be shewn that
AX: AB=™m:Mm+N;
and henve, by Theorem 22, that
AY :AC=™:m-i2.
*, AX: AB=AY: AG,
Conversely, if AX: AB=AY: AG
it may be proved as above that XY 1s par' to BC
256 GEOMETRY.

‘THEOREM 61. [Euclid VI. 3 and A.}

if the vertical angle of a triangle is bisected internally or exter
nally, the bisector divides the base internally or externally tnio
segments which have the same ratio as the other sides of the triangle.
Conwersely, if the base is divided internally or externally ints
segments proportional to the other sides of the triangle, the line joining
the point of section to the vertex bisects the vertical angle internally
or externally.
E R’

B’

it Paces
B X C B C x
Fig.1. Fig.2.

In the A ABC, let AX bisect the 4 BAC, internally in Fig. 1,

ard externally in Fig. 2; that is, in the latter case, let Ax
bisect the exterior £ B’AC.
It is required to prove in both cases that
BX : XC=BA: AC.
Let CE be drawn through C par' to XA to meet BA (prodaced,
if necessary) at E. In Fig. 1 let a point B’ be taken in AB
Proof. Because XA and CE are par’,
.., in both Figs., the 2 BAX =the int. opp. 2 AEG
Also, by hypothesis,
the 2 BAX = the 2 XAC
= the alt. 2 ACE,
. the AEC =the LACE:
.. AC= AE,
Again, because XA is par’ to CE, a side of the 4 BCE,
’., in both Figs., BX : XC = BA: AE;
that is, BX : XC=BA: AC.
Q E.D.
 INTERNAL AND EXTERNAL DIVISION. 257

Conversely, let BC be divided internally (Fig. 1) or exter.

nally (Fig. 2) at X, so that BX : XC=BA: AC.
It is required to prove that the L BAX=the ZL XAC.

Proof. For, with the same construction as before,

because XA is par' to CE, a side of the A BCE,
| ByexC
= BA SAE,

But, by hypothesis, BX :XC=BA:AC;

*, BA: AC=BA:
AE;
“. AC=AE.
. the L AEC=the LACE
= the alt. 2 XAC.
And in both Figs.,
the ext. L B‘AX=the int. opp. LAEC;
.. the 2 BAX =the Z XAC.
Q.E.D.

DEFINITION.

When a finite straight line is divided internally and exter-

nally into segments which have the same ratio, it is said to be
cut harmonically.
Hence the following Corollary to Theorem 61.
The base of a triangle is divided harmonically by the internal and
eaternal bisectors of the vertical angle :
for in each case the segments of the base are in the ratio of
the other sides of the triangle.

[For Theorems and Examples on Harmonic Section see p. 323.]

8.3.4. s R
258 GEOMETRY.

EXERCISES ON THEOREM 60.

(Numerical and Graphical.)
1. Ona base AB, 3°5” in length, draw any triangle CAB ; and from
re age AX 2:1” long. Through X draw XY parallel to BC to meet
at Y.
Measure AY, YC; and hence compare the ratios

gos of
XB’ YC’ AX’
eG
AY’
(iii) AB AC
XB’ YC
2. ABC isa triangle, and XY is drawn paralle: to BC, cutting the
other sides at X and Y.
(i) lf AB=3°6”, AC=2-4”, and AX=2'1”, calculate the length
ot AY.
gi) If AB=2:0"”, AC=1°5", and AY=0°9”, calculate the length
of
(iii) If X divides AB in the ratio 8:3, and if AC=8°'8 em., find
Y, YG;
3. ABC is a triangle, and XY is drawn parallel to BC, cutting the
other sides produced at X and Y.
(i) If AB=4°5 cm., AC=3'5cm., and AX=7'2 cm., find by cal-
culation and measurement the length of AY.
(ii) If X divides AB externally in the ratio 11; 4, and if AC=4°9 cm.,
tind the segments of AC.

( Theoretical. )
4. Three parallel straight lines cut any two transversals proportion-
ally.
5. The straight line which joins the middle points of the oblique
sides of a trapezium is parallel to the parallel sides.
6. Two triangles ABC, DBC stand on the same side of the common
base BC ; and from any point E in BC lines are drawn parallel to BA,
BD, meeting AC, DC in F and G. Shew that FG is parallel to AD.
7. Ina triangle ABC a transversal is drawn to cut the sides BC, CA,
AB (produced if necessary) at D, E, and F respectively, and it makes
eel angles with AB and AC; prove that
BD: CD=BF : CE,
 PROPORTIONAL DIVISION OF STRAIGHT LINES. 259

EXERCISES ON THEOREM 61.

(Numerical and Graphical.)
1, Draw a triangle ABC, making a=1°'5", b=2°4", and c=3°6".
Bisect the angle A, internally and externally, by lines whicn meet BC
and BC produced at X and Y.
Measure BX, XC; BY, YC; hence evaluate and compare the ratios

BX BY BA
XC’ YC’ AC
2. In the triangle ABC, a=3°5 cm., }=5°4 cm., c=7'2 cm.; and
the internal and external bisectors of the 2A meet BC at X and Y.
Calculate the lengths of the segments into which the base is divided
at X and Y respectively ; and verify your results graphically.
3. Frame constructions, based upon Theorem 6],
(i) to trisect a straight line of given length ;
{ii) to divide a given line internally and externally in the ratio 3: 3.

( Theoretical.)
4. AD isa median of the triangle ABC; and the angles ADB, ADC
are bisected by lines which meet AB, AC at Eand F respectively. Shew
that EF is parallel to BC.
5. ABCD is a quadrilateral: shew that if the bisectors of the angles
A and C meet in the diagonal BD, the bisectors of the angles B and D
will meet on AC.
6. Employ Theorem 61 to shew that in any triangle
(i) the internal bisectors of the three angles are concurrent ;
(ii) the external bisectors of two angles and the internal bisector of
the third angle are concurrent.
7. If 1 is the in-centre of the triangle ABC, and if Al is produced to
meet BC at X, shew that
Al: IX=AB+AC: BC.
8. Given the base of a triangle and the ratio of the otaer sides, find the
locus of the vertex.
9. Construct a triangle, having given the base, the ratio of the
ether sides, and the vertical angle.
260 GEOMETRY

EQUIANGULAR TRIANGLES.

THEOREM 62. [Euclid VI. 4.]

If two triangles are equiangular to one another, their correspond
ing sides are proportional.
A

Pi
G D

B H C EB P

Let the A* ABC, DEF have the 2* A and B respectively equal

to the Z* D and E; and consequently the 2C equal to the 2 F.
It 1s required to prove that
AB: DE=BC: EF=CA: FD.

Proof. Apply the A DEF to the AABC, so that E falls om

8, and EF along BC;
then since the 2 E=the 2B, ED will fall along BA.
Let G and H be the points at which D and F fall respectively ;
so that GBH represents the A DEF in its new position.
Now, by hypothesis, the 2 D=the 2A;
that is, the ext. 2 BGH=the int. opp. 2 BAC ;
.. GH is par' to AC.
Hence BA: BG=BC: BH; Theor. 60, Cor.
that is, AB : DE=BC;EF. ,
Similarly, by applying the ADEF to the AABC, so that F
falls on C, and FE, FD along CB, CA, it may be shewn that
BC: EF=CA: FD.
Hence finally, AB: DE=BC: EF=CA: FD. Q.E.D,
 EQUIANGULAR TRIANGLES, 261

THEOREM 63. (Euclid VI. 5.]

Tf two triangles have their sides proportional when taken tn order,
she triangles are equiangular to one another, and those angles are
agual which are opposite to corresponding sides.
re D

B C c
{n the A’ ABC, DEF, let
AB: DE=BC: EF=CA: FD,
It is required to prove that the A* ABC, DEF are equiangular ta
one another.
At E in FE make the 2 FEG equal to the 2B;
and at F in EF make the 2 EFG equal to the 2C.
.. the remaining 2 EGF =the remaining 2A.
Proof. Since the A*ABC, GEF are equiangular to one
another,
G (EMCIE S1sG 3 [E|= Theor. 62.
But, by hypothesis, AB: DE=BC:EF;
.. AB: GE=AB: DE.
a. GE—DE.
Similarly GF = DF.
Then in the A* GEF, DEF,
GE= DE,
because GF = DF,
and EF is common ;
. the triangles are identically equal; Theor. 7
*. the . DEF=the 2 GEF
=the 2B;
and the 2 DFE=the 2 GFE
=the LC.
.". the remaining 4 D=the remaining LA;
that is, the A DEF is equiangular to the AABC. Q.E.D.
363 GEOMETRY,

EXERCISES ON EQUIANGULAR TRIANGLES.

(Numerical and Graphical. The results are to be obtained by calculation

and checked graphically.) ;

1. Ina triangle ABC, XY is drawn parallel to BC, cutting the other

sides at X and Y:
(i) If AB=2°5", AC=2°0’, AX=1°5"; find AY.
(ii) If AB=3°5", AG=27, AY=12"5 find AX,
(iii) If AB=4°2 cm., AX=3‘6cm., AY=6'6 cm.; find AC,
2. In the figure of the last example :
(i) If AB=2°4’, BC=3°6", AX=1°4"; find XY.
(ii) If BC=7'7 cm., XY=5'5 cm., AX=4‘5 cm.; find AB.

3. In the triangle ABC, a=3:0", b=3°6’, c=4:2”; and QR, drawn

parallel to AC, measures 3°0’. Find the remaining sides of the
triangle QBR.

4. ABC is a triangle in which a=8 cm., b=7 cm., and c=10 cm,
in AB a point P is taken 4 cm. from A, and PQ is drawn parallel te
BC. Find the lengths of PQ and QC.

5. The sides of a triangular field are 400 yards, 350 yards, and
300 yards respectively. In a plan of the field the greatest side
measures 2°4”; find the lengths of the other sides.

6. XY is drawn parallel to BC, the base of the triangle ABC. If

AX=83 ft., XY=34 ft., AY=6 ft. 2in., and XB=4} ft.; calculate the
sides of the triangle ABC.

7. The triangle ABC is right-angled at C; and from P, a point in

the hypotenuse, PQ is drawn parallel to AC.
If AC=1}”, BC=3”, and PQ=4”; find BQ, BP, and AP.

8. In a triangle ABC, AD is the perpendicular from A on BC; and

through X, a point in AD, a parallel is arawn to BC, meeting the other
sides in P, Q.
If BC=9 om., AD=8em., DX=30m.; find PQ.

9. In the triangle ABC, a=2‘0cm.,b=3'50m.,c=4*5cem. BD and

CE are drawn from the ends of the base to the opposite sides, and they
intersect in P.
If EP: PC=DP: PB=2:65,
find the lengths of ED, AD, and DC.
 EQUIANGULAR TRIANGLES. 263

EXERCISES ON EQUIANGULAR TRIANGLES.

( Theoretical. )
1. Shew that the straight line which joins the middle points of two
‘sides of a triangle is
(i) parallel to the third side; (ii) one-half the third side.
2. In the trapezium ABCD, AB is parallel to DC, and the diagonala
intersect at O: shew that
OA : OC=OB: OD,
If AB=2DC, shew that O is a point of trisection on both diagonals.
3. If three concurrent straight lines are cut by two parallel trans-
versals in A, B, C, and P, Q, R respectively ; prove that
AB: BC=PQ:
QR.
4. ABCD isa parallelogram, and from D a straight line is drawn to
cut AB at E, and CB produced at F. In this figure name three triangles
which are equiangular to one another ; and shew that
DAL AEB sBE=-C2eD.
5. In the side AC of a triangle ABC any point D is taken: shew
that if AD, DC, AB, BC are bisected in E, F, G, H respectively, then
EG is equal to HF.
6. AB and CD are two parallel straight lines; E is the middle point
of CD; AC and BE meet at F, and AE and BD meet at G: shew that
FG is parallel to AB.
7. AB isa diameter of a circle, and through A any straight line is
drawn to cut the circumference in Cand the tangent at B in D: shew that
(i) the As CAB, BAD are equiangular to one another;
(ii) AC,.AB, AD are three proportionals;
(iii) the rect. AC, AD is constant for all positions of AD.
8. If through any point X within a circle two chords AB, CD are
drawn, and AC, BD joined ; shew that
(i) the As AXC, DXB are equiangular to one another;
(ii) AX: DXK=XC: XB.
Hence obtain an alternative proof of Theorem 57.
9. If from an external point X a tangent XT and a secant XAB are
firawn to a circle, and AT, TB joined; shew that
(i) the As AXT, TXB are equiangular to one another ;
(ii) XA: XT=XT : XB.
Hence obtain an alternative proof of Theorem 58.
 i
264 GEOMETRY.

DEFINITIONS.
1. Two rectilineal figures are said to be equiangular to
one another when the angles of the first, taken in order, are
equal respectively to those of the second, taken in order.
2. Rectilineal figures are said to be similar when they are
equiangular to one another, and also have their corresponding
sides proportional.
Thus the two quadrilaterals ABCD,
EFGH are similar if the angles at
A, B, C, D are respectively equal to F
those at E, F, G, H, and if the follow- A
ing proportions hold : = \
AB BC CD_DA
EF =FA-an
FG GH HE: D Cc H G

3. Similar figures are said to be similarly described with

regard to two sides, when these sides correspond.

NOTE ON SIMILAR FIGURES.

Similar figures may be described as those which have the same shape.
For this, ¢wo conditions are necessary :
(i) the figures must have their angles equal each to each, taken in order ;
(ii) their corresponding sides must be proportional.
In the case of triangles we have learned that these conditions are not
independent, for each follows from the other: thus
(i) if the triangles are equiangular to one another, Theorem 62
proves that their corresponding sides are proportional ;
(ii) if the triangles have their sides proportional, Theorem 63 proves
that they are equiangular to one another.

This, however, is not necessarily the case with

rectilineal figures of more than three sides, For
example, the first diagram in the margin shews
twe fgnres which are equiangular to one another,
but which clearly have not their sides propor-
tional; whiie the figures in the second diagram
have their sides proportional, but are not equi-
angular to one another.
 SIMILAR TRIANGLES. 265

THEOREM 64. [Euclid VI. 6.]

If two triangles have one angle of the one equal to one angle of
the other, and the sides about the equal angles proportionals, the
triangles are similar.
A

B Cc E

In the A° ABC, DEF, let the ~A=the ZD,

and let AB : DE=AC: DF.

It is required to prove that the A* ABC, DEF are similar.

Proof. Apply the A DEF to the A ABC, so that D falls on &

and DE along AB;
then because the 2 EDF =the 4 BAC, DF must fall along AC.
Let G and H be the points at which E and F fall respectively ;
go that AGH represents the A DEF in its new position.
Now, by hypothesis, AB: DE=AC: DF;
that is, AB: AG=AC:AH;
hence GH is par' to BC. Theor. 60, Cor
.. the ext. AGH, namely the LE, =the int. opp. 2 ABC;
and the ext. LAHG, namely the 2 F, =the int. opp. 2 ACB.
Hence the A* ABC, DEF are equiangular to one another,
so that their corresponding sides are proportional ; Theor. 62
that 13, the A* ABC, DEF are similar.
"4 Q.E.D.
366 GEOMETRY.

*THEOREM 65. (Euclid VI. 7.]

If two triangles have one angle of the one equal to one angle of
the other, and the sides about another angle in one proportional to the
corresponding sides of the other, then the third angles are either
equal or supplementary ; and in the former case the triangles are
sumilar.
A

2H at
B Cc E FE E F’ FE
Fig.t. Fig.2. Fig. 3.
In the A* ABC, DEF, let the 2 B=the 2E; and let the sides
about the 2’ A and D be proportional,
namely AB : DE=AC: DF.
It is required to prove that
either the .C=the -F, [as in Figs. 1 and 2];
or the LC=the supplement of the LF. [Figs. 1 and 3.}
Proof. (i) If the .A=the 2D, [Figs. 1 and 2],
then the .C=the ZF; Theor. 16.
and the A’ are equiangular, and therefore similar.
(ii) But if the 2A is not equal to the LEDF [Figs. 1 and 3}
let the 2 EDF’ =the 2A.
Then the A* ABC, DEF’ are equiangular to one another;
.. AB: DE =AC: DF.
But by hypothesis, AB: DE =AC: DF;
.. AC: DF’ =AC: DF.
*, DF’=DF.
.. the 2 DFF’ =the 2 DF'F.
But the .C=the 2 DF'E Proved
=the supplement of the 2 DF'F
=the supplement of the 2 DFE.
“Q.E.D.
 SIMILAR TRIANGLES, 267

EXERCISES ON SIMILAR TRIANGLES.

(Theoretical. )
1. Ina triangle ABC, prove that any straight line parallel to the
base BC and intercepted by the other two sides is bisected by the
median drawn from the vertex A.
2. Two triangles ABC, A’B’C’ are equiangular to one another ;
if p, p’ denote the perpendiculars from A, A’ to the opp. sides,
Feo wens cock ee circum-radii;
(aah co Back See neal in-radii ;

prove that each of the. ratios

@, LS T is equal to the ratio of any pair
p r
of corresponding sides.

3. Prove that the radius of the circle which passes through the
mid-points of the sides of a triangle is half the circum-radius.
4. If two straight lines AB, CD intersect at X, so that
XA: XC=XD:
XB;
(i) shew by Theorem 64 that the A* AXD, CXB are similar;
(ii) hence prove the points A, D, B, C concyclic.

5. A, B, C are three collinear points, and from B and C two parallel}

lines BP, CQ are drawn in the same sense, so that
PB: QC=AB: AC;
shew by Theorem 64 that the points A, P, Q are collinear.

6. If intwo triangles ABC, A’B’C’, the LB=the LB’, and 5=2,

what conclusion may be drawn? ots b
Shew by diagrams how this conclusion is affected, if it is also given
that (i) c is less than b,
(ii) c is equal to b,
(iii) c is greater than b.

7. ABCD isa parallelogram ; P and Q are points in a straight lins

parallel to AB; PA and QB meet at R, and PD and QC meet at S-
shew that RS is parallel to AD.

8. Ina triangle ABC the bisector of the vertical angle A meets the
base at D and the circumference of the circum-circle at E, if EC is
joined, shew that the triangles BAD, EAC are similar; and hence
prove that
AB.AC=AE.AD.
268 GEOMETRY.

THEOREM 66. [Euclid VI. 8.1

In a right-angled triangle, if a perpendicular is drawn from the
right angle to the hypotenuse, the triangles on each side of it are
similar to the whole triangle and to one another.
A

B D C
Let BAC be a triangle right-angled at A, and let AD be
drawn perp. to BC.
It is required to prove that the A* BDA, ADC are similar to the
A BAC and to one another.
In the A* BDA, BAC,
the 2 BDA=the 2 BAC, being rt. angles,
and the 2 B is common to both ;
.. the remaining 2 BAD=the remaining LBCA; Theor. 16
hence the A BDA is equiangular to the A BAC ;
.. their corresponding sides are proportional ;
*, the A* BDA, BAC are similar.
In the same way it may be proved that the A*ADC, BAG
are similar.
Hence the A* BDA, ADC, having their angles severally equal
to those of the A BAC, are equiangular to ove another;
.. they are similar.
Q.E.D.
Coro.uary. (i) Because the A*DBA, DAC are similar,
. DB: DA=DA:DC;
that is, DA is a mean proportional between OB and DO;
and DA?=DB. DC.
(ii) Because the A* BCA, BAD are similar,
-. BC: BA=BA:
BD;
hence BA?=BC. BD.
(iii) Because the A*CBA, CAD are similar,
.. CB: CA=CA:CD:
hence CA?=CB. CD.
 MISCELLANEOUS EXAMPLES. 269

EXERCISES.

(Miscellaneous Examples on Theorems 62-66.)

1. ASC is an equilateral triangle of which each side=a. In

BC, produced both ways, two points P and Q are taken, such that
BP=CQ=a, and AP, AQ are joined. Shew that
(i) PQ: PA=PA: PB,
{ ii) PA? =3a?,
2. ABC is a triangle right-angled at A, and AD is drawn perpen-
dicular to BC: if AB, AC measure respectively 4” and 3’, shew that the
seginents of the hypotenuse are 3°2” and 1:8”.

3. ABC is a triangle right-angled at A, and a perpendicular AD is

drawn to the hypotenuse BC; shew (i) by Theorem 25, (ii) by Theorem
66, that
BC. AD=AB. AC.

4. ABC is a triangle right-angled at A, and AC’ is drawn perpen-

dicular to the hypotenuse, also C’A’ is drawn parallel to CA. If
AC=15cm., and ARB=20 cm., shew that AC’=12 cm., and C’A’=9°6 cm.

5. At the extremities of a diameter of a circle, whose centre is C

and radius 7, tangents are drawn: these are cut in Q and R by any
third tangent whose point of contact is P. Shew that
(i) QR subtends a right angle at C;
(i) PQ). PR=77.

6. Two circles of radii r and 7’ respectively have external contact

at A, and a common tangent touches them at Pand Q. Shew that
(i) PQ subtends a right angleat A; [Ex. 9. p. 187]
(ii) PQ2=4rr’,
[Produce PA, QA to meet the circumferences at X and Y, and prove
the triangles PAY, XAQ right-angled and similar.]

7. Two circles touch one another externally at A, and a common

tangent PQ is produced to meet the line of centres at S. Shew that,
if PA, AQ are joined,
(i) the triangles SAP, SQA are similar;
(ii) SA2=SP. SQ.

8. Two circles intersect at A and B; and at A tangents are drawn,

one to each circle, to meet the circumferences at C and D: shew that if
BC, BO joined,
hin BC : BA=BA: BD.
270 GEOMETRY

THE TRIGONOMETRICAL RATIOS.

Pp
1. Let PAQ be any acute angle ; in AP, one of B
the arms of the angle, take a point B, and draw
BC perp. to AQ.
Then with reference to the ZA in the right-
angled A BAC, the following definitions are used.
A Cc Q
The ratio ad or opposite side is called the sine of the ZA.
hypotenuse

The ratio AC or adjacent side ee: cosine of the LA.

AB hypotenuse
-
Thehe ratio
BC opposite side :
rat AG ; or a a aeae as Hay egPee tangent tof
atime of the the LA.

The reciprccals of these ratios are known respectively as the cosecant,

the secant, and the cotangent of A.
These six ratios are called the trigonometrical ratios of the LA, and
are usually expressed in the following shorter form.
: BC AC _BC
sinA=3_5) cos A= AR; tan A=76>

cosec A= ad sec A= a cot A= oA

Nors. The squares of these ratios, namely (sin A)*, (cos A)’, ... ace
usually written in the form sin?A, cos?A, ....
p

2. In the adjoining figure, let BC, DE B

be perps. to AQ from points in AP, and let
FG be perp. to AP from a point F in AQ.

A CG FQ
Then the A* BAC, DAE, FAG are similar, so that
BC _DE FG
AB’ AD AF’
But these ratios ae the value of sin A according as it is determined
from the A BAC, the A DAE, or the A FAG.
Thus sinA is unaltered so long as the LA remains the same. A
similar proof holds for each of the trigonometrical ratios, shewing that
they depend only on the size of the angle and not upon the lengths of
ite arms
 THE TRIGONOMETRICAL RATIOS. 271

EXERCISES.
1, In a triangle ABC, Seca at C, a=8, b=15; find c, and
write down the values of sinA, cosA, and tan A.

2. In a right-angled triangle, the sides containing the right angle

are 35 and 12: find the hypotenuse, and write down all the trigono-
metrical ratios of the smallest angle.

3. If A is any acute angle, shew that Theorem 29 may be made to

assume either of the forms:
(i) sin?A+cos?A=1; (ii) sec?A=1+ tan?A.

4. ABCD is a quadrilateral in which the diagonal AC is at right

angles to each of the sides AB, CD. If AB=1-‘5 cm., AC=3°6 cm.,
AD=8°5 cm., draw the figure, and find sinABC, tanACB, cosCDA,
tan DAC.

5. IfA is any acute angle, shew that

(i) sin(90°-A)=cosA; (ii) tan(90°-A)=cotA.

§. Construct an acute angle whose sine is0°6, [See Prob. 10, p. 83.}
Measure the angle with your protractor and give its value to the nearest
degree.

7. Construct an acute angle A from each of the fcllowing data :

(i) tamnA=0°7; (ii) cosA=0°9;° (iii) sinA=0-71.
In each case measure the angle to the nearest degree.

8. Construct an acute angle A such that tanA=1°6. Measure tke

angle A, and ascertain by measurement and by calculation the value of
cos A.

9. Prove the following results

(i) sin 45°=cos 45°= ude (ii) sin 60° = cos 30° =*5.
My

V2"
[See Ex. 11, p. 123, and Ex. 14, p. 124.]

10. Construct a triangle ABC, right-angled at C, having the hypo-

tenuse 10 cm. in length, and tan A= 0°81. Measure AC and the angle A;
and find the values of sin A and cosA.

11, Draw a right-angled triangle ABC from the following data :

tan A=0°7, LC=90°, 6=2'8 cm.
Measure c and the LA.
via GEOMETRY.

3. The definitions ou page 270 may

be extended to obtuse angles as follows :
Let XOX’ be a straight line, and let
OY be perp. to it.
Let the angle A be traced by the re-
volution about O of the line OP which
starts from the position OX.
Draw PM perp. to X’OX, thus form-
ing a right-angled triangle POM. Then
whatever the position of OP, the trigono-
metrical ratios of the angle A through
which OP has turned are thus defined :

sin A= SP cosh=2N, tanA=EN

with the wnderstanding that OM is to be considered positive when it ts ta

ihe right of OY, and negative when to the left of OY. [Compare p. 133.]
For example, in the above figure,
: PM 8 :
sin A= 65 ig — lo

OM -6_ ,
cos A= Gp = "17g =~ 6.

PM 8 4
tan A=oqj=—6= i

ExampLye. To express trigonometrically

(i) the perpendicular from the vertex of a criangle on the base ;
(ii) the projection of one side on another.

(i) In both Figs., AG=s8in C,

the sine of C being positive in each case,

“ p=bsinC,

(i) InFig. 1, — FP=cos. Fig. 1.

‘

In Fig. 2, oe also represents cos ©,

¢f CD is considered negative. a

¢, numerically
UD = +bcosC in Fig. 1. O,--2--sg>
GD = ~ boos in Fig. 2, Fig.2.
 THE TRIGONOMETRICAL RATIOS. 273

SOME GEOMETRICAL RESULTS EXPRESSED IN

TRIGONOMETRICAL FORM.

(The diagrams referred to are those of the preceding example. }

1. In both Figs., p=bsinC.

Similarly it may be proved that p=csin B.
: ee ae, Dee, ac
Hence bsinC=csinB; ~. <nB an

Similar] eea b eec

Peg oy sinA sinB sinC
that is, the sides of a triangle are proportional to the sines of the opposite
angles.

2. From this property of a triangle deduce Theorem 62

3. In both Figs.
area of A ABC=3BC.AD=}ap;
and p=bsinC; «=~ A=tadsin€
Similarly A=bc sin A=4ca sin B=34ab sin C,

4, Express in trigonometrical form the area of

(i) a parallelogram, given two adjacent sides and the included angle,
(ii) a rhombus, given one side and one angle.

5. Shew that the circum-radius: of a triangle is given by the

formula _ a _abe
QsinA 4A”
6. In Fig. 1, we have
AB?=BO?+CA?-2BC. CD. Theor. 55.
In Fig. 2, we have
AB?=BC?+CA?+2BC. CD. Theor. 54.
Now in Fig. 1, CD=+b6cosC;
and in Fig. 2, CD=-ocosC.
Hence in both cases we have, on substitution,
c?=a?+ b? —2ab cos C.
Similarly it may be shewn that
a?=b? +c? —2bc cos A.
b?=c? + a? - 2ca cos B.

5.3.6 8
274 GEOMETRY.

PROBLEMS.

PROBLEM 35.
To find the fourth proportional to three given straight lines.
K
Let A, B, C be the three given E,
at. lines, to which the fourth pro- ys)
portional is required. |
AGC'iB’ D & = ois
Construction. Draw two st. lines DL, DK of indefinite
length, containing any angle.
From DL cut off DG equal to A, and GE equal to B;
and from DK cut off DH equal to C.
Join GH.
Through E draw EF pat' to GH.
Then HF is the fourth proportional to A, B, C.
Proof. Because GH is par’ to EF, a side of the A DEF;
os DQ) GEDAhr.
But DG=A, GE=B, and DH=C;
« AL pee: APs
that is, HF is the fourth proportional to A, B, C.

PROBLEM 36.
To find the third proportional to two given straight lines.
K
Let A, B be the two lines to
which the third proportional is H
required.
AB OD
Construction. Draw two st. lines DL, DK.
From DL cut off DG equal to A, and GE equal to B;
and from DK cut off DH also equal to B.
Join GH.
Through E draw EF par' to GH.
Then HF is the third proportional to A, B.
Proof. As above, in Problem 35,
 INTERNAL AND EXTERNAL DIVISION. 275

PROBLEM 37,
To dinde a gwen straight line internally and externally in a
given ratio.

MN A ¥
Let AB be the st. line to be divided internally and externally
in the ratio M:N.
Construction. From A draw a st. line AH at any angle with
AB.
From AH cut off AP equal to M.
From PH and PA cut off PC and PC’, each equal to N.
Join BC, BC’.
Through P draw PX par' to BC, and PY par' to BC’.
Then AB is divided internally at X, and externally at Yin
the ratio M:N.
Proof. (i) Because PX is par' to BC, a side of the A ABC,
.. AX: XB=AP: PC
=M:N.
(ii) Because PY is par’ to BO’, a side of the A ABC’,
oe AY OYE —=AP PG
=M:N.
Bags Rr
COROLLARY. By a similar process a Q
st. line AB may be divided internally p
into segments proportional to three lines
L, M, N.
A a ak =|
Construction. Draw AH, and from it
Ee
cut off AP, PQ, QR equal respectively to
i
L, M,N. Join RB; and through P and N-
~Q draw PX, QY par’ to BR.
Then evidently AX: L=XY:M=YB:N.
276 GEOMETRY.

PROBLEM 38,
To find the mean proportional between two given straight lines.
D

B A Cc
Let AB, AC be the two given st. lines between which the
mean proportional is to be found.
Construction. Place AB, AC in a straight line, and im
opposite senses ; and on BC describe the semi-circle BDC.
From A draw AD at rt. angles to BC, to cut the O” at D.
Then AD is the mean proportional between AB and AC.
Proof. Join BD, DC.
Now the 2 BDC, being in a semi-circle, is a rt. angle.
And because in the pny inesABDC, DA is drawn from
the rt. angle perp. to the hypotenuse,
*. the A*ABD, ADC are similar; Theor, 66.
‘, AB: AD=AD:AC;
that is, AD is the mean proportional between AB and AC.

D
Nore. If the given lines AB, AC are placed in
the same sense, the mean proportional between
them may be cut off from AB by the following
useful construction. A CX 8B
On AB draw a semi-circle; and from C draw CD perp. to AB to cut
the Oat D. From AB cut off AX equal to AD.
Then AX is the mean proportional between AB and AC.
For the A*ABD, ADC are similar, Theor, 66.
* AB:AD=AD :AC;
that is, AB :AX=AX : AC.
 EXTREME AND MEAN RATIO. Pale f

GRAPHICAL EVALUATION OF A QUADRATIC SURD.

EXAMPLE. Find the approximate value of (i) V5, (ii) N21.

(i) V5=V5x1. Hence take AB, AC respectively to represent 5 and
1 in terms of any convenient unit, and find AD the mean proportional
between them.
Then since AB:AD=AD: AC,
“. AD?=AB.AC
=5x1=5.
AD=N5.
Thus by measuring AD, the value of V5 is roughly found to be 2°24,
(ii) V21=V7x3. Here take AB, AC equal to 7 cm. and 3 cm.
respectively, and proceed as before.

Nort. Factors should be chosen so as to give convenient lengths for

AB, AC.
e.g. N23=N2°3x10; VIl=N22x5.

DEFINITION.

A straight line is said to be divided in extreme and mean

ratio, when the whole is to the greater segment as the greater
segment is to the less.
A Xx B

Thus AB is divided at X in extreme and mean ratio,

when AB: AX=AX : XB;
from which it follows that
AB. BX=AX?;
or, the rectangle contained by the whole line and one part is equal to the
square on the other part.

Hence a straight line may be divided in extreme and mean

ratio by Problem 33. For Construction and Proof see page
240.
278 GEOMETRY.

EXERCISES.

1. Find graphically, testing your results by arithmetic:

{i) The 4% proportional to 2°4”, 1:5”, 1°6”.
(ii) The 3" proportional to 2°5” and 1:5”.
(iii) The mean proportional between 7°2 cm. and 5°0 cm.
2. Divide a line, 2°0” in length, internally and externally in the
ratio 7:3; and in each case find the segments by measurement and
calculation.

3. Obtain graphically the unknown term in the following statements

of proportion ; and check your result by arithmetic :
(i) 1°25:a% =1°0:1°6. [Take 1” as the unit of length.]
(ii) #:4°2=42:6°3. [Take 1 cm. as the unit of length.]
(iii) 2:16 = 25:a. [Let 1” represent 10.]

4. Divide a line, 7:2 cm. in length, into three parts proportional to

the numbers 2, 3, 4. Test your construction by méasurement and
palculation.
5. Divide a line, 3°9” in length, into three parts, so that the second
= § of the first, and the third
=4 of the second.
6. On a side of 1°5” draw a pha equal in area to a squareon
a side of 2”. Measure the other side of the rectangle.
7. Wiad graphically the approximate values of
(i} V8; (ii) NTO; (iii) VTA.
8. Determine by geometrical constructions the approximate values
of the following expressions, in each case verifying your drawing
arithmetically :
. 35x24 ” 6°84 271x126
(i) 3” (ii) a73° (iii) AN
151
9. Draw a triangle ABC from each of the following sets of data,
«nd in each case calculate and measure the lengths of the sides:

(i) The perimeter 8";

=4°8” ; and a37475
eee
(ii) The perimeter=11'l cm.; anda=b, b= $e.

(iii) The perimeter=11°8 cm, ; and a 4

{iv) a=4°0”, A=90°; and b:c=5:3.
 HEIGHTS AND DISTANCES, 279

EXERCISES,

(Proportion applied to the calculation of Heights and Distances.)

1. A field is represented in a plan by a triangle ABC, in which

a=8cm., b=5'6cm.,c=6'4cm. If the greatest side of the field is 200
metres, find the lengths of the other sides.
A fence, run across the field, is represented in the plan by a line PQ
pea to BC drawn from a point P in AB distant 4:0 cm. from A.
ind the length of the fence.

2. A’s speed is to B’s in the ratio 8:7; find graphically by how

much A would beat B in a 100 yards’ race, supposing each man to run
throughout at a uniform rate.

3. Ona map in which 1” represents 25 miles, three places A, B, and

Care marked. Of these, B appears N.W. of A at a distance 0°8”; and
C appears N.E. of Aatadistance 1‘5”. Find the actual distance between
B and C.
4. A man, whose height is 6 feet, standing 32 feet from a lamp-post,
observes that his shadow cast by the light is § feet in length: how
high is the light above the ground, and how long would be the shadow
of a boy 5 feet in height standing 20 feet from the post?

5. Aman 6 feet in height, standing 15 feet from a lamp-post, observes

that his shadow cast by the light is 5 feet in length: how high is the
light, and how long would his shadow be if he were to approach 8 feet
nearer to the post?

6. To find the width of a canal a rod is fixed vertically on the bank

so as to shew 44 feet of its length. The observer, whose eye is 5 ft. 8 in.
above the ground retires at right angles from the canal until he sees the
top of the rod in a line with the further bank. If his distance from the
nearer bank is now 20 feet, what is the width of the canal?

7. Aman, wishing to ascertain the height of a tower, fixes a staff

vertically in the ground at a distance of 27 ft. from the tower. Then,
retiring 3 ft. farther from the tower, he sees the top of the staff in line
with the top of the tower. If the observer’s eye and the top of the staff
are respectively 5 ft. 4 in. and 12 ft. above the ground, find the height
of the tower.

8. A person due S. of a lighthouse observes that his shadow cast

by the light at the top is 24 feet long. On walking 100 yards due EB.
he finds his shadow to be 30 feet long. Supposing him to be 6 feet
high, find the height of the light from the ground.
280 GEOMETRY,

SIMILAR FIGURES. —

THEOREM 67.
Similar polygons can be divided into the same number of similar
triangles; and the lines joining corresponding vertices in each figure
are proportional.
D

A B fe G
Let ABCDE, FGHKL be similar polygons, the vertex A
corresponding to the vertex F, B to G,and so on. Let AC, AD
be joined, and also FH, FK.
It is required to prove that
(i) the A* ABC, FGH are similar; as also the A* ACD, FHK,
amd the A* ADE, FKL.
(ii) AB: FG@=AC: FH=AD: FK.
Proof. (i) Since the polygons are similar,
the 2 ABC =the 2 FGH,
and AB: F@=BC:GH;
.*. the A* ABC, FGH are similar. Theor, 64
. the 2 BCA=the 2 GHF;
but because the polygons are similar,
the £ BCD =the 2 GHK;
.. the .ACD=the 2 FHK.
Also AC: FH = BC: GH, for the A* ABC, FGH are similar,
=CD:HK, for the polygons are similar.
That is, the sides about the equal 2" ACD, FHK are. pro
portional,
.". the A* ACD, FHK are similar. Theor. 64
In the same way the A* ADE, FKL are similar.
(ii) And AB; F@=AC:FH, from the similar A* ABC, FGH ;
=AD:FK, from the similar A* CAD, HFK.
Q.E.D.
 SIMILAR FIGURES. 28)

Nore. In the last Theorem the polygons D

have been divided into similar triangles by
lines drawn from a pair of corresponding vertices. we ee C
But this restriction is not necessary. E
For take any point O in the polygon ABCDE, Va
and join it to each of the vertices.
In the polygon FGHKL, make the LGFO’ A B
equal to the 2 BAO,
and make the L FGO’ equal to the LABO.
Join O’ to each vertex of the polygon FGHKL. LE
We leave as an exercise to the student the
proof that the two polygons are thus divided
into the same number of similar triangles.

PROBLEM 39. [First Method.]

On a side of given length to draw a figure similar to a given
rectilineal figure.
M
E

Let ABCDE be the given figure, and LM the length of the

given side; and suppose that this side is to correspond to AB
Construction. From AB cut off AB’ equal to LM.
Join AC, AD.
From B’ draw B’C’ par' to BC, to cut AC at C’.
‘ From C’ draw C’D’ par' to CD, to cut AD at D’.
From D’ draw D’E’ par' to DE, to cut EA at E’.
Then AB‘C’D’E’ is the required figure.
Outline of Proof. (i) By construction the figure AB'C’D'E
is equiangular to the figure ABCDE.
(ii) From the three pairs of similar triangles it may be shewn
that AB’_BC_CD_DeE’_E'A.
AB BC CD DE EA
that is, corresponding sides of the polygons are proportioned.
282 GEOMETRY,

THEOREM 68,
Any two similar rectilineal figures may be so placed that the lines
joining corresponding vertices are concurrent.
A

Fig’. 1, Fig. 2.
Let ABCD, A’B'C'D’ be similar figures.
Then since the 2 B’=the 2B, the figures can be so placed
that A’B’, B'C’ are respectively par' to AB, BC. It follows, since
the figures are equiangular to one another, that C’D’ is par' to
CD, and D’A’ par' to DA.
It is required to-prove that when corresponding sides of the given
Agures are parallel, then AA’, BB’, CC’, DD’ are concurrent.
Join AA’, and divide it externally at $ in the ratio AB: A’B..
Join SB and SB’: it will:be shewn that SB and SB’ are in
one straight line.

Proof. In the A*SAB, SA’B, since AB and A’B’ are par’,

. the 2SAB=the 2SA’B’;
and, by construction, SA:SA’=AB:A‘B’;
.*. the A* SAB, SA’B’ are equiangular to one another; Theor. 64
.. the ASB=the z A’SB’.
Hence SB, SB’ are in the same st. line ;
that is, BB’ passes through the fixed point 8.
Similarly CC’ and DD’ may be shewn to pass through §.
That is, AA’, BB’, CC’, DD’ are concurrent. .E.D. |

Norr. Observe that the joining lines AA’, BB’, CC’, DD’ are all —
divided externally at S in the ratio of any pair of corresponding sides
of the given figures,
 SIMILAR FIGURES. 383

Notr. In piacing the given figures so that A’B’, B’C’ are respectively
parallel to AB, BC, two cases arise :
(i) A’B’ and AB may have the same sense, as in Figs. 1 and 2;
Gi AB atid AB <0. e..cccss0s opposite senses, as in the Fig. below.

C D
In the latter case it follows also that C’D’ is par! to CD, and D’A
par! to DA, and it may be proved as before that AA’, BB’, CC’, DD’ are
concurrent ; but here S divides AA’ internally in the ratio of corre-
sponding sides, and the position of the figures is transverse.
Tn each case the point S is called a centre of similarity, or homothetie
sentre ; and similar figures so placed are said to be homothetic.

PROBLEM 39. [Second Method.]

On a given side to draw a figure similar to a given figure.

Let ABCD be the given figure, and A’B’ the given side;
and let A’B’ correspond to AB.
Construction. Place A’B’ par’ to AB; and join AA’, BB’ by
lines meeting at S.
Join SC, SD.
Through B’ draw B’C’ par’ to BC, to meet SC at C’;
through C’ draw C’D’ par’ to CD, to meet SD at D’.
Join A’D’.
Then A’B'C'D’ is the required figure.
The student should prove (i) that A’B’C’D’ is equiangular to
ABCD, (7i) that corresponding sides of these figures are pro
portional. The proof is the converse of Theorem 68.
284 GEOMETRY.

EXERCISES ON SIMILAR FIGURES.

(Numerical and Graphical.)

1. On a base AB, 6°5 cm. in length, draw a quadrilateral ABCO
from the following data :
LA=8v0°, LB=70°", AD=4:4cm., BC=3:2 cm.
Taking any convenient point as centre of similarity, make
(i) A reduced copy of ABCD, such that the ratio of each side to the
corresponding side of ABCD is 3: 4.
(ii) An enlarged copy of ABCD, such that the ratio of each side to
the corresponding side of ABCD is 5: 4.
2. Draw a semi-circle on a given diameter AB, and inscribe a square
in it, so that two vertices may be on the arc, and the other two on AB.
If AB=2r, and the side of the inscribed square=a, shew that
5a?
= 4r?,
3. Draw a sector of a circle of radius 2°4”, the central angle being
60°; and inscribe a square in it.
If the radius of the sector=7, and the side of the square=a, calculate
from measurements the ratio a: r.

4. Inasector of which the radius =5 cm., and the central angle=455

inscribe a rectangle having its adjacent sides in the ratio 2: 1.
Prove that two such rectangles can »e drawn, and compare by
measurement their greater sides.

5. Draw a triangle ABC, making a=8 cm., b=7 cm., and c=6 cm,
Working from the vertex A as centre of similarity, inscribe a squares
in the triangle, so that two of its angular points may be in the base BO,
and the other two in AB, AC.

6. Draw a triangle ABC, making a=2°6”, B=110°, C=35°.

In the triangle ABC inscribe an equilateral triangle, having
(i) one side parallel to BC;
(ii) one side parallel to any given straight line.

7. Ina ” ine triangle ABC inscribe a triangle similar to a given

triangle DEF.
In how many ways may this be done?

% Draw a regular hexagon ABCDEF on a side of 1°2’, and in it

inscribe a square having two sides parallel to AB and DE, and ite
vertices on the remaining sides of the hexagon,
 PROPORTIONAL AROS AND ANGLEA 235

THEOREM 69. [Euclid VI. 33.]

In equal circles, angles, whether at the centres or circumferences

a
bave the same ratio as the arcs on which they stand.

Let ABE, CDF be equal circles; and let the 2" AGB, CHD at
the centres, and the 2* AEB, CFD at the O“, stand on the arcs
AB, CD.
It is required to prove that
(i) the LAGB: the LCHD=the arc AB: the are CD;
(ii) the L AEB: the 1 CFD =the arc AB: the arc CD.
Proof. Suppose the are AB: the are CD=™:n;
s0 that, if the arc AB is divided into m equal parts, then the
arc CD may be divided into n such equal parts.
in each circle let radii be drawn to the points of division of
the arcs AB, CD. .
Then the 2* AGB, CHD, in equal circles, are divided inte
angles which stand on equal ares, and are therefore all equal.
And of these equal angles the AGB contains m,
and the CHD contains 2 ;
.. the LAGB: the ~CHD=m:n.
Hence the .AGB:the ~CHD=the arc AB: the arc CD.
And since the 2 AEB=one half of the AGB; Theor. 38.
and the 2 CFD=one half of the ~CHD;
.. the AEB: the 2 CFD=the arc AB: the arc CD.
Q.E.D.

COROLLARY. Since in equal circles, sectors which have equal

angles are equal {Theor. 42, Cor.], it may be proved as above
that the sector AGB: the sector CHD = the arc AB: the arc CD.
S86 GEOMETRY,

PROPORTION APPLIED TO AREAS.

THEOREM 70, [Euclid VI. 1.]

The areas of triangles of equal altitude are to one another as their
bases.
A ; D

B Cc E

Let ABC, DEF be two triangles of equal altitude, standing

on the bases BC, EF.
It is required to prove that
the A ABC: the ADEF=BC: EF,
Proof. Let the triangles be placed so that the bases BO,
EF are in the same st. line, and the triangles on the same side
of the line.
Join AD; then AD is par'to BF. Def. 2. p. 9%
Suppose the base BC ; the base EF=m:n;
so that, if BC is divided into m equal parts, then EF may te
divided into n such equal parts.
In each triangle let st. lines be drawn from the vertex to
the points of division in the bases BC, EF.
Then the A* ABC, DEF are divided into triangles which
stand on equal bases, and have the same altitude, and are
therefore all equal.
And of these equal A‘, the A ABC contains m ;
and the A DEF contains n.
. the A ABC: the A DEF=m:n.
Hence the A ABC : the A DEF =BC: EF.
Q.E.D
 PROPORTION APPLIED TO AREAS, 287

CoroLiary. The areas of parallelograms of equal altitude are

$0 one another as their bases.
For let DB, EG be par™ of the same 0D OSes 16
altitude, standing on the bases AB, EF.
Join AC, HF.
Then
since the par” DB= twice the ACAB;
and the par™ EG=twice the AHEF; 4 B EF F
.. the par™ DB: the par™ EG=the A CAB : the A HEF
= AB ; Et.

ALTERNATIVE PROOF OF THEOREM 70.

Let p represent the altitude of each of the A* ABC, DEF.

Then the area of the AABC=}. base x altitude=4.BCxp;
qc Hhesared Ob CoG A DEF oo yay oe on ns =4.EF xp.
AABC 4.BCxp BC
"ADEF 4.EFx» EF

EXERCISES.

(Numerical. }
1, Two triangles of equal altitude stand on bases of 6°3”ay and 5:4”
respectively ; if the area of the first triangle is 12} square inches, find
she area of the second.
2. The areas of two triangles of equal altitude have the ratio
- 24:17: if the base of the first is 4°2 cm., find the base of the second to
the nearest millimetre.
3. Two triangles lying between the same parallels have bases of
16:20 metres and 20°70 metres; find to the nearest square centimetre
the area of the second triangle, if that of the first is 50°1204 sq. metres.
4. Two parallelograms whose areas are in the ratio 2:1: 3°5 lie
between the same parallels. If the base of the first is 6-6” in length,
find the base of the second.
5. Two triangular fields lie on opposite sides of a common base;
and their altitudes with respect to it are 4°20 chains and 3°71 chains.
if the first field contains 18 acres, find the acreage of the whole quadii-
lateral.
388 GEOMETRY.

THEOREM 71.
If two triangles have one angle of the one equal to one angle of
the other, their areas are proportional to the rectangles contained by
the sides about the equal angles.

In the A* ABC, DEF, let the 2* at B and E be equal.

It is required to prove that
the A ABC
: the A DEF =AB. BC: DE. EF. .
Let AG and DH be drawn perp. to BC, EF respectively, and
denote the lengths of these perp* by p and p.
Proof. The AABC=43BC.p; and the ADEF=4EF .p’
. AABC TER
DER BC. pgh cine (i)
. But since the 2B=the ZE, and the .G=the ZH,
”. the A*ABG, DEH are equiangular to one another; Theor. 16

os Set me, sessscvcteonnell) Tiers Glee

Substituting for 7 in (i),

AABC BC.AB.
ADEF
EF. DE’
or the A ABC: the ADEF=AB.
BC: DE.EF..
Q.E.D.
CoROLLARY. Similarly it may be shewn that parallelograms
having one angle of the one equal to one angle of the other are pro
portional to the rectangles contained by the sides about the equat
angles.
 PROPORTION APPLIED TO AREAS, 283

EXERCISES ON AREAS.

(On Theorem 70.)

i. Assuming the area of a triangle=X base xaltitude, prove that

triangles on equal bases are proportional to their altitudes.
Also deduce this result geometrically from Theorem 70.
2. XY is drawn parallel to BC, the base of the triangle ABC, cutting
the sides AB, AC in X and Y.
doin BY and CX, and prove, by Theorem 70, that
(i) AX: XB=AY: YC.
(ii) AB: AX =AC: AY.
3. Shew that every quadrilateral is divided by its diagonals inte
four triangles whose areas are proportionals.
4. If two triangles are on equal bases and between the same
parallels, any straight line parallel to their bases will cut off equal areas
om the two triangles.

(On Theorem 71.)

5. In two triangles ABC, DEF, the 2B=the LE. If AB, BC are
2-7” and 3°5” respectively, and DE, EF are 2°1” and 1°8’, shew that
A ABC: ADEF =5:2.

6. The As ABC, DEF are equal in area, and the 4B=the LE.
a7 AB =56cm., BO=6:°3cm., DE=7'2 cm., find EF.
7. In two parallelograms ABCD, EFGH, the LB=the LF,
and the areas have the ratio 3:4. If AB=4°8 cm., BC=13°5 cm.,
EF =10°8cm., find FG.
If » and p’ denote the perpendiculars drawn from A and E to BC
and FG respectively, shew that p: p’=4: 9.
8. Prove the formula
area of A=} ab sinC;
and deduce Theorem 71.
9. The A ABC=the A DEF in area; andAB: DE=EF: BC; shew
that the L* B and E are equal or supplementary.
10. The sides AB, AC of the triangle ABC are cut by any straight
line at P and Q respectively. By joining PC, and twice applying
Theorem 70, shew that
A APQ: A ABC=AP. AQ: AB. AC.
Hence obtain an alternative proof of Theorem 71.
H.8,G tT
230 GEOMETRY.

THEOREM 72. [Euclid VI 19.j

The areas of similar triangles are proportional to the squares on
corresponding sides.
A
D

B G Cc = H

Let ABC, DEF be similar triangles, in which BC and EF are

sorresponding sides.
Ji is required to prove that
the AABC : the A DEF = BC?: EF?,
' Let AG and DH be drawn perp. to BC, EF respectively ;and
aenote these perp* by p and p’.

Proof. The AABC=43BC.p; and the A DEF=4EF.p’.

AABC BC. Did (i
ae 5 a gone

But since the B= the 2 E from the similar A* ABC, DEF,

and the .G=the ZH, being right angles;
”. the A*ABG, DEH are equiangular to one another ; 7heor. 16
. » AB
4 r = Theor. 62.

=e0 from the similar A* ABC, DEF

Substituting forP4 in (i),
Pp e

AABC BC.BC_ BC,

ADEF EF.EF EF?’
or, the A ABC: the A DEF = BC?: EF*,
Q. EIA
 AREAS OF SIMILAR TRIANGLES. W3E

EXERCISES ON THE AREAS OF SIMILAR TRIANGLES.

(Numerical and Graphical.)

i. In any triangle ABC, the sides AB, AC are cut by a line XY
drawn parallel to BC. If AX is one-third of AB, what part is the
triangle AXY of the triangle ABC?
2. Two corresponding sides of similar triangles are 3 ft. 6 in. and
2 ft. 4in. respectively. Ifthe area of the greater triangle is 45 sq. ft.,
find that of the smaller.
3. The area of the triangle ABC is 25°6 sq. cm., and XY, drawn
arallel to BC, cuts AB in the ratio 5:3. Find the area of the triangle
KY.
4. Two similar triangles have areas of 392 sq. em. and 200 sq. em.
respectively ; find the ratio of any pair of corresponding sides.
5. ABC and XYZ are two similar triangles whose areas are re-
spectively 32 sq. in. and 60°5 sq. in. If XY=7-7’, find the length oi
the corresponding side AB.
6. Shew how to draw a straight line XY parallel to BC the base of
s triangle ABC, so that the area of the triangle AXY may be nine-
sixteenths of that of the triangle ABC.

(Theoretical. )
7. ABC is a triangle, right-angled at A, and AD is drawn per:
pendicular to BC, shew that
3 ABAD : AACD=BA? : AC.
8. A trapezium ABCD has its sides AB, CD parallel, and its
diagonals intersect at O. If AB is double of CD, find the ratio of the
- triangle AOB to the triangle COD.
9. XY is drawn parallel to BC the base of the triangle ABC, if
F AAXY :fig. XBCY=4: 5,
shew that AX eK,
10. Prove that the areas of similar triangles have the same ratio as
the squares of
(i) corresponding altitudes ;
(ii) corresponding medians ;
(iii) the radii of their in-circles ;
(iv) the radii of their circum-circles.
992 GEOMETRY.

THEOREM 73. [Euclid VI. 20.}

The areas of similar polygons are propor tional to the squares oa
norresponding sides.

Let ABCDE, FGHKL be similar polygons, and let AB, FG be

sorresponding sides. ;
Ti is required to prove that
tie polygon ABCDE : the polygon FGHKL = AB?: FG®.
Join AC, AD, FH, FK.

Proof. ‘Then the A* ABC, FGH are similar ; Theor, $3.

also the A* ACD, FHK are similar ;
and the A* ADE, FKL are similar.
“2 the AABC:the AFGH= = AC? : FH? Theor. 7%.
= the ~ ACD: the A FHK.
Similarly,
the AACD:the AFHK= AD? : _ FK?
=the A ADE: the A FKL.
oo A ABC _ A ACD SADE
AFGH AFHK AFKL

And in this series of equal ratios, the sum of the ante

cedents is to the sum of the consequents as each antecedent
‘s to its consequent ; Theor. V. p. 251,
»*. the fig. ABCDE :the fig. F@HKL=the A ABC: the A FGH
= AR? ; FQ’
Q. B.D.
 AREAS OF SIMILAR POLYGONS. 293

COROLLARY 1. Let a, 6, ¢ represent thrée lines in pro

portion, so that
= 2; and consequently b?=ae.

a b Cem
Now suppose similar figures P and Q to be drawn on a and
6 as corresponding sides,
a?
then Hig. P_a?_@?_a@
Fig. Q 6% acc
Hence if three straight lunes are proportionals, and any similar
figures are drawn on the first and second as corresponding sides, then
the fig. on the first : the fig. on the second = the first : the third.

K
COROLLARY 2. Let ak -
AB :CD=EF :GH; ee
A B: 76 D
and let similar figures KAB, LCD
be similarly described on AB, CD, M
and also let similar figures MF, N
NH be similarly described on
EF, GH. eX
a = 2 G H
‘ AB: EF AB]; EF
h c fe = SS
eo CD GH’ CD* GH?
But the fig. KAB : the fig. LCD =AB?: CD?; Theor. 73
and the fig. MF: the fig. NH=EF?: GH?
.*. the fig. KAB : the fig. LCD =the fig. MF : the fig. NH.
Hence if four straight lines are proportional, and a pair of
similar rectilineal figures are similarly described on the first and
second, and also a pair on the third and fourth, these figures are
proport ional.
394 GROMETRY.

EXERCISES ON THE AREAS OF SIMILAR FIGURES.

(Numerical and Graphical.)

1, Shew how to draw a straight line XY parallel to the base BC of

a triangle ABC, so that the area of the triangle AXY may be four-ninths
cf the triangle ABC.

2. The sides of a triangle are 2°0’, 2°5”, 3:°2”; find the sides of a
similar triangle of three times the area. ’
(The results are to be given to the nearest hundredth of an inch.]

3. Two similar triangles have areas in the ratio 13°69: 16°81, and
an altitude of the greater is 10 ft. 3 in. Find the corresponding
altitude of the other.

4. ABC is a triangle whose area is 16 sq. cm.; and XY is drawn

parallel to BC, dividing AB in the ratio 3:5; if BY is joined, find the
grea of the triangle BXY.

5. One-fifth of the area of the triangle ABC is cut off by a line X¥

drawn parallel to BC. If BC=10cm., find XY to the nearest millimetre,

6. The area of a regular pentagon on a side of 2°5” is approximately

10% sq. in. ; find the area of a similar figure on a side of 3-0’.

7. The length of a rectangular area is 10°8 metres, and the ratio of

the length to the breadth is 12:5, Find the length and breadth of
similar rectangle containing one-ninth of the area,

8. In the plan of a certain field, 1” represents 66 yards ; if the area

of the plan is found to be 100 sq. in., find the area of the field in acres.
Explain why in this example the shape of the field is immaterial.

9. An estate is represented on a plan by a quadrilateral ABCD

drawn to the scale of 25” to the mile. If AC=20’, and the offsets from
wohooB and D measure 24” and 26” respectively, find the acreage of the
estate.
10. A field of 1:89 hectares is represented on a plan by a trian
whose sides measure 18 cm., 14 cm., and 15 em, On what scale is the
plan drawn?
 AREAS OF SIMILAR FIGURES. 29 i|

EXERCISES ON THE AREAS OF SIMILAR FIGURES,

(Theoretical.)
1, Hf ABC is a triangle, right-angled at A, and AD is drawn OM
@ieular to BC, shew that ; : eit
(i) BC?; BA?7=BC: BD; [Theor. 73, Cor. 1.]
(ii) BC? : CA?=BC: CD.
lence deduce BC?=BA?+ AC*.
2. A triangle ABC is bisected by a straight line XY drawn paraile)
to the base BC. Determine the ratio AX : AB.
Hence shew how to bisect a triangle by a straight line drawn parz)e!
to the base.
3. Two circles have external contact at A, and s common tangent’,
touching them at B and OC, meets the line of centres at S. Tf AS, AC
are joined, shew that
4 SBA: A SAC=SB:
SC.
4, Two circles intersect at A and B, snd at A tangents are draws,
sme to each circle, meeting the circumferences at C and D. Hf AB, CE
und BD are joined, shew that
A CBA: A ABD=CB:
BD.
5. DEF is the pedal triangle (see p. 207] of the triangle ABC; prov
Yhat 4 ABC: A DBF =AB?: BD3.
Hence shew that
fig. AFDC: A DBF =AD?: BD*

6. Ina given triangle ABC a second triangle is inscribed by joining

the middle points of the sides. In this inscribed triangle a third is
inscribed in like manner, and so on. What fraction is the fourtl:
triangle of the triangle ABC?

7. A regular hexagon is drawn on a side of @ centimetres, and

second hexagon is inscribed in it by joining the middle points of the
sides in order. In like manner a third hexagon is inscribed in the
second, and soon. Find the ratio of the first hexagon to the fifth.

8. Shew that the areas of two similar cyclic figures are proportion, to
the squares of the diameters of their circum-circies. [Huclid xu. 1.3
296 GEOMETRY.

TneorEM 74. [Euclid VI. 31.)

In a right-ungled triangle, any rectilineal figure described on tha
hypotenuse is equal to the sum of the two similar and similarly
described figures on the sides containing the right angle.

Let ABC be a right-angled triangle of which BC is the

hypotenuse ; and let P, Q, R be similar and similarly described
figures on BC, CA, AB respectively.
It is required to prove that
the fig. R + the fig. Q =the fig P.
Proof. Since AB and BC are corresponding sides of the
similar figs. R and P,
fig. RAB? ,
Ss hobo aSee (i) Theor. 3
g.

: fig. Q AC? a
I n ]like
k manner, fig. is
po P~ Pa BO wee ccereeees (ii) \ .

Adding the equal ratios on each side in (i) and (ii)

fig R+fig.Q AB?+AC?
igP 6=6=—SCOUBO
But AB? + AC? = BC? ; Theor, 2%
.. the fig. R+the fig. Q=the fig. P. Q.E.D,

COROLLARY. The area of a circle drawn on the hypotenuse of

a right-angled triangle as diameter is equal to the sum of the cirches
similarly drawn on the other sides.
For the areas of circles are proportional to the squares ov
their diameters. [Page 203.]
 MISCELLANEOUS EXERCISES. 297

EXERCISES.

(Miscellaneous. )

1. Ina triangle ABC, right-angled at A, AD is drawn perpendicular

to the hypotenuse. Shew that
(i) BA?-=BC...BD; (ii) CA2=CB. CD.
Hence deduce Theorem 29, namely,
BC?= BA? + AC?.

2. In the diagram of Theorem 74, draw AD perpendicular to BO;

hence prove that, if the fig. F=the A ABC, then
(i) the fig. Q=the AADC; (ii) the fig. R=the A ADB.

3. In the diagram of Theorem 74, if AB:AC=8:5, and if the

fig. P=8°9 sq. cm., find the areas of the figs. Q and R.

4. BY and CZ are medians of the triangle ABC, and YZ is joined.

Find the ratio of the triangle BOC to the triangle YOZ. [See p. 97.]

5. ABC is an isosceles triangle, the equal sides AB, AC each

measuring 3°6’. From a point D in AB, a straight line DE is drawn
cutting AC produced at E, and making the triangle ADE equal in area
to the triangle ABC. If AD=1°8", fina AE.

6. AB is a diameter of a circle, and two chords AP, AQ are produced

to meet the tangent at B in X and Y.
Shew that (i) the A®*APQ, AYX are similar;
(ii) the four points P, Q, Y, X are concyelic.

7. In the triangle ABC, the angle A is externally bisected by a line

which meets the base produced at D and the circum-circle at E: shew that
AB. AC=AE. AD.

8. When is a straight line said to be divided in extreme and mean

ratio?
If a line 10 cm. in length is so divided, find the approximate lengths
of the segments, and check your work graphically.

9. Draw an isosceles triangle equal in area to a triangle ABC,

and having its vertical angle equal to the angle A.

10. On a given base draw an isosceles triangle equal in area to &

given triangle ABC.
298 GEOMETRY.

PROBLEM 40.
To draw a figure similar to a given rectilineat figure, and equal
to a given fraction of it in area.

Let ABCDE be the given figure, to which a similar figure is

to be drawn, having its area a given fraction (say rae beer
of that of the fig. ABCDE.

Construction. Make AF three-fourths of AB. Prob. 7.

From AB cut off AB’ the mean .proportional between AF
ond AB. Prob, 38 Note.
On AB’ draw the fig. AB’C’D’E’ similar to the fig. ABCDE.
Prob. 39.
Then the fig. AB’C’D’E’
= } of the fig. ABCDE.
Proof. By construction, AB’? =AF . AB.
Now the figs. ABCDE, AB'C’D’E’ are similar, and AB, AB’ are
corresponding sides ;
fig. AB'C’D'E’ AB? Theor. 3%
" “fig. ABCDE AB?
AF .AB
 SIMILAR FIGURES, 299

EXERCISES.

1. Divide a triangle ABC into two parts of equal area by a line X¥

drawn parallel to the base BC and cutting the other sides at X and Y,
Find (i) by calculation, (ii) by measurement, the ratio AX : AB.

2. Divide a triangle ABC into three parts of equal area by lines PQ,
XY drawn parallel to the base BC. If P and X lie on AB, prove that
AP _AX_AB
If See Ne.
_ Hence shew how a triangle may be divided into m equal parts by
lmes drawn parallel to one side.

3. Draw a rectangle of length 8cm., and breadth 5cm. Then draw

% similar rectangle of one-third the area.
Measure its length to the nearest millimetre, and verify your resuli
by calculation.

4, Draw a quadrilateral ABCD from the following data:

the £A=90°; AB=BC=8ecm.; AD=DC=6 cm.
Draw a similar quadrilateral to contain an area of 36 sq. cm., and
find to the nearest millimetre the length of the side corresponding ta
AB.

&. Divide a circle of radius 3’ into three equal parts by means of

“wo concentric circles.

6. Draw a rectilineal figure equal im area to a given figure E, ana

similar to a given figure S. [Euclid vi. 25.]

[First replace the given figures E and S by equivalent squares

(see Problems 19 and 32). Let the sides of these squares be a and 4
respectively, and let s be one of the sides of S.
Find p, a fourth proportional, to b, a, s, so that b: a=s: p.
On p draw a figure P similar to the figure S, so that p and s are
corresponding sides. Then P is the figure required ;
Papa. E
a S- 2" S
a*e the fig. P=the fig. E.]
300 GEOMETRY

RECTANGLES IN CONNECTION WITH CIRCLES.

Norr. We here give a simple proof of a al 57 ana 58 brought
ander a single enunciation. [See Nore p. 234

THEoREM 75. [Euclid III. 35 and 36.]

If any two ‘chords of a circle cut one another internally or
eaternally, the rectangle contained by the segments of one is equal
to the rectangle contained by the segments of the other.
Cc C
D

A B A B x

D
Fig. 1. Fig. 2.
In the © ABC, let the chords AB, CD cut one another at X,
internally in Fig. 1, and externally in Fig. 2.
It is required to prove in both cases that
the rect. XA, XB=the rect. XC, XB. .
Join AD, BC
Proof. In the A* AXD, CXB,
the LAXD=the 2 CXB, being opp. vert. 2’ in Fig. 1, and the
same angle in Fig 2;
and the 2A= the 2th being 2" at the O”, standing on the
same are BD;
‘. the remaining angles are equal ; Theor. 16.
hence the A* AXD, CXB are equiangular,
.
XA XD.
XC XB’
}
“. XA.XB=XC.
XD:
that is, the rect. XA, XB=the rect, XO, XO.
Q. 1B.
 RECTANGLES IN CONNECTION WITH CIRCLES. 304

Corotnary. Jf from an external point a secant and a tangent

are drawn to a circle, the rectangle contained by the whole secant and
the part of wt outside the circle ts equal to the square on the tangent
C

luet XBA be a secant, and XT a tangent drawn to the

© ABC from the point X.
It 1s required to prove that XA.XB= XT?
Let XDC be a second secant;
$hen XA. XB=XC XD; Theor, 75. Fig. 2
and this is true for all positions of the line XDC.
Now let XDC turn about X away from the centre, so that
the points C and D continually approach one another and
ultimately coincide at T ;
then XDC becomes the tangent XT,
and XC ..XD becomes XT. XT, or XT2,
.., ultimately, XA. XB=XT2

EXERCISES FOR SQUARED PAPER.

{. From the point (1°7, 0) as centre, a circle is drawn to touch OY

at O, and cutting OX at A. If any line is drawn from A to cut OY at @
and the circle at P, shew that AP. AQ is constant, and find its value
when 1” is taken as the unit of length.
2, Acircle of radius 10 is drawn from centre C (5, 6). If TT’ is
the chord of contact of tangents from P (29, 16), and if PC meets TT’ in
@ find the value of
(i) CQ.CP; (ii) PQ.CP; and (iii) the length of TT’.
3. From centres (°3, 0) (2, 0) circles of radii 2°6 and 2°5 respectively
are drawn. Find the coordinates of their common points, and the
length of their common chord. Also find the length of a tangent to
each circle from the point (1°3, 3°4). .Verify your results by measure
ment.
B02 GEOMETRY,

* ')HEOREM 76.
if the vertical angle of a triangle is bisected by a straight line
which cuts the base, the rectangle contained by the sides of tha
triangle is equal to the rectangle contained by the segments of tha
base, together with the square on the straight line which bisects
the angle,

Let ABC be a triangle, having the 2 BAC bisected by AD.

It is required ta prove that
the rect, AB, AC =the rect. BD, DC + the sg. on AD.
Suppose a circle circumscribed about the 4 ABC; and let AD
»e produced to meet the O” at E.
Join EC.
Proof. ~ Then in the A’ BAD, EAC,
because the 2 BAD = the 2 EAG,
and the LABD=the 2 AEC in the same segment ;
, the remaining 2 BDA=the remaining 2 ECA;
that is, the A* BAD, EAC are equiangular to one another ;
AB AD
ee AE = AG’ Theor. 6%,

dence AB.AC=AE.AD
= (AD + DE) AD
=AD?+ AD. DE.
But AD .DE=BD.DC; Theor. 7,
.. the rect. AB, AC=the rect. BD, DC+the sq. on AD.
QED.
EXERCISE.
“E the vertical angle BAC is bisected externally by AD, shew thst
AB.AC=BD. DC - AD*
 THE DIAMETER OF THE CIRCUM-CIRCLE. 303

* THEOREM 77,
if from the vertical angle of @ triangle a straight line is draum
perpendicular to the base, the rectangle contained by the sides of the
triangle is equal to the rectangle contained by the perpendicular ane
bhe deameter of the circum-circle.

In the AABC, let AD be the perp. from A to the base BC:

and let AE be a diameter of the circum-circle.
dé ts required to prove that
the rect. AB, AC=the rect. AE, AD.
Join EC.
Proof. Then in the A* BAD, EAC,
the rt. angle BDA=the rt. angle ECA, in the semicircle ECA,
and the LABD=the Z AEC, in the same segment ;
.. the remaining 2 BAD = the remaining Z EAC ;
that is, the A* BAD, EAC are equiangular to one another.
ne Be Theor, 6%
AE AC
_ Hence AB. AC=AE.AD;
or the rect. AB, AC =the rect. AE. AD
Q ED,
Notr. Let a, 6, ¢ denote the sides of the AABC, R its cirenm
recius, and p the perp. AD.
Then since AE. AD=AB. AC
2R.p=cb
A
2 weDp
abe
_abo
$04 GEOMETRY.

THEOREM 78. [Ptolemy’s Theorem. |

The rectangle contained by the diagonals of a quadrilaterat in-
scribed in a circle is equal to the sum of the two rectangles contained
by its opposite sides.

Let ABCD be a quadrilateral inscribed in a circle, and let

AC, BD be its diagonals.
Tt is required to prove that
the rect. AC, BD =the rect. AB, CD+ the rect, BO. DA.
Make the 2 DAE equal to the 2 BAC:
to each add the 2 EAC,
then the 2 DAC=the 2 EAB
Proof. Then in the A* EAB, DAC,
the 2 EAB=the z DAC,
and the 2 ABE=the Z ACD in the same segment;
*, the A’ EAB, DAC are equiangular to one another ; Theor. 14.
BA BE Th —
oe —=>=>
GA.
—
‘CD’
5 a
7
6:

hence AB AOD = ACy BE tyinetnseutaaeseceeeens (i)

Again in the A* CAE, CAB,
the 2 DAE = the 2 CAB,
and the L ADE = the 2 ACB, in the same segment ;
.". the A* DAE, CAB are equiangular to one another;
DA_ DE,
Rh os"
hence BO. CA eA ES ccacseisy dueoneiins wooed}
Adding the equal vectangles on each side in (i) and (ii)
AB .CD+BC.DA=AC.BE+AC. DE
~ AC (BE + DE)
™AC . BD.
QB.
 EXERCISES ON THEOREMS 76-78. 305

EXERCISES,

1. ABC is an isosceles triangle, and on the base, or base produced,

any point X is taken: shew that the circumscribed. circles of the
triangles ABX, ACX are equal.

2. From the extremities B, C of the base of an isosceles triangle

ABC, straight lines are drawn perpendicular to AB, AC respectively,
and intersecting at D: shew that
BC. AD=2AB. DB.

3. If the diagonals of a quadrilateral inscribed in a circle are at

right angles, the sum of the rectangles contained by the opposite sides
is double the area of the figure.

4, ABCD is a quadrilateral inscribed in a circle, and the diagonal}

8D bisects AC: shew that
AD . AB=DC. CB.

5. If the vertex A of a triangle ABC is joined to any point m the

base, it will divide the triangle into two triangles such that their
circumscribed circles have radii in the ratio of AB to AC.

6. Construct a triangle, having given the base, the vertical angle,

and the rectangle contained by the sides.

7. Two triangles of equal area are inscribed in the same circle: shew
that the rectangle contained by any two sides of the one is to the rect-
angle contained by any two sides of the other as the base of the second
is co the base of the first.

8. Pisa point on the are BC of the circum-circle of an equilateral

triangle ABC. If P is joined to A, B, and C, shew that
PB+PC=PA.
9. ABCD is a quadrilateral inscribed in a circle, and BD bisects the
angle ABC: if the points A and C are fixed on the circumference of the
circle, and B is variable in position, shew that
AB+BC: BD isa constant ratio.
ij -

10. From the formula R=7 (see Norn, p.303) find the value of R
when the sides of the triangle are as follows:
(eZ), 2075157 3 (i) BOft., 25 ft-, Vth.
Draw the triangles to a convenient scale and check your work by
measurement.
"a.S8.G U
£06 GECMETRY

MISCELLANEOUS THEORETICAL EXAMPLES

ON PARTS I.—-V.

1. Two circles whose centres are C and D intersect at A and B;

and a straight line PAQ is drawn through A and terminated by the
circumferences : prove that
(i) the 2 PBQ=the ACAD;
(ii) the 2 BPC =the 2 BQD.
2. AB is a given diameter of a circle, and CD is any chord parallel
to AB; if any point X in AB is joined to the extremities of CD, shew
that
XC?+ XD?= XA? + XB.
3. The opposite sides of a cyclic quadrilateral are produced ¢¢
meet: shew that the bisectors of the two angles so formed are per
pendicular to one another.

4. Given the vertical angle, one of the sides containing it, and the
sngth of the perpendicular from the vertex on the base : construct the
triangle.
5. A, B, C are three points in order in a straight line: find a point
? in the straight line so that PB may be a mean proportional between
PA and PC.
6. Through D, any point in the base of a triangle ABC, straight
lines DE, DF are drawn parallel to the sides AB, AC, and meeting the
sides at E, F: shew that the triangle AEF is @ mean proportional
between the triangles FBD, EDC.

7. PQ is a fixed chord in a circle, and PX, QY any two parallel

chords through P and Q; shew that XY touches a fixed concentric
circle,

8. Two circles touch each other at C, and straight lines are drawn
through C at right angles to one another, meeting the circles at P, P’
and Q, Q’ respectively: if the straight line which joins the centres is
verminated by the circumferences at A and A’ shew that
P’P? + Q’Q?=A’A?.
9. AE bisects the vertical angle of the triangle ABC and meets the
base in E. If d, d’ are the diameters of the circum-circles of the
sriaugles ABE, ACE, shew that
d:d’=BE:EC.,

5
 MISCELLANEOUS #&AMPLES. 303

0, AB, AC are chords of a circle ; a line parallel to the tangent at

& eats AB, AC in D and E respectively: shew that
AB.AD=AC.AE.
ii. If & straight line is divided at two given points, determine a
third point such that its distances from the extremities may be propor-
tional to its distances from the given points.

12. Given the feet of the perpendiculars drawn from the vertices on
the opposite sides: construct the triangle.

13. Ifa quadrilateral. can have one circle inscribed in it, and another
circumscribed about it, shew that the straight lines joining the opposits
points of contact of the inscribed circle are perpendicular to one another.
14. Two equal circles move between two straight lines placed at
right angles, so that each line is touched by one circle, and the twa
vircles touch one another: find the locus of the point of contact.

15. AB is a diameter of a given circle; and AC, BD, two chords oz

the same side of AB intersect at E: shew that the circle which passes
through D, E, C cuts the given circle orthogonally. [See Def. p. 330.]

16. If four circles are described to touch every three sides of a

quadrilateral, shew that their centres are concyclic.

17. AB is a straight line divided at C and D so that

AB:AC=AC:AD;
from A a line AE is drawn in any direction and equal to AC; shew that
BC and CD subtend equal angles at E.
18. Given the vertical angle, the ratio of the sides containing it,
and the diameter of the circumscribing circle, construct the triangle.

i9. O is a fixed point, and OP is any line drawn to meet a fixed

straight line in P; if on OP a point Q is taken so that OQ to OP is
constant ratio, find the locus of Q. ;

20. O is a fixed point, and OP is any line drawn to meet ths

circumference of a fixed circle in P; if on OP a point Q is taken so that
OQ to OP is a constant ratio, find the locus of é

21. Two equal circles intersect at A and B ; and from C, any point
en the circumference of one of them, a perpendicular is drawn to AB,
meeting the other circle at O and O’; shew that either O or O’ is the
erthocentre of the triangle ABC. Distinguish between the two cases.
BOR GEOMETRY.

22. Three equal circles pass wesine the same point A, and their
other points of intersection are B, C, D: shew that of the four pointe
A, B, C, D, each is the orthocentre of the triangle formed by joining
the other three.
23. From a given point without a circle draw a straight line to the
concave circumference so as to be bisected by the convex circumference.
When is this problem impossible ?
24. Given the base, the altitude, and the radius of the circum:
circle: construct the triangle.

25. Given the base of a triangle and the sum of the remaining sides:
find the locus of the foot of the perpendicular from one extremity of
the base on the bisector of the exterior vertical angle.
26. Construct a triangle having given either the three ex-centres, or
the in-centre and two ex-centires.

27. If O is the orthocentre of a triangle ABC, shew that

AO?+ BC? = BO? + CA?= CO? + AB?=d?,
where d is the diameter of the cireum-circle.

28. If C is the middle point of an arc of a circle whose chord is AB,

and D is any point in the conjugate arc; shew that
AD+DB: DC=AB: AC.
29. D isa point in the side AC of the triangle ABC, and E is a point
in AB. If BD, CE divide each other into parts in the ratio 4:1, then
D, E divide CA, BA in the ratio 3:1.

30. If the perpendiculars from two fixed points on a straight line

passing between them are in a given ratio, the straight line must pass
through a third fixed point.

31. From the vertex A of any triangle ABC draw a line meeting BC
produced in D so that AD may be a mean proportional between the
segments of the base,

32. Two circles touch internally at O; AB a chord of the larger

circle touches the smaller in C which is eut by the lines OA, OB in the
points P, Q: shew that OP: OQ=AC: CB.

33. AB is any chord of a circle; AC, BC are drawn to any point ©

fn the circumference and meet the diameter perpendicular to AB at
D, E: if O is the centre, shew that the rect. dp, OE is equal te the
square on the radius.
 MISCHLLANEOUS #XAMPLES. 309

34. YD is a tangent to a circle drawn from a point Y in the diameter

AS produced ; from D a perpendicular DX is drawn to the diameter;
_ shew that the points X, y divide AB internally and externally in the
same ratio.

35. Determine a point in the circumference of a circle, from which

lines drawn to two other given points shall have a given ratio.

36. Given the base, and the position cf the bisector of the vertical
angle: construct the triangle.
’

37. EA, EA’ are diameters of two circles touching each othez
externally ‘at E; a chord AB of the former circle, when produced,
touches the latter at C’, while a chord A’B’ of the latter touches the
tormer at C: prove that
AB. A’B’=4B0C’. BC.

38, From a given external point draw a straight line to cut off «
quadrant from a given circle.
39. Shew that the straight lines joining the circum-centre cf 4
triangle to its vertices are perpendicular to the corresponding sides of
the pedal triangle.

40. P is any point on the circum-circle of a triangle ABC; and

perpendiculars PD, PE are drawn to the sides BC, CA. Find the locus
of the circum-centre of the tr iangle PDE.

41, P is any point on the circum-circle of a triangle ABC: shew

that the angle between Simson’s Line for the point P and the side BC
is equal to the angle between AP and that diameter of the circum-circle
which passes through A.

42, Given the base, the vertical angle, and the difference of the
angles at the base: construct the triangle.

43, Shew that the circles circumscribed about the four triangles
formed by two pairs of intersecting straight lines meet in a point.

44. Shew that the orthocentres of the four triangles pornied by two
pairs of intersecting straight lines are collinear.

45. Of all polygons of a given number of sides, which can be

iuscribed in a given circle, that which is regular has the maximum
area and the maximum perimeter.
S10 GEOMETRY.

46. On straight line PAB, two points A and B are marked and the
line PAB is made to revolve round the fixed extremity P. C is a fixed
point in the plane in which PAB revolves ; prove that if CA and CBare
\oined, ane the parallelogram CADB is completed, the locus of D will
be a circle.
47. Describe an equilateral triangle equal to a given isosceles
triangle,

48. Given the vertical angle of a triangle in position and magnitude,

snd the sum of the sides containing it: find the locus of the circum-
centre.

49, ABC is any triangle, and on its sides equilateral triangles ars
feseribed externally: ifX, Y, Z are the centres of their in-cire!ss
chew that the triangle XYZ is equilateral.
50. In a given circle inscribe a triangle so that two sides may pes:
hrough two given points and the third side be parallel to & giver
straight line.
51. In a given circle inscribe a triangle so that the sides may pass
through three given points.

52. A, B, X, ¥ are four points in a straight line, and O is such 4

»oint in it that the rectangle OA, OY is equal to the rectangle OB, OX;
fs circle is described with centre O and radius equal to a mean propor-
“ional between OA and OY, shew that at every point on this circle At
and X¥ will subtend equal angles.

83. Find the locus of a point which moves so that its distances from
two intersecting straight lines are in a given ratio,
64, IfS, |, |,, are the circum-centre, in-centre, and an ex-centre of
® triangle, and R, r, 7, the radii of the corresponding circles, and
if N is the centre of the nine-points circle, prove that
(i) S27=R?-2Rr; — (ii) SI2=R®+2QRv,3
(ii) NI=hR-r; fiv) Nh =4R er,
 MISCELLANEOUS THEOREMS AND EXAMPLES, 31}

MISCELLANEOUS THEOREMS AND EXAMPLES.

I, SOME CONSTRUCTIONS OF CIRCLES.

: Exampte 1. Draw a circle to touch a given circle (C), and also

touch a given straight line PQ at a given point A.

Construction. At A draw AF perp.

te PQ:
then the centre of the required ©
must lie in AF.
Take C the centre of the given ©,
sad draw the diam. BD perp. toe PQ.
Join A to one extremity D, of the
iameter, cutting the O® at E.
CE, and produce it to cut AF
ab

Then F is the centre, and FA the radius of the required circle.

nee, the proof; and shew that a second solution is obtained by
Soiming AB, and producing it to meet the O°.]

EXAMPLE 2. Draw a circle to pass through two given Bovis Aand B.

and to touch a given straight line CD.

Construction. Join BA, and produce

ib to meet CD at P.
Find PX the mean proportional be-
tween PA and PB. Prob. 38, Note.
From PD (or PC) cut off PQ equal
to PX.

Then the circle drawn through A, B, and Q [Prob. 25] will touch GD
Re

(Supply the proof ;and shew that there are in general two solutions,
Modify the construction to meet the case when AB is parallsl to OD,}
$12 GEOMETRY

Examee 3. Draw a circle to pass through two given points A and B.

and to touch a given circle (C).

Construction. Through A
and B draw any circle to cut
the given circle at P and Q.
Join AB, PQ and produce
them te meet at D.
From D draw a tangent DE
to the given circle.

Then the circle drawn through A, B, E will touch the given circle at E.
[Supply the proof from Theorems 58 and 59; and shew that there
are in general two solutions.
Modify the construction to meet the case when the straight line
bisecting AB at right angles passes through C.]

Exampie 4. Draw a circle to pass through a given point P, and te

louch two given straight lines AB, AC.
Construction. Draw AX bi-
secting the LBAC. Then all
circles touching AB, AC have
their centres in AX.
From any point d in AX
draw de perp. to AB; hence
with d as centre, draw a circle
touching AB and AC,
Join AP, cutting the © (d)
at p, p’.
Join pd; and through P
draw PD par! to pd, cutting
AX at D.
Then D is the centre, and DP the radius of the required circle touching
ABand AC. .
Peg DE perp. to AB. The proof is obtained by shewing that
OE=DP, by means of the similar A* ADE, Ade, and the similar
A* ADP, Adp.
Shew that a second solution may be obtained by joing dp’, and
proceeding as before.
Modify the construction to meet the case when the given lines are
parallel. }
 SONSTRUCTION OF CIRCLES. 312

EXERCISES FOR SQUARED PAPER.

1. Given a circle of radius 10 having its centre at the origin, draw

a circle to touch the given circle and also touch the x-axis at the point
(20, 0).
Shew that two equal circles can be so drawn. Calculate the radius
of that in the first quadrant and the coordinates of its point of contact
with the given circle.

2. Given a circle of radius 10 having its centre at the origin, draw

a circle to touch the given circle at the point (6, 8), and also to touch
the y-axis.
Shew that two such circles can be drawn. Find their radii and
points of contact with the y-axis.

3. Draw a quadrant of a circle of radius 2”, and inscribe a circle in

it. Shew that the radius of the inscribed circle is the positive root of
the equation r?+4r-—4=0.
Obtain the radius by calculation and by measurement.
4, Shew that two circles can be drawn to touch both axes of
soordinates and to pass through the point (2’, 2”); and prove that their
radii are given by the quadratic r2+ 4r/2 - 8=0.
Draw the smaller of these circles and obtain its radius by measure-
ment. :

5. Join the points (2’, 0) and (0, 3”); also join the points (3”, 0) and
(0, 2”); then draw a circle to touch the joining lines and to pass through
the origin.

6. Within an equilateral triangle on a side of 3:0” draw three equal

circles each to touch two sides of the triangle and the other two circles,
If 7 is the radius of one of these circles, shew that

r (tan 60° + 1) =3

Hence find x to the nearest hundredth of an inch.

J. Within a circle of radius 2:0’ draw three equal circles each ts

touch the other two and the given circle.
{f r is the radius of one of these equal circles shew that
r(1+cosec 60°)=2.
Hence find r to the nearest hundredth of an inch.
314 ANOMETRY.

Yi. MAXIMA AND MINIMA.

‘When a line, angle, or figure varying under specified conditions,

gradually changes its position and magnitude, we may be required te
note if any situations exist in which, after increasing, it begins to
decrease: or, after decreasing, to increase. In such sitnations ths
magnitude is said to have reached a maximum or minimum value. We
propose here to deal with problems in which the variable magnitude
sdmits of only one transition from an increasing to a decreasing state—
and vice versi: so that for our present purpose the maximum is
actually the greatest, and the minimum actually the least value that
the variable magnitude can take.
Two hints towards the solution of such problems may be given.
(i) Since a variable geometrical magnitude reaches its maximum or
minimum value at a turning point, towards which the magnitude may
mount or descend from either side, it is natural to expect a maximum
or minimum value when the magnitude assumes a symmetrical form o
position ; and this is usually found to be the case.

Examete 1. Divide a straight line AB internally so that the rectangle

contained by the two segments may be a maximum.
Bisect AB at C, and on AB draw a semi-circle. D
Take any point X in AB; and draw XP perp. -
to AB to cut the Ov at P.
Then AX.XB=PX2, Prob. 32.
Now PX is greatest when it coincides with AK cx RA
the radius CD ;
. AX. XB isa maximum, when X is the mid-point of AB.
Observe that in this case the maximum is reached when PX occupies
the symmetrical position in which it bisects AB at right angles.

(ii) Again we can find when a geometrical magnitude assumes its

maximum or minimum value, if we can discover a construction for
drawing the magnitude so that it may have an assigned value: for we
may then examine between what limits the assigned value must lie is
order that the construction may be possible ; and the higher or lowaer
‘amit will give the maximum or minimum sought for.
 MAXIMA AND MINIMA. 315

ke has been pointed out that if under certain conditions existing

among the data, ¢wo solutions of a problem are possible, and under
ather conditions, no solution exists, there will always be some inter-
mediate condition under which the two solutions combine in a single
solution. [See page 94.] ;
in these circumstances this single solution will be found to correspond
to the maximum or minimum value of the magnitude to be constructed.

Exampte 2. To find at what point in OD, a given straight line of

sadejinite length, the angle subtended by a finite line AB is a maximum.
First find at what point in CD a given angle is subtended by AB.
This is done as follows :
On AB draw a segment of a circle containing an angle equal to the
given angle. ‘Problem 24.
if the are of this segment intersects CD, two points in CD are found
at which AB subtends the given angle: but if the arc does not meet CB,
so solution is given. ~ .
In accordance with the principles explained above, we expect that
the maximum angle is determined when the arc touches CD, that is,
meets it at two coincident points. This we shall prove to be the case.
A
Draw a circle through A and B to touch
CD; and let P be the point of contact.
Ex. 2. p. 311.
Then the LAPB is greater than any other
ungle subtended by AB at a point in CD on the
same side of AB as P.
CQ P D
Proof. For take any other point Q in CD on the same side of AB ar
FP; and join AQ, QB.
Let BQ meet the circle at K. Join AK.
Then the 2AKB=the ZAPB, in the same segment.
But the ext. LAKB is greater than the int. opp. LAQB;
.. the APB is greater than the LAQB.
Hence the LAFB is a maximum.

Notre. Two circles may be described to pass through A and B, and

‘te touch CD, the points of contact being on opposite sides of AB; hence
two points in CD may be found such that the angle subtended by AB at
zach of them is greater than the angle subtended at any other point in
GD on the same side of AB.
316 ; GEOMETRY.

Exampue 3. In CD, a straight line of indefinite length, find a point P

such that the sum of its distances from two points A, B on the same side of
DD is a minimum. :
Draw AF perp. to CD; and produce
AF to E, making FE equal to AF.
Join EB, cutting CD at P. Join AP.
Then AP +PB is a minimum.

Proof. Take any other point Q in

OD, and join AQ, BQ, EQ.
Now the As AFP, EFP are con-
gruent,
- AP=EP.
Similarly AQ=EQ.
And in the AEQB, £Q+QB is greater than EB;
hence AQ+QB is greater than EB,
that is, greater than AP + PB.
Thus AP+ PB is a minimum.

Norr. It follows thatthe 2APF=the LEPF Theor. 4

=the BPD. Theor. 3
Thus AP + BP is a minimum, when AP, PB are equally inclined to CO.

EXAMPLE 4. Given two intersecting straight lines AB, AC, and & point
© between them ; shew that of all straight lines which pass through P and
are terminated by AB, AC, that which is bisected at P cuts off the triangle
of miniinum area.
Let EF be that st. line, terminated by
AB, AC, which is bisected at P.
Then the A FAE is of minimum area.
Proof. For let HK be any other st. line
passing through P.
Through E draw EM par! to AC.
Then the A* HPF, MPE are evidently
congruent, Theor, Vi. E K
and are therefore equal in area.
. the A HPF is less than the A KPE.
To each add the fig. AHPE ;
then the A FAE is less than the A HAK,.
That is, the A FAE is a minimum.
 MAXIMA AND MINIMA. 317

EXERCISES ON MAXIMA AND MINIMA.

h. ‘Two sides of a triangle are given in length; how must they be
placed in order that the area of the triangle may be a maximum ?
Find the area of the greatest triangle in which a=6'8 cm., and
b=4°5 cm.
2. Of all triangles of given base and area, shew that that which is
isosceles has the least perimeter. [See Ex. 3, p. 316.]
Calculate the minimum perimeter of a triangle of which the base =2°0",
and the area=3°12 sq. in.
8. Construct a triangle of maximum area on a base of 10 cm., and
having a vertical angle of 60°. Calculate its area.
~ 4. With the origin as centre draw a circle of radius 1°5”, and draw
AB joining the points (3”, 0), (0, 3”). Find a point in AB such that the
tangents drawn from it to the circle contain the maximum angle.
Measure the angle, and account for the result.
5. A straight rod slides between two straight rulers placed at right
angles to one another; in what position is the triangle intercepted
between the rulers and rod a maximum ?
6. Divide a given straight line into two parts, so that the sum of
the squares on the segments
(i) may be equal to a given square ;
(ii) may be a minimum.
7. Through a point of intersection of two circles draw a straight
lime terminated by the circumferences,
(i) so that it may be of given length ;
(ii) so that it may be a maximum.
8. Draw a circle to touch the axes of x and y at two points A and B,
each 2” distant from the origin,
Find a point on the major arc AB.such that the sum of its coordinates
is a Maximum.
Also find a point on the minor are AB such that the sum of its
coordinates is a minimum.
Tn each case calculate the sum, and test by measurement.
9. Straight lines are drawn from two given points to meet one
another on the convex circumference of a given circle: prove that their
sum is a minimum when they make equal angles with the tangent at
the point of intersection.
10. Shew that of all triangles having a given vertical angle and
altitude, that which is isosceles has the least area.
What is the least area a triangle can have if its vertical angle
=60°,
and altitude=6 ¢em.? Find its perimeter,
518 GEOMETRY.

11, Given two intersecting 1g ee: to a circle, draw a tangent te

the convex arc so that the triangle formed by it and the given tangents
may be of maximum area.
12. Find graphically (to the nearest degree) the greatest vertical
sngle which a triangle may have, when its base=1°6’, and its ares
=1:2 sq. in.
13. A and B are two points both within, or both without, a given
circle. Find a point on the circumference at which AB subtends ths
greatest angle. [See Ex. 2, p. 315.]
14, Aand B are two points on the a-axis distant 0°8” and 1°8” from
the origin O. Find graphically, a point P on the y-axis, such that the
angle APB is a maximum.
Calculate the length of OP, and measure the maximum angle.
15. <A bridge consists of three arches, whose spans are 49 ft.,
22 ft. and 49 ft. respectively :how far from the bridge is the point on
either bank of the river au which the middle arch subtends the greatest
angle?
16. From a given point P without a circle whose centre is C, draw
a straight line to cut the circumference at A and B, so that the triangle
ACB may be of maximum area.
Find the area of the greatest triangle that can be so drawn, wher
the radius=6 cm., and shew that the area is independent of the
position of P.
17. Find the area of the greatest rectangle which can be inscribed
in a circle of radius 5°5 cm,
18. A and B are two fixed points without a circle: find a point P
on the circumference, such that AP? + PB? may be a minimum.
[See Theor. 56.)
19. <A segment of a circle is described on the chord AB: find «
point C on its arc so that the sum of AC, BC may be a maximum.
20. Of all triangles that can be inscribed in a@ circle that which has
the greatest perimeter is equilateral.
21. Ofall triangles that can be inscribed in a given circle that which
has the greatest area is equilateral,
22. Of all triangles that can be inscribed in a given triangle that
sohich has the least perimeter is the pedal triangle.
23. Of all rectangles of given area, the square has the lesst
perimeter,
24. Describe the triangle of maximum area, having its angles equal,
to those of # given triangle, and its sides passing through three given
pointa.
 GRAPHS, 319

Wf. GRAPHS. APPLICATION TO MAXIMA AND MINIMA.

Problems dealing with the maximum or minimum values of some
variable magnitude may often be conveniently treated by exhibiting ite
gradual changes by means of a graph. For details of graphical work
the student may consult Hall’s Introduction to Graphical Algebra. It
will be sufficient here to explain the following general method of pro-
cedure. The variable magnitude whose values we have to examine
may be denoted by y, and the quantity, in terms of which it is
expressed, by x. By plotting a series of corresponding values of # and
y on the coordinate axes OX, OY, a series of points is determined. If
a continuous curve is drawn through them, the ordinate of each point
denotes the value of the magnitude in question corresponding to a given
value of the quantity x.
The advantage of this method is that it exhibits a visual picture of
continuous change, so that the graph enables us to read off the value ci
y corresponding to any given value of x; and in particular the positions
of maximum and minimum values are seen at a glance.

1.|S) al
Ee aesaaeoe aaah oon
5 HCN ERIE Hugues 7 pace
ee22 Se Sauuun ane
fen Mists
ileUe| .
ineEEE Paeaceacen
I
a

sumanM, Ssiseeenee
o*M
PEE EC¥
in this figure the continuous curve ABCDEF represents the graph of
# wariable quantity Q. As «increases gradually, the ordinate y travels
psrallel to OY, and its value at any point gives the value of Q for the
corresponding value of x. At P, the value of y is greater than that
at B or C on either side, and here Qiisamaximum. Similarly at P, the
value of y is less than that at D or E, and here @ is a minimum.
It will now be evident that maximum and minimum values occur
the turning points where the ordinates are algebraically greatest and
east respectively in the immediate vicinity of such points.
The following points should also be noticed:
(i) In any continuous curve maximum and minimum values occur
alternately.
(ii) There will always be a maximum or a minimum value between
any two equal values of the ordinate.
(ili) The slope of the curve at any point indicates the rate of change
at that point of the quantity under discussion, and at each
point of maximum or minimum value the tangent to the
curve is parallel to the axis of a.
320 GEOMETRY.

Exame.e 1. OACB is a parallelogram in which OA=8 om, and

OB=6cm. Jf OB rotates about O, trace the changes in the value of OC
as the angle AOB increases from 0° to 180°. Illustrate the changes by
means of a graph. Read off the value of OC for an angle of 72°, and find
the value of the angle when OC=5°6 cm.
B Cc
By drawing a series of figures, increas-
ing the angle AOB by increments of 30°, 6
we shall find by measurement the corre-
sponding values of OC and the LAOB to
be as in the following table. O - FS

£a0B | 0
o° |
|30°
30°|
|60°
60°|90°
0°||12
120° |150° |180°|
oF
oc |14-0
aS |13:5 72|
2] 41
41||20]

Denoting the angle by x, let its successive values be plotted on the

x-axis, and let the corresponding values of OC be taken as ordinates,
If each division on OX is taken to represent 6°, and each division on O'¥
to represent 1 cm. we obtain the adjoining graph.

Ae
Rae
ie.

From this we see that when #=72°, y=11°5 cm., and when y=5'6 cm.
+= 136°.
The student should plot the graph for himself on a much larger scale
than is possible on this page. He should also continue the values of OC
for a few angles greater than 180°. He will then find that a minimum
point has been reached when the angle AOB is 180°. Also it should bs
evident that at 360° the value of OC is again equal to 14, which is ite
maximum value.
 GRAPHS, 32k

Bxameue 2. ABC is a triangle in which BC, BA have constant lengths

6 cm. and 5 cm. Jf BC is fixed, and BA revolves about B, trace the
changes in the area of the triangle as the angle B increases from 0° to 180°.
liustrate these changes by a graph, and determine for what values of the
angle B the area is 10 sq. em. Also find for what value of B the area ta
® maximum. A

Proceeding as in Ex. 5, p. 110, the corre-

sponding values of the area and the angle will
be as in the following table : \
B 6 Cc
7}
| Angle 0° :30° 60° | 90° |120° )150° 180°

| Area in sq. om, |0 \7°5| 13:0 |150 |13.0| 75 | 0 |

Plot the values of the angle on the x-axis, and let the successive
values of the area be taken as corresponding ordinates. Then with the
same units as in the last example we obtain the adjoining graph.

120° 150° 180°

It is easily seen that the maximum value of the ordinate is 16,

corresponding to an angle of 90°.
Thus the area of the triangle is a maximum when the angle included
by the given sides is a right angle.
Also the curve is symmetrical with regard to its maximum ordinate,
so that there are two values of the angle which furnish a given area,
other than the maximum. When the area is 10 sq. em. the two values
of the angle are 42° and 138°.
Notr. The areaof AABC=45.6sinB=15sinB. Hence the graph
may be plotted from a Table of sines. [See Graphical Algebra, p. 29.)
H.S.G, x
322 GEOMETRY.

EXERCISES ON GRAPHS.

1. PQ is a perpendicular, 8 cin. in length, to a straight line XY,

and PR is an oblique making an angle a with PQ. By giving to a the
values 15°, 30°, 45°, 60°, 75°, find by measurement the correspouding
values of PR, and tabulate the results. Illustrate the changes in PR by
means of a graph, and find from it (i) the length of PR when a=63",
(ii) the value of a when PR=8°8 cm.
_ 2. In a trianglea=4 cm. b=5 cm. Plot a graph to shew the
changes in the area of the triangle for different values of C. “Find from
the graph (i) the area of the triangle when C=63°; (ii) the values of
C when the area is 9°5 sq. em.; (iii) how the sides must be placed when
the area is a maximum.
3. A straight rod AB of length 5 em. slides between two straight
rulers CD, CE placed at right angles to each other. Draw a graph to
shew the variations of the area of the triangle BCA for different values
of the length CA.
Point out the position of AB when the area is a maximum.
4. AB is a straight line 10 em. in length divided internally at P.
As P moves from A to B illustrate graphically the variations of
(i) AP.PB; (ii) AP?+PB?.
In each case determine from the graph the position of P which gives
a maximum or minimum value.
5. In a triangle c=6 cm. and A=60°. Trace the changes of a
graphically for different values of b. Find from the graph the minimum
value of a. Draw the triangle for this value, and hence check your
result. i
6. Through A, the extremity of the diameter AB of a semi-circle, a
line AP is drawn to.the cireumference. Trace graphically the variations
of the area of the triangle BAP for different values of the angle PAB.
Find the value of this angle when the area is greatest.
7. By means of Theorem 73, shew that the graph of the equation
y=m«", where m is constant, exhibits the changes in area of any series
of similar rectilineal figures similarly placed on sides of varying length.
Draw a graph to shew the changes in the area of a square as its side
varies, and from the graph find approximately the side of a square
whose area is 11°8 sq. in.
8. Draw the graphs of the curves represented by

(i) y=2e~™ (ii) y=5-4e-22

Find the maximum value of 5 — 4a —2*,
 HARMONIC SECTION. 323

f¥. HARMONIC SECTION.

DEFINITIONS.
_&. Three quantities are said to be in Arithmetical Progression whe:
tas difference between the last pair is equal to that between the firss
pair.
hus a, 6, ¢ are in A.P. when °
¢-b=b-a,
and 6 is said to be an Arithmetic Mean between o and ¢.

2. Three quantities are said to be in Geometrical Progression when

ri ratio of the third to tha second is the same as that of the second $-
the first.
Thus a, 0, ¢ are in G.P. when

and © fs said to be a Geometric Mean between a and e.

% Three quantities are said to be in Harmonical Progressies when

the first bears to the third the same ratio as the difference between the
Girst and second bears to the difference between the second and third
Thus a, 0, c are in H.P. when
a a-b
c b-c
and 5 is said to be a Harmonic Mean between @ and &.

Norr. Since, by definition,

it follows that

. the reciprocals ip = 2are in A.P., a result which is often uszein.

a .

4. If A, G, H denote the arithmetic, geometric, and harmomiz

means respectively between two given quantities a and 6, it easily
follows from the above definitions that
om 2ab
A=—>-> G= ab, a
ce
324 GEOMETRY.

DEFINITION. A finite straight line is said to be cut bax,

monically when it is divided internally and externally inte
segmants which have the same ratio.
——

A P B Q

Thus AB is divided harmonically at P and Q, if

AP : PB=AQ: QB.
P and Q are said to be harmonic conjugates of A and B.
Now by taking the above proportion alternately,
we have PA:AQ=PB:BQ;
from which it is seen that if P and Q divide AB internally and sz-
ternally in the same ratio, then A and B divide PQ externally and
internally in the same ratio; hence A and B are harmonio conjugates
of P and Q.
In other words: if AB {fs divided harmonically at P and Q, then PQ
ts divided harmonically at A and B.

Exampee l. Jf AB ts divided internally at P and externally at Q in

the same ratio, then AB is the harmonic mean between AQ and AP.

———-—_—--——.
A P B Q

For, by hypo'hesis, AQ:QB=AP:PB;

“., alternately, AQ: AP =QB: PB,
that is, AQ: AP=AQ--AB: AB-~AP;
“. AQ, AB, AP are in Harmonie Progression.

Examp.Le 2. If AB is divided harmonically at P and Q, and O is the

middle point of AB;
then OP. OQ=OA.
———EEE———

A OF) 4k» a
For since AB is divided harmonically at P and Q.
“, AP: PB=AQ: QB;
*, AP-PB: AP+PB=AQ - QB: AQ+QB,
or, 20P :20A=20A: 20Q;
. OP, OQ=O0A?,
Conversely, if OP. OQ=OA’,
it may be shewn that
AP: PB=AQ: QB;
that is, that AB is divided harmonically at P and Q.
 WARMONIC SECTION. 325

EXAMPLE 38. The Arithmetic, Geometric and Harmonc means of two

straight lines may be thus represented graphically.
Let AP, AQ be the given lines, whose H
Arithmetic, Geometric, and Harmonic
Means are to be found. ay S
On PQ as diameter draw a circle;
and from A draw the tangents AH, AK. A BO Q
Draw the chord of contact HK, cut-
ting AQ at B.
Join OH. K
Then (i) AO is the Arithmetic mean between AP and AQ:
tor clearly AO=3(AP+AQ).
(ii) AH is the Geometric mean between AP and AQ:
for AH?=AP. AQ. Theor. 58.
{iii) AB is the Harmonic mean between AP and AQ:
for, from the similar rt.-angled As AOH, HOB,
OA .OB=OH? Theor. 66, Cor.
(0)
. PQ is cut harmonically at Aand B; Ex. 2. p. 324.
’, also AB is cut harmonically at P and Q.
That is, AB is the Harmonic mean between AP and AQ.

And from the similar rt.-angled triangles OAH, HAB,

AO. AB=AH?; Theor. 66, Cor.
2, the Geometric mean between two straight lines is the mean proportional!
hetween their Arithmetic and Harmonic means.

Exampue 4. Given the base of a triangle and the ratio of the other
sides, to find the locus of the vertex.
Let BC be the given base, and let A
BAC be any triangle standing upon
it, such that BA: AC=the given ratio,
Tt is required to find the locus of A.
Bisect the LBAC internally and B Pia Q
axternally by AP, AQ.
Then BC is divided internally at P, and externally at Q,
go that BP: PC=BQ : QC=the given ratio :
. Pand Q are fixed points.
And aie AP, AQ bisect the 2 BAC internally and externally,
*, the LPAQ is a rt. angle;
°
or the locus of A is the circle described on PQ as diameter.
326 GEOMETRY.

EXERCISES ON HARMONIC SECTION,

§. If AB is divided harmonically at X and Y, shew that

Ag) 1 1
) AB=AX TAY
4 1 1
(ii) Xv ~By tay

2. Xand Y are harmonic conjugates of A and B;

fi) if AB=2-4”, and AX=1°5", find AY;
(ii) if XY=1°5 cm., and AY=2em., find BY.

3. Any straight line is cut harmonically by the arms of an aagle

and its internal and external bisectors.

4. Given three points B, P, C in a straight line: find tne locus of

points at which BP and PC subtend equal angles.

5. If through the middle point of the base of a triangle any lin»

is drawn intersecting one side of the triangle, the other produced,
and the line drawn parallel to the base from the vertex, it is divided
harmonically.

6. If from either base angle of a triangle a line is drawn intersec?-

ing the median from the vertex, the opposite side, and the line draws
oarallel to the base from the vertex, it is divided harmonically.
7. P, Q are harmonic conjugates of A and B, and C is an external!
point; if the angle PCQ is a right angle, shew that CP, CQ are the
vmterna) and external bisectors of the angle ACB.
8 ABisa eo straight line, bisected at O, and divided harmoni-
cally at X and Y.
Trace the change of position of Y as X moves from O to B.
Taking AB equal to 20 cm. draw a graph to illustrate the variations
of OY as OX changes,
9. Justify the following construction for finding the harmonic
mean between two straight lines of given length.
Let AB and CD be the given lines, and let them be placed so as to
be parallel. Join their ends towards the same parts by AC and BD,
and tewards opposite parts by AD and BC, cutting at O. Then if
POQ is drawn parallel to the given lines and terminated by AC and
BO. PQ is the required Savniaale mean,
 MARMONIO SECTION 327

DEFINITIONS.

1. A series of points in a straight line is cailed a range

lf the range consists of four points, of which one pair ars
harmonic conjugates with respect to the other pair, it Is said
to be a harmonic range,
2, A series of straight lines drawn through a point is
called a pencil.
The point of concurrence is called the vertex of the pencil,
and each of the straight lines is catled a ray.
A pencil of four rays drawn from any point to a harmonic
range is said to be a harmonic pencil.
3. A straight line drawn te cut a system of lines is called
8 transversal.
4. A system of four straight lines, no thres of which are
concurrent, is called a complete quadrilateral.
These straight lines will intersect two and two in sz
roints, called the vertices of the quadrilatera!; and each of,
the three straight lines which join the opposite vertices is
ealled a diagonal.

THEOREMS ON HARMONIC SECTION.

l. Ifa transversal 1s drawn parallel to one ray of @ harmente pens,

the other three rays intercept equal parts upon it; and conversely.
2. Any transversal is cut harmonically by the rays of @ harmonic
pencil.
3. Ina harmonic pencil, if one ray bisect the angie between the other
pair of rays, it ts perpendicular to its conjugate ray. Conversely, tf one
pair of rays form a right angle, then they bisect internally and externailss
the angle between the other pair.
4, If A, P, B, Q anda, p, b, q are harmonic ranges, one on each of
two given straight lines, and vf Aa, Pp, Bb, the straight lines which join
three pairs of corresponding points, meet at S; then will Qq also pas:
through §S.
5. If two straight lines intersect at A, and if A, P, B, Q and A,
p, b, q are two harmonic ranges one on each straight line (the pointe
corresponding as indicated by the letters), then Pp, Bh, Qq will be com
current : also Fq, Bb, Qp will be concurrent.
6. Use the last result to prove that in a@ complete quadrilatered cach
diagonal és cut harmonically by the other two.
328 GEOMETRY.

Vv. CENTRES OF SIMILITUDE.

EXAMPLE 1. Jn two circles if any two parallel radii are drawn (owe om
each circle), the straight line joining their extremities cuts the line of centres
2m one or other of two fixed points.

Take two circles whose centres are C and C’, and radii 7 and »
respectively ; and let CP, C’P’ be any two par' radii drawn in the same
sense in Fig. 1, and in opposite senses in Fig. 2. Let PP’ cut CC’ at S
Ii is required to prove that (whatever be the direction of CP, CP’) S ¢&
an one or other of two fixed positions,

Proof. In both Figs. the A*SCP, SC’P’ are equiangular;

* 8C:80’=CP: CP’
a et oe

Hence $ divides CO’ (rere inig, 2} in the fixed ratio r+,

“ in each Fig., S is a fixed point for all directions of CP, C’P’.
 CENTRES OF SIMILITUDE. 39,9:

_Corotiary. Let TT’ be a common tangent to the two circles,

zamect in Fig. 1, and transverse in Fig. 2.
Then in both cases the radii CT, C’T’ are par’;
.. TT’ cuts the line of centres at S.

DEFINITION. In the figure given below the points S and $'

which divide the line of centres of two circles externally and
internally in the ratio of their radii are called Centres of
Similitude, the former being the centre of direct and the
‘atter of fransverse similitude.

SCG _r Zoe
COROLLARY. Since
SO = SC"
the centres of the circles and the centres of similitude form an harmonic
range.
Hence the transverse and direct common tangents intersect on the line
af centres at points which divide that line harmonically.

EXERCISES.

i. From centres C and C’, 5:5 cm. apart, draw two circles of radi
3°2.cm. and 1‘2 cm. respectively, and determine (i) graphically (ii) by
calculation the distances of their centres of similitude from C.
2. Two circles whose centres are C and C’ respectively have radii
1:8” and 1:0’, and their direct centre of similitude is 2°7” distant
from C. Find the distance (i) between their centres (ii) between their
centres of similitude.
3. In Fig. 1 of the preceding page, if SP cuts the circles (C) and
Glsagain at Q and Q’ respectively, shew that
S@SP'=sSP,SQ/=ST .:Si’.
$36 GEOMETRY.

EXERCISES.

(On Centres of Similitude. Continued.}

4, In the triangle ABC, | is the in-centre, and |, the ex-cenire
pga to A. If Al, cuts BC at Y, shew that A and Y are the centres
of similitude of the two circles.

5. Shew that the orthocentre and centroid of a triangle are respec-

tsvely the external.and internal centres of similitude of the circumscribed
and nine-points circle.

6. If a variable circle touches two fixed circles, the line joining the
points of contact passes through a centre of similitude. Distinguish
between the different cases.

7. Describe.a circle which shall touch two given circles and pass
through a given point.

8. Describe a circle which shall touch three given circles,

9. C,, C,, Cy are the centres of three given circles; Sj, S,, are the
internal and external centres of similitude of the pair of circles whose
centres are Cy, Cy, and S’,, S,, 5’3, 83, have similar meanings with regara
to the other two pairs of circles: shew that
(i) S’,C,, S’,C,, S’,C, are concurrent ;
{ii) the six points S,, S,, 83, S’,, S’,, 8’, lie three and three on form
straight lines. [See Theorems Ix. and x., pp. 344, 345.]

ORTHOGONAL CIRCLES.
Derrition. Circles which intersect at a point, so that the
two tangents at that point are at right angles to one another,
are said to be orthogonal, or to cut one another orthogonally
1. If tsvo circles cut one another orthogonally, the tangent to each
circle at a point of intersection will pass through the centre of the
other circle.
2. If two circles cut one another orthogonally, the square on the
distance between their centres is equal to the sum of the squares on
their radii.
3. Find the locus of the centres of all circles which cut a given
arele orthogonally at a given point.
4. Describe a circle to pass through a given point and cut a given
circle orthogonally at a given point, :
 POLE AND POLAR. 33h

Vi. POLE AND POLAR.

: DEFINITIONS.

i. If in any straight line drawn from the centre of a circle

{wo points are taken such that the rectangle contained by
their distances from the centre is equal to the square on the
radius, each point is said to be the inverse of the other.
_ Thus in the figure given below, if O is the centre of the circle, and if
OP. OQ=(radius)*, then each of the points P and Q is the inverse
of the other. -
It is clear that if one of these points is within the circle the other
must be without it.

2. ‘The polar of a given point with respect to a given circl¢

is the straight line drawn through the inverse of the given
noint at right angles to the line which joins the given point te
the centre: and with reference to the polar the given point is
called the pole.

Thus in the adjoining figure, if OP .OQ=(radius)?, and if through

P and Q, LM and HK are drawn perp. to OP; then HK is the polar of
the point P, and P is the pole of the st. line HK with respect to the
given circle; also LM is the polar of the point Q, and Q the pole of LM
It is clear that the polar of an external point must intersect the circis,
and that the polar of an internal point must fall without it: also that
the polar of a point on the circumference is the tangent at that point.
333 GEOME7RY.

_ Examptel. The polar of an external point with reference to a circhs

is the chord of contact of tangents drawn from the given point to the circle.
From the external point P let two tangents
PH, PK be drawn to a circle of which O is the
centre.
Join HK.
It is required io prove that HK is the polar
of P,
Now OP evidently cuts the chord of contact
HK at right angles at Q.
Join OH.
Then from the similar rt.-angled A* POH,
by
a ’
OP: OH=OH: OQ;
“, OP. OQ =(radius)? ;
sence HK is the polar of P.
Q.B.D.
Exampitn2. If Aand P are any two points, and if the polar of A with
respect to any circle passes through P, then the polar of P must pase
through A.
Let BC be the polar of the point A with respect to a circle whose
centre is O, and let BC pass through P.
It is required to prove that the polar of
® passes through A, ;
Join OP; and from A draw AQ perp.
to OP. We shall shew that AQ is the
polar of P.
Now since BC is the polar of A,
‘, the LABP is a rt. angle;
Def. 2, page 331,
and the L AQP is a rt. angle: Constr,
.. the four points A, B, P, Q are concyclic;
, OQ. OP=0A. OB Theor, 58,
=(radius)®, for CB is the polar of A,
And since AQ is perp. to OP,
“. AQ is the polar of P.
That is, the polar of P passes through A.
QE. D,

Norg. A similar proof applies to the case when the given poin'
‘is without the circle, and the polar BC cuts it.
The above Theorem is known as the Reciprocal Property of Pole and
Polar,
 POLE AND POLAR. By

EXAMPLE 3. The lovis of the intersection of tangents drawn to a circle

at the extremities of all chords which pass through a given point within
the circle is the polar of that point.
Let A be the given point within the
circle. Let HK be any chord passing
through A; and let the tangents at H and
K intersect at P.
Tt 1s required to prove that the locus of P is
the polar of the point A.
(2) To shew that P lies on the polar
of A.
Since HK is the chord of contact of
tangents drawn from P,
“. HK is the polar of P. Ex. 1, p 332.
But HK, the polar of P, passes through A;
. the polar of A passes through P: Ex. 2, p. 382.
that is, the point P lies on the polar of A.
(8) To shew that any point on the polar of A satisfies the given
sonditions.
Let BC be the polar of A, and let P be any point on it.
Draw tangents PH, PK, and let HK be the chord of contact.
Now from Ex. 1, p. 332, we know that the chord of contact HK is
the polar of P,
and we also know that the polar of P must pass through A; for P is on
BC, the polar of A: Hx. 2, p, 332.
- that is, HK passes through A.
“. P is the point of intersection of tangents drawn at the extremities
vf a chord passing through A.
From (a) and (8) we see that the required locus is the polar of A.
Nors. If A is outside the circle the theorem (a) still holds good;
but the converse theorem (8) is not true for all points in BC. For if A
is without the circle, the polar BC will intersect it; and no point on
that part of the polar which is within the circle can be a point of
intersection of tangents.

We -now see that

(i) The Polar of an external point with respect to a circle is the
chord of contact of tangents drawn from tt.
(ii) The Polar of an internal point is the locus of the intersee-
tions of tangents drawn at the extremities of all chords which pass
rough wt.
(iii) The Polar of a point on the circumference is the tangent
at that point.
334 GEOMETRY,

Examete 4, Any chord of a cirele through a jixed point P is divided

ikarmonically by P and the polar of P.

Fig. 1. Fig. 2.
Let O be the centre of the given circle, and let QR be a chord passing
through the given point P,
(i) When P is aa external point (Fig. 1).
Draw a tangent PT, and let the polar of P cut PO in L, and Q#
Ss.
it is required to prove that QR is divided harmonically at P and S.
Draw OM perp. to QR, and join OT.
Thea PQ. PR=PT?
=P. PO) since PTO is'art. Z,
=PM.PS, since S, L, O, M are concyctia
. 2PQ.PR=2PM.PS
=(PQ+PR)PS; Ex. 9, p. 66
2PQ.PR, :
es PS=50>PR ;

“, PQ, PS, PR are in Harmonical Progression;

that is, PS is divided harmonically at Q and R,
*. also QR is divided harmonically at P and S.
(li) When P is an internal point (Fig. 2).
Let SL be the polar of P.
Then since the polar of P passes through S, the polar of S paases
through P. :
*. by the former case QR is divided harmonically at S and P.
The above theorem is known as the Harmonic Property of Pole and
Polar

DEFINITION.

A triangle so related to a circle that each side is the polar

of the opposite vertex is said to be self-conjugate with respect
to the circle.
 POLE AND POLAR. 335

EXERCISES ON POLE AND POLAR.

i, The straight line which joins any two points is the polar with
respect to a given circle of the point of intersection of their polars.

2. The point of intersection of any two straight lines is the pole of the
straight line which joins their poles.

3. Find the locus of the poles of all straight lines which pass through
m given point, ;
4. Find the locus of the poles, with respect to a given circle, of tangents
drawn to a concentric circle.

5. If two circles cut one another orthogonally and PQ be any diameter

of one of them; shew that the polar of P with regard to the other circle
passes through Q.

6. Iftwo circles cut one another orthogonally, the centre of each circle
is the pole of their common chord with respect to the other circle.

7. Any two points subtend at the centre of a circle an angle equal to

one of the angles formed by the polars of the given points.

8. Os the centre of a given circle, and AB a fixed straight line.

P is any point in AB; find the locus of the point inverse to P with
respect to the circle.

9. Given a circle, and a fixed point O on its cirewmference: P 3s any

point on the circle: find the locus of the point inverse to P with respect to
any circle whose centre is O.

10. Given two points A and B, and a circle whose centre is O; shew
that the rectangle contained by OA and the perpendicular from B on the
volar of A is equal to the rectangle contained by OB and the perpendicular
from A on the polar of B.

1l. Four points A, B, C, D are taken in order on ihe circumference of

a etrcle ; DA, CB intersect at P, AC, BD at Q, and BA, CD in R: shew
that the triangle PQR is self-conjugate with respect to the circle.

12. Give a linear construction for finding the polar of a given point
utth respect to a given circle. Hence find a linear construction for
drawing a tangent to a circle from an external point.
13. Ifa triangle is self-conjugate with respect to a circle, the centre aj
she circle is at the orthocentre of the triangle.
14. The polars, with respect to a given circle, of the four pornts of é
harmonic range form a harmonic pencil : and conversely.
336 GEOMETRY

VII. THE RADICAL AXIS,

Example l. To find the locus of points from which the tangents draw.
vo two given circles are equal.

Fig. 2.
Let A and B be the centres of the given circles, whose radii are «
and 6; and let P be any point such that the tangent PQ drawn to the
wircle (A) is equal to the tangent PR drawn to the circle (B).
It is required to find the locus of P.
Join PA, PB, AQ, BR, AB;
from P draw PS perp. to AB.
Then because PQ=PR, .. PQ?=PR2
But PQ?= PA?- AQ?; and PR?=PB?-BR?: Theor. 28
“. PA?-AQ?=PB?- BR?;
shat is, PS? + AS? - a?= PS?+ SB?-}?; Theor. 2
or, AS? - a?= SB? - b?.
Hence AB is divided at S, so that AS?- SB?=a?-?:
“. Sis a fixed point.
Hence all points from which equal tangents can be drawn to the two
sircles lie on the straight line which cuts AB at rt. angles, so that the
difference of the squares on the segments of AB is equal to the difference
of the squares on the radii.
Again, by simply retracing these steps, it may be shewn that in
Fig. 1 every point in SP, and in Fig. 2 every point in SP exterior to
the circles, is such that tangents drawn from it to the two circles are
equal.
Hence we conclude that in Fig. 1 the whole line SP is the required
locus, and in Fig, 2 that part of SP which is without the circles.
An either case SP is said to be the Radical Axis of the two circles,
 RADICAL AXIS. RADICAL CENTRE. 337

CoroLLary. If the circles cut one another as in Fig. 2, tt is clear

that the Radical Axis is identical with the straight line which passes
through the points of intersection of the circles; for it follows readily
from Theorem 58 that tangents drawn to two intersecting circles from
aay point in the common chord produced are equal.

ExampLe 2. Whe Radical Ames of three circles taken in pairs are

concurrent.

Let there be three circles whose centres are A, B, C.

Let OZ be the radical axis of the © (A) and (B);
amd OY the Radical Axis of the ©8 (A) and (C), O being the point of
their intersection,
It is required to prove that the radical axis of the ©® (B) and (C}
passes through O.
{t will be found that the point O is either without or within all ihe
circles.
I. When O is without the circles.
From O draw OP, OQ, Of tangents to the ©* (A), (B),
Then because O is a point on the radical axis of (A) and on3
5) AO S10[88
And because O is a point on the radical axis of (A) and (C),
— O)P=Olm :
. OQ=OR;
. Oisa point on the radical axis of (B) and (C):
that is, the radical axis of (B) and (C) passes through O.
fi. If the circles intersect in such a way that O is within them aj ;
¢he radical axes are then the common chords of the three circles taken
twe and two; and it is required to prove that these common chords
wre concurrent. This may be shewn indirectly by Theorem 57.

Derinirion. ‘The point of intersection of the radical axes

af three circles taken im pairs is called the radical centre.
H.S.G Y¥
338 GEOMETRY,

Exame.e 3, 7'o draw the radicai axis of two given circles.

Let A and B be the centres of the given circles.

It is required to draw their radical azve.

fa) If the given circles intersect, then the st. line drawn through their
oints of intersection will be the radical axis.

(8) But if the given circles do not intersect,

describe any circle so as to cut them in E, F and G, H.
Join EF and HG, and produce them to meet in P,
Join AB; and from P draw PS perp. to AB.
Then PS is the radical axis of the ©* (A), (B).

Proof. From the © EFGH, PE.PF=PH. PG.

Now the sq. on the tangent from P to the © (A) =PE. PF ;
and the sq. on the tangent from P to the © (B)=PH. PG.
Hence the tangents from P to the ©* (A) and (B) are equal:
. Pisa point on the radical axis.
And since PS is perp. to the line of centres,
. PS is the radical axis, Ex. 1. p 336.

DEFINITION. If each pair of circles in a given system have

tihe same radical axis, the circles are said to be co-axal.
 YHE RADICAL AXIS. 339

EXERCISES ON THE RADICAL AXIS.

h. Shew that the radical axis of twe circles bisects any one of their
common tangents.
2. Jf tangents are drawn to two circles from any point on their
radical axis; shew that a circle described with this point as centre and
any one of the tangents as radius, cuts both the given circles orthogonally.
[See Def. p. 330.]
_ 3. Os the radical centre of three circles, and from O a cangent OT
is drawn to any one of them: shew that a circle whose centre is O and
radius OT cuts all the given circles orthogonally.
4. If three circles touch one another, taken two and two, shew thas
“her convmon tangents at the points of contact are concurrent.
5. If circles are described on the three sides of a triangle as diameter,
their radical centre is the orthocentre of the triangle.
6. All circles which pass through a fixed point and cut a given circle
orthogonally, pass through a second fixed point.
7. Find the locus of the centres of all circles which pass through a
psven point and cut a given circle orthogonally.
8. Describe @ circle to pass through two given points and cut a given
corele orthogonally.
9. Find the locus of the centres of all circles which cut two given
oircles orthogonally.
10, Describe a circle to pass through a given point and cut two given
eurcles orthogonally.
il. The difference of the squares on the tangents drawn from any
woint to two circles is equal to twice the rectangle contained by the straighi —
“ane joining their centres and the perpendicular from the given point on
ther radical axis.
12, In a system of co-axal circles which do not intersect, any point is
taken on the radical aais; shew that a circle described from this point as
centre, with radius equal to the tangent drawn from it to any one of the
circles, will meet the line of centres in two fixed points.
{These fixed points are called the Limiting Points of the system.|
13. In a system of co-axal circles the two limiting points and the
points in which any one circle of the system cuts the line of centres form a
harmonic range. .
14, Ina system of co-axal circles a limiting point has the same polar
unth regard to all the circles of the system.
15. If two circles are orthogonal any diameter of one ia cut harmons
cally by the other.
340 GROMETRY

Vill, INVERSION

DEFINITIONS.

i, If from any fixed point O a straight

line OP is drawn, and a point P’ is taken on
OP, or OP produced, such that OP. OP’ =k?,
where & is constant, then each of the
points P and P’ is said to be the inverse of
the other with respect to the circle whose
centre is O and radius &.

2. The point O is called the origin of inversion, and & iz

called the radius of inversion. Also /? is sometimes referred
to as the constant of inversion.

3. If P traces out a locus, to every position of P there is »

corresponding position of P’, The locus of P’ is called the
inverse of the locus of P.
From Definition 1 it is clear that any straight line passing
through the origin is its own inverse.

Exampir 1. To find the inverse of a straight line not passing through

the origin of inversion.
Let P be any point on the given st. line AB, O
the origin, and & the radius of inversion.
Draw OQ perp. to the given line. Take P’
and Q/ the inverses of P cont Q,
Join P’Q’.
Then OP. OP’=/2
=0Q.0Q’.
* the pts. P, P’, Q, Q’ are concyclic ;
*. the LOP’Q’=the LOQP
=art. 4.

Hence the locus of P’ is a circle which passes through O, such thai

the diameter OQ’ is perp. to the giver line
 (NVERSION 341

EXAMPLE 2. To Jind the inverse of a circle with respect to a point

mm ste curcumference.
Let OQ be the diameter of the given ;
viurele which passes through the origin 6.
Take any point P on this circle; and with
& as radius of inversion, let Q’ and P’ be the
_ unverses of Q@ and P. O °
Join PQ, PQ’. Q |@
then © OPO =
=0Q. OQ’.
. the pts. P, P’, Q, Q’ are concyclic;
. the 2OQ/P’=the _OPQ
=O) ig, 74.
“. P’Q’ is perp. to OQ’.
Hence the locus of P’ is a st. line perp. to the diameter through the
origin.
Exame.e 3. To find the inverse of a circle with respect to a pons
not om the carcumference. ~

Let O be the origin and P any point on the given circle whose centre

Let P’ be the inverse of Pysothat OP OPS,

Let OP meet the given circle again inQ. Join QC.
Draw OT a tangent to the circle, and let OT =¢.
Then OP. OP’=#?, and OP. OQ=¢%,
OP.OP’ 12.
OP.0Q?#?’
OF + OQ sea
Draw P’C’ par! to QC to meet OC produced in C’,
Then OC’: OC=OP’: 0Q
. alee eG
. C’ isa fixed point.
Also C’P’: CQ=OP’:0Q
=e.
- C’P’ is constant, and the locus of P’ is a circle whose centre is ©’,
CoRoLLARY. The origin is a centre of similitude of the circle and its
inverse.
342 GEOMETRY.

Exampie 4. Any line through the origin of inversion cuts two inverse
loci at the same angle on opposite sides of the line.

Fis. 3, Fig. 2.

Let P and Q be two points on a locus, and let P’, Q’ be their inversey
with respect to O.
Then OP
UP =?
=0Q. 0Q’.
*. the pts. P, P’, Q, Q’ are concyclic:
LOQP=the LOP’Q’.
Now let Q move up to P, so that the st. line QP ultimately becomes
the tangent at P to the locus of P. Then at the same time the st. ling
(’P’ becomes the tangent at P’ to the locus of P’,
Hence in Fig. 2, if PR and P’R’ are the tangents at P and P’,
the LOPR=the LOP’R’,
that is, OPP’ cuts the loci of P and P’ at the same angle on opposws
sides of OPP’.

CornoLuaRy. At any pownt of intersection two curves cut at the some

ingle as their inverses at the inverse point. Also if two curves touch at \
their rnverses touch at the inverse point P’.

_ Exampie 5. To express the distance between two points in terms of

the distance between their inverses and the distances of these points from
the origin,
If P’, Q are the inverses of P, Q, [Fig. 1 of Example 4.]
OP. OP’=?=0Q . OQ’;
and fron: the similar triangles OPQ, OQ’P’,
PQ’ OF OP.OP ad
PQ ~0Q~ OP. 0Q~ OP. 0@
. oer
* PQ’=op-
oq
 INVERSION. 343

EXERCISES ON INVERSION,

!. If O, P, Q, R are collinear points and P’, Q’, R’ the inverses of

P, Q, R with respect to O, prove that
(i) if OP, OQ, OR are in Arithmetical Progression then OP’, OQ’,
OR’ are in Harmonical Progression.
(ui) if OP, OQ, OR are in Geometrical Progression, then OP’, OQ’,
OR’ are also in Geometrical Progression.
2. Find the inverse of the circum-circle of an isosceles triangle with
respect to the vertex of the triangle as origin.

3. Shew that a circle can be inverted into itself with respect to any
point O as origin.
[Take & equal to the length of the tangent from O.]

4. Shew that a circle inverts into itself with respect to the centre
of any orthogonal circle.
5. AB is a chord of a circle bisected at O. Shew that, with O as
origin, and OA as radius of inversion, the circle inverts into itself.

6. Shew that any two circles can be inverted into themselves.

[See Ex. 1. p. 336. Take the origin O on SP, and take & equal t«
the length of the tangent from O.] ;

7. Shew that any three circles can be inverted into themselves.

[See Ex. 2. p. 337.]

8. Shew that if a straight line cuts a circie, each may be inverted

imto the other by suitable selection of the origin and constant of
imversion.

9. Shew that any three circles may be inverted into three circles
whose centres are collinear.
[See Ex. 3, p. 339, and take the origin of inversion on the orthogonal
circle.]

10. Shew that the diameters of a circle may be inverted into a

geries of co-axal circles orthogonal to the inverse of the given circle,

11. P, Q, R are ‘three points taken in order on a straight line

Find the inverse of the statement
PQ +QR=PR.
Bad @ROMETRY.

iX. CEVA’S THEOREM,

If three concurrent straight lines are drawn from the angwar points af
a triangle to meet the opposite sides, then the product of three alternate
segments taken in order is equal to the product of the other three segments.

A A

B
ZS D Cc B Cc D

Let AD, BE, CF be drawn from the vertices of the AABC ta

intersect at O, and cut the opposite sides at D, E, F
It is required to prove that
BD.CE.AF=DC. EA. FB.
Now the A* AOB, AOC have a common base AO; and it may be
shewn, by drawing perpendiculars from B and C to AD, that
BD; DC=the alt. of A AOB: the alt. of A AOC:
. BD_AAOB
‘ DOA AOC’
ae} CE. ABCC.
similarly, EA=ABOA?

and AF _ ACOA
FB ACOB
Multiplying these ratios, we have
BD CE AF a
DG”.EA FR: **
or, BD.CE.AF=DC.
EA. FB.

Norg. The converse of this theorem, which may be proved

directly, is very important ; it may be enunciated thus:
If three straight lines drawn from the vertices of a triangle cut the
opposite sides so that the product of three alternate segments taken 1m
order is equal to the product of the other three, .then the three straight
snes are concurrent,
That is, ifBD. CE.
AF =DC. EA. FB,
then AD, BE, CF are concurrent.
 MENELAUS’ THEOREM. 345

X. MENELAUS’ THEOREM.

Lf a transversal is drawn to cut the sides, or the sides produced, of a

ivsangle, the product of three alternate segments taken wn order 1s equal te
te product of the other three segments.

Fig. 1. locas

Let ABC bea triangle, and let a transversal meet the sides BO, GA,
AB, or these sides produced, at D, E, F.
It ws required to prove that
BDACE AF=DOZEACrEB:
Draw AH par! to BC, meeting the transversal at *.
Then from the similar A* DFB, HAF,
A SAGs
FB BD"
aad from the similar As DCE, HAE,
CE CD
EA AH"
r Pe Ne eae CE; AF
-~CD.
*.5 by muitiplication, EA’ FR-BD?

' BD.CE.AF
EBA DC. EA. FB"
or, BD.CE.AF=DC.
EA. FS

Norr. In this theorem the transversal must either meet two sides
and the third side produced, as in Fig. 1; or all three sides produced,
as in Fig. 2. ;
The converse of this theorem may be proved indirectly :
If three points are taken in two sides of a triangle and the third side
produced, or in all three sides produced, so that the product of thres
alternate segments taken in order is equal to the product of the other thres
segments, the three points are collinear.
346 GEOMETRY.

DEFINITIONS.

1. If two triangles are such that the three straight lines

joining corresponding vertices are concurrent, they are said
to be co-polar.
2. If two triangles are such that the three points of inter-
section of corresponding sides are collinear, they are said te
06 co-axial.

EXERCISES.

1, By means of Ceva’s Theorem prove the following properties of 4

triangle.
(i) The perpendiculars to the sides from their middle points are
concurrent.
(ii) The bisectors of the angles are concurrent.
(iii) The medians are concurrent.

2. D,E, F are the points of contact of the in-circle of a triangls

with the sides BC, CA, AB respectively. If EF, FD, DE meet these
sides respectively in P, Q, R, shew that P, Q, R are collinear.
3. With the same notation as in Example 2, shew that the pointes
%, D, C, P form a harmonic range.
4. If the tangents at A, B, C of the circum-circle of the triangle
ABC meet the opposite sides in D, E, F respectively, shew that
BD : CD=BA?: AC’,
Hence prove that D, E, F are collinear.
5. The straight lines which join the vertices of a triangle to the pointe
of contact of the inscribed circle (or any of the three escribed circles) are
concurrent.

6. The middle points of the diagonals of a complete quadrilateral are

collinear, [See Def. 4, p. 327.]
7. Shew that each diagonal of a complete quadrilaieral is divided
varmonically by the other two diagonals.
8. Co-polar triangles are also co-axial ; and conversely co-axial tri-
angles are also co-polar,
9. The six centres of similitude of three circles lie three by three ow
Jour straight lines.
 PART’ VI.

SOLID GEOMETRY.

LINES AND PLANES.

DEFINITIONS AND FIRST PRINCIPLES. ©

1. From the Definitions of Part I. it will be remembered

that
(i) A point has no magnitude; that is to say neither length,
breadth, nor thickness.
(ii) A line has length, without breadth or thickness.
(i) A surface has length and breadth, without thickness.
(iv) A solid has length, breadth, and thickness.
Thus a point is said to be of no dimension ;
RIGO Mage hy cae cae 0c of one dimension ;
ASULTACE) wceoserone of two dimensions ;
BeSOUG wees cares of three dimensions ;

2. Solids, surfaces, lines, and points are thus related to

one another:
(i) Solids are bounded by surfaces.
(ii) Surfaces are bounded by lines; and surfaces meet in
lines.
(iii) Lines are bounded by points ; and lines meet in points.
It must also be noticed that a line which intersects a surface cuts it
in one or more points.

3. A plane is a surface such, that if any two points are

taken in it, the straight line joining them lies wholly in the
surface.
Unless the contrary is stated, the straight lines treated in this Section
are supposed to be of infinite length, and the planes of infinite extent.
348 GEOMETRY.

4. Lines which are drawn on a plane, or through which a

plane may be made to pass, are said to be co-planar.

5. Lines through which a plane cannot be made to pass aré

said to be skew.

6. Planes are said to be parallel when they'do not meet

though indefinitely extended.

7. <A straight line and a plane are said to be parallel when

they do not meet though both are indefinitely extended.

8. A straight line is perpen-

dicular to a plane when it is
perpendicular to every straight
line which meets it in that plane.
Such a straight line is said to be
normal to the plane.

AXIOMS.

1. A straight line drawn through two points on a plane must,

if indefinitely produced, lie wholly in that plane.

2. Through a given straight line, or through two given points,

an infinite number of planes may pass; for we may suppose a
plane to turn about any straight line that lies in it, and thus
to pass in succession through an infinite number of positions.

3. If a plane of unlimited extent turns about a straight line that

lies in it, it may be made to pass through any point in space outside
the given line.
 DEFINITIONS AND FIRST PRINCIPLES, 349

From these principles it follows that a straight line may be

related to a plane in three ways:
(i) It may be parallel to the plane, in which case it has no
point in common with the plane.
(ii) It may cut the plane, in which case it has one (and only
one) point in common with the plane.
(iii) It may lie in the plane, in which ease it has an indefinite
number of points in common with the plane.

Again, two straight lines may be related to one another in

three ways:
If the lines are co-planar, they may
(i) intersect one another,
or (11) be parallel.
If the lines are not co-planar,
p that is, if they yh are skew,
(iii) they neither intersect nor are parallel.

D (ej
For example, in the solid figure
represented in the margin, (i) the
edges AB, BC, in the plane ABCD,
intersect; (ii) the edges AB, DC in
the same plane are parallel ; (iii) the
edges AB, A’D’, through which no
plane can be made to pass, are skew,
and neither intersect nor are parallel,
350 GEOMETRY.

THEOREM 79. [Euclid XI. 2.}

One, and only one, plane may be made to pass through any two
intersecting straight lines.

Let the two given st. lines AB, CD intersect at E.

It is required to prove that one, and only one, plane may be made
to pass through AB, CD.
Proof. ‘Take any plane XY passing through AEB, and let
this plane turn about AB until it passes through C, taking the
position X’Y’. The position of the revolving plane is then
fixed, so that only one plane can pass through the st. line AB
and the point C.
And since E and C are points in this plane, the whole
at. line CED lies in it;
. one, and only one, plane can pass through AB and CD.
Q.E.D.

COROLLARY. <Any three straight lines, of which each pair cut

one another, must be co-planar.

INFERENCES,
Hence the position of a plane is fixed,
(i) if it passes through a given straight line and a given
point outside that line;
(ii) if it passes through two intersecting straight lines;
(iii) if it passes through three points not collinear ;
(iv) if it passes through two parallel straight lines.
 LINES AND PLANES. 351

THEOREM 80. [Huclid XI. 3.]

Two intersecting planes cut one another in a straight line, and in
no point outside it. . .

It is required to prove that the two intersecting planes PQ, XY cut

in a straight line, and in no povnt outside it.
Proof. Let A and B be points in both the planes PQ, XY;
then the straight line joining A and B lies wholly in both
planes ; that is, the planes intersect along the st. line AB.
And since the planes pass through the st. line AB, they can
have no point in common outside AB; for otherwise they
would coincide. Q.E.D.

Nore. It will now be seen that (i) if three or more concurrent

straight lines cut a given straight line, they are co-planar ;
(ii) if three or more parallel straight lines cut a given straight line,
they are co-planar.

’ The generation of a plane. Thus a plane may be generated by

(i) a straight line turning about a fixed point and sliding over a
fixed straight line ;
(ii) a straight line sliding over two fixed intersecting straight lines,
or two fixed parallel lines ;
(iii) a straight line moving parallel to itself and sliding over a fixed
straight line.

Triangles and Quadrilaterals in space. ‘The sides of every triangle

- must lie in one plane (Theor. 79); but the sides of a quadrilateral need
not lie in one plane, as may be seen by folding a plane quadrilateral
about either diagonal. A quadrilateral so formed is called skew or
gauche, two adjacent sides lying in one plane and the other two in
another plane.
352 GEOMETRY.

TuHorem 81. [Euclid XI. 4.]

If a straight line is perpendicular to each of two intersecting
straight lines at their point of intersection, it is also perpendicular
to the plane in which they lie. ;

Let AD be perp. to each of the st. lines AB, AC.

It is required to prove that AD is perp, to the plane XY which

passes through AB, AC.

In the plane XY draw any line AE through A; and in the

same plane draw BC to cut AB, AE, AC at the points B, E, C.
Produce CA to F, making AF equal to AD.
Join DB, DE, DC; and FB, FE, FC.

Proof. In the A* BAD, BAF,

because BA bisects DF at rt. angles,
’. BD= BF. Theor. 12. Cor. 2.
Similarly CD = OF.

Hence if the A BFC is turned about its base BC, until the ©
vertex F comes into the plane of the A BDC, then F must
coincide with D, for the A* BFC, BDC are identically equal.
.. EF coincides with ED; that is, EF =ED.
 LINES PERPENDICULAR TO A PLANE. S08

Hence in the A* DAE, FAE,

since DA, AE, ED=FA, AE, EF respectively,
”. the 2 DAE=the Z FAE;
«» DA is perp. to any line AE which meets it in the plane XY; -
that is, DA is perp. to the plane of AB, AC.
Q.E.D.

QUESTIONS AND EXERCISES.

1. ‘*Straight lines are parallel when they do not meet though

sndefiuitely produced.” Supply what is wanting to this definition,
justifying your answer by an illustration.

2. Give instances from the walls and edges of a room of

(i) parallel planes ;
(ii) lines parallel to a plane ;
(iii) lines perpendicular to a plane ;
(iv\ pairs of skew lines.

3. ‘Surfaces cut in lines.” Need these lines be straight? Give

instances to the contrary from any surfaces known to you.

4. How many straight lines may be perpendicular toa given straight

line at a given point
(i) in space of two dimensions?
(ii) in space of three dimensions?

5. Place two pencils so as to illustrate the fact that a straight line

is not necessarily perpendicular to a plane, if it is perpendicular to one
line in the plane.

6. Shew that three straight lines drawn from a point in space may
be so placed that each is perpendicular to the other two.
Prove that in this case each line is perpendicular to the plane of the
other two, and illustrate from the walls and edges of a room.

7. From O the centre of a circle a perpendicular OA is erected to

the plane of the circle. Shew that all points on the circumference are
equidistant from A.
B.8.G. :
354 ’ GEOMETRY.

THEOREM 82. [Euclid XI. 5.]

All straight lines drawn perpendicular to a given straight line at
a given point are co-planar.

Xx

Let each of the st. lines BC, BD, BE be perp. to the st. line
AB at the point B.
It is required to prove that BC, BD, BE are in one plane.

Proof. Let XY be the plane which passes through BC, BD;

and let HF be the plane which passes through AB, BE.
Suppose the planes HF, XY to intersect in the st. line BF.
Because AB is perp. to BC and BD,
*. AB is also perp. to BF which meets it in the plane of
BC, BD; Theor. 81.
Hence the 2* ABE, ABF are both right angles and in the
same plane HF ;
‘, BE coincides with BF.
That is, BC, BD BE are in the same plane XY.
Q.E.D.

CoroLiary. Jf a right angle revolves about one of its arms,

the other arm generates a plane. +

HyporueticAL Construction. Through any point in a

straight line a plane may be supposed to be constructed perpendicular
to the given line.
 LINES PERPENDICULAR TO A PLANE. 355

DEFINITIONS.

1. The direction of a plumb line hanging freely at rest is

said to be vertical.
2. Any plane perpendicular to a vertical line is said to be
horizontal.
-3. Any straight line drawn in a horizontal plane is said to
be horizontal.

QUESTIONS AND EXERCISES.

1. How many vertical straight lines can pass through a given point,
and how many horizontal lines?

2. A sheet of note-paper partly opened out is placed with two of its

short edges on a horizontal table ; why is the crease vertical ?

3. Shew that two observations with a spirit-level are sufficient to

determine if a plane is horizontal, provided that the two positions are
not parallel.

4 <A circle of radius 4°2 em. is drawn on a horizontal plane, and a

straight line OP of length 5°6 cm. stands vertically at the centre O.
Find the distance of P from any point on the circumference, shewing
that the distance is the same for all such points.

5. Two straight lines AB, CD bisect one another at O, and OP is

perpendicular to them both ; shew that
()}-PA=PBi5 Gi) sPE=RDis
and that this result is independent of the angle at which AB and CD
intersect.
If AB=3°6’, CD=1°4", and OP=2-4’,
find PA and PC.

6. ABCD is a horizontal square, and at O its mid-point (that is, the

intersection of its diagonals) a vertical rod OP is fixed, the extremity P
being joined by threads to the angular points of the square.
(i) Prove that PA, PB, PC, PD are all equal.
(ii) If each side of the square is 20 cm., and the height of the rod is
40 cm., find PA to the nearest millimetre.
(iii) If PA=85 cm., and the height of the rod is 75 cm., find a side
ot the square to the nearest millimetre.
356 GEOMETRY.

THEOREM 83. [Euclid XI. 8.]

If two straight lines are parallel, and if one of them is per-
pendicular to a plane, then the other is also perpendicular to the
same plane.

Let AB, FE be two par' st. lines, cutting the plane XY at

B and E; and let AB be perp. to the plane.
It is required to prove that FE is also perp. to the plane XY.
Join AE, BE; and through E draw CED in the plane XY
perp. to BE, making EC and ED of any equal lengths.
' Join BC, BD; also AC, AD.
Proof. Since EB bisects CD at right angles,
*. BC=BD. Theor. 12. Cor. 2,
And in the A* ABC, ABD,
since BC= BD, AB is common, and 2 ABC =z ABD,
for AB is perp. to the plane of BC, BD,
~ AC=AD. Theor. 4.
Again, in the A* CEA, DEA,
since CE=DE, EAis common, and AC=AD,
’, the ~CEA=the 2 DEA;
that is, CE is perp. to EA.
But CE is perp. to EB by construction ;
’, CE is perp. to the plane containing EA and EB.
And EF is in this plane; for both EA, EB lie in the plane
of the par" AB, FE.
’. CE is also perp. to EF.
 LINES PERPENDICULAR TO A PLANE. 357
Again since AB, FE are par’, and since by hypothesis the
L ABE is a rt. angle,
’, the 2 FEB is a rt. angle. Theor, 14.
Thus FE, being perp. both to EB and EO, is also perp. to the
plane XY which contains them.
Q.E.D.

Conversely, 7f AB and FE are both perpendicular to the plane XY,

they are parallel to one another.
With the same construction, prove, as before, that CE is
perp. to the plane of EA, EB.
Now it follows from the hypothesis that CE is perp. to EF;
. EF is in the plane of EA, EB.
But AB is also in the plane of EA, EB;
. ’, AB, FE are co-planar.
And since, by hypothesis, each of the Z°ABE, FEB is a rt.
angle,
[SAB AS pat tolbe
Q.E.D.

CoROLLARY. Jf AB is perpendicular to a plane XY, and iffrom

B, the foot of the perpendicular, a line BE 1s drawn perpendicular
to any line CD in the plane, then the join AE is also perpendicular
to CD.

Make EC, ED of any equal length

Join BC, BD; also AC, AD.
The proof follows as in the last
Theorem.

This important result is known as ‘' The Theorem of the Three

Perpendiculars.”
358 GEOMETRY.

THEOREM 84.

Through any point in or outside a plane there can always be one,

and only one, straight line perpendicular to the plane.

(i) Let A be the given point in

the plane XY.
Take BC any st. line through
A in the given plane.
Let AP, making a rt. angle with /
BC, revolve about BC as axis, X Fig.r.
Then AP traces out a plane perp.
to BC. Let this plane cut the plane XY in the st. line DAE.
Now as AP turns from AD to AE, it must pass through one
position in which it is perp. to DE, ic. perp. both to BC and
DE, and consequently perp. to the plane XY.

(ii) Let A be the given point

outside the plane XY.
Take BC any st. line in the
xy ; and let AD be the perp.
rom A on BC.
Draw DE perp. to BC in the
plane XY. Xx

Let AP be the perp. from A on DE.

Then AP must be perp. to the plane XY.

‘Proof. Through P draw PF par' to BC.

Now BC, being perp. to DA and DE, is perp. to the plane ADE;
hence FP is also perp, to the plane ADE. Theor. 83.
’, the LAPF is a rt. angle.
that is, AP is perp. both to PF and DE, and consequently to the
plane XY.
 LINES PERPENDICULAR TO A PLANE. > 359

(iti) There can be only one perpendicular to the plane XY from

the point P, whether this point is in the plane or outside it.

Fig.z.

For, if possible, suppose two perp* PA, PB to be drawn from

P to the plane XY, and let the plane through PA, PB cut the
plane XY in the straight line MN.
Then PA, PB are both perp. to MN and in the same plane
with it, which is impossible.
Q.E.D.

EXERCISES.
1. Given two set-squares, shew how a straight rod may be placed
perpendicular to a plane at a given point in it.

2. Shew how, with a set-square and straight rod, the construction

of Fig. 2. p. 358 may be practically ee to finding the position of a
straight line perpendicular to a plane XY from a point A outside it.

3. BC isa straight line drawn ona plane surface of a solid, and Aan
external point [see Fig. 2, p. 358]: it is required to find the position of
a line through A perpendicular to BC. Why is the usual method
(Problem 4, p. 74) here inapplicable as a practical construction?
Devise a feasible construction, with ruler and compasses only, for
finding the foot D of the required perpendicular AD. Hence shew how
a perpendicular may be drawn from A to the plane.
[For another construction of a line perpendicular to a plane from
® point outside it, see p. 361, Ex. 3.]
 860 - GEOMETRY.

THEOREM 85.
(i) Of all straight lines drawn from an external point to a plane,
the perpendicular ts the shortest.
(ii) Of obliques drawn from the given point, those which cut the
plane at equal distances from the foot of the perpendicular are equal,
A

(i) Let AB be the perpendicular, aiid AC any oblique, drawn

from the external point A to the plane XY.
It is required to shew that AB is less than AC.
Join BC.
Proof. Since AB is perp. to the plane XY, it is also
perp. to BC which meets it in that plane.
Hence in the A ABC, the 2 ACB is less than the 2 ABC ;
*, AB is less than AC. Theor. 10.

(ii) Let the obliques AC, AD cut the plane XY at equal

distances BC, BD from the foot of the perpendicular AB.
It is required to prove that AC = AD.

Proof. Since AB is perp. to the plane XY, it is also perp to

BC, BD which meet it in that plane.
Hence the A* ABC, ABD are congruent ;
for AB is common, BC = BD, and 2 ABC =Zz ABD,
*, AC=AD.
Q.E.D.
 LINES PERPENDICULAR TO A PLANE, 36)
’

EXERCISES.

1. Find the locus of the feet of equal obliques drawn from an external
point to a plane.

2. Pisa point outside the plane of the triangle ABC and equidistant
from its vertices. If the perpendicular from P to the plane cuts it at S,
shew that S is the circum-centre of the triangle ABC.
If the A ABC is rt. angled at C, and a=4°8", b=3°6", and SP=4'0";
find PA.

3. Deduce from the last example a practical method of drawing

a perpendicular from a given external point to a plane, having given
ruler and compasses, and a straight rod of any length greater than the
required perpendicular.

4. Give a geometrical construction for drawing a straight line

equally inclined to three straight lines which meet in a point, but are
not in the same plane. .

5. AB is a straight line in a plane XY, and PQ is the perpendicular

to the plane from a point P outside it.
(i) If QR is perpendicular to AB, shew that PR is also perpendicular
to AB.
Gi) If PR is perpendicular to AB, shew that QR is also perpendicular
to AB.

6. ABCD isa square on a side of 0°96 metre, and from Q its mid-
point a rod QP of length 1°40 m. is fixed perpendicular to the plane,
If R is the mid-point of the side BC, calculate the value of cos PRQ
correct to the third decimal figure.

7. AB is the line of section of two planes, and from P any point in

AB two lines PQ, PR are drawn perpendicular to AB, one in each plane,
Shew that the line drawn from any point in PQ perpendicular to the
plane containing it lies in the plane of PQ, PR.
8. Prove that:
(i) All points in space equidistant from two given points lie in a
plane.
(ii) All points in space equidistant from three non-collinear points
lie in a straight line (viz. the intersection of two planes).
(iii) There is only one point equidistant from fowr non-co-planer
points (viz. the intersection of two straight lines),
362 GEOMETRY.

THEOREM 86. [Euclid XI. 9.] f

Straight lines which are parallel to a given straight line are
parallel to one another.

Let AB, CD be each par' to the st. line PQ.

It is required to prove that AB, CD are par’ to one another.
Proof. Suppose XY to be any plane perp. to PQ.
Then because AB is par' to PQ,
‘, AB is perp. to the plane XY. Theor. 83,
And because CD is par' to PQ,
*, CD is perp. to the plane XY.
*, AB and CD, being both perp. to the plane XY are par' to
one another. : Theor. 83. Converse.
Q.E.D
Norz. This Theorem has already been proved [Page 40] in the
special case where AB, CD, and PQ are co-planar.

EXERCISES.
1. AB, CD, EF are three equal and peony straight lines not in one
plane: shew that the triangles ACE, BDF are congruent.
2. If the middle points of adjacent sides of a skew quadrilateral are
joined, prove that the figure so formed is a parallelogram,
3. If a triangle revolves about its base, shew that the vertex
describes a circle.
4. At O the mid-point of a regular hexagon drawn on a horizontal
lane, a normal OP of length 9°6 cm. is erected, and one side AB is
isected at X. If AB=4°0 cm., find the values of PA, OX, PX,
cos OAP, cos OXP; and shew that AB is perpendicular to PX.
 PARALLEL LINES IN SPACE. 363

‘THEOREM 87. [Huclid XI. 10.]

Lf two intersecting straight lines are respectively parallel to twa
other intersecting straight lines not in the same plane with them,
then the first pair and the second pair contain equal angles.

Let the st. lines AB, BC be respectively par’ to the st. lines
DE, EF, which are not in the same plane with them.
Lt is required to prove that ihe L ABC =the 2 DEF.
Make BA equal to ED; and make BC equal to EF.
Join AD, BE, CF, AC, DF.
Proof. Because BA is equal and par' to ED, |
", AD is equal and par' to BE. Theor. 20.
Similarly CF is equal and par' to BE.
_ Hence AD and OF, being each equal and par' to BE, are equal
and par’ to one another ; Theor. 86.
*, AC is equal and par' to DF.
Then in the A* ABC, DEF,
pecause AB, BC, AC=DE, EF, DF, respectively,
*, the .ABC=the Z DEF. Theor. 7.
Q.E.D.
364 GEOMETRY.

THEOREM 88. [Euclid XI. 14.]

Planes to which the same straight line is perpendicular are
parallel to one another.

Let the straight line AB be perp. to each of the planes XY, PQ.
It is required to prove that the planes XY, PQ are parallel.
Proof. Since AB is perp. to the plane XY, it is perp. to the
line joining A to any point in that plane.
Similarly AB is perp. to the line joining B to any point in
the plane PQ.
Hence if the planes XY, PQ had a point in common, then,
by joining this point to A and B, two perpendiculars could be
drawn from it to AB, one in each plane ; which is impossible.
Thus the planes XY, PQ have no point in common, and
are therefore parallel.
Q.E.D.

EXERCISES.

1. AB,CD are normal to a plane, cutting itat Band D. If AB,CD

are equal in length and on the same side of the plane, shew that ABDC
is a rectangle.

2. Use the last Example to find the locus of points equidistant from
a given plane.

3. Find the locus of points equidistant from two given points.

 PARALLEL PLANES. 365

THEOREM 89. [Euclid XI. 16.]

Tf two parallel planes are cut by a third plane, their lines of
section with it are parallel.

Let the par' planes AB, CD be cut by the plane XY in the

lines of section EF, GH.
It is required to prove that EF, GH are parallel.
Proof. Now EF and GH cannot meet, since they lie respec-
tively in the planes AB, CD, which have no point in common.
Moreover EF and GH are co-planar, since they are in the
given plane XY.
EF and GH are parallel. Q.E.D.

EXERCISES.

1. Through a given point there can only be one plane parallel to a

given plane.
2. Prove that a straight line which is normal to one of two parallel
planes is also normal to the other.
3. Shew that planes which are parallel to the same plane are
parallel to one another.
4, Prove that intercepts of parallel lines between parallel planes
are equal.
5. Given two pairs of parallel] planes, how many lines of section will
they have? Shew that these lines are parallel.
366 GEOMETRY.

THEOREM 90. [Euclid XI. 15.]

If two intersecting straight lines are parallel respectively to two
other intersecting straight lines which are not in the same plane with
them, then the plane containing the first pair is parallel to the vlane
containing the second pair.

Let the st. lines AB, BC be respectively par' to the st. lines
DE, EF, which are not in the same plane with them.
It is required to prove that the plane of AB, BC is parallel to the -
plane of DE, EF.
From B let BG be drawn perp. to the plane of DE, EF, and
meeting it at G.
Draw GH, GK par' respectively to ED, EF.
Proof. Because BG is perp. to the plane of DE, EF,
", each of the 2° BGH, BGK is a rt. angle.
Now by hypothesis BA is par' to ED,
and by construction GH is par' to ED ;
‘, BA is par' to GH. Theor. 86,
And since the 2 BGH is a rt. angle ;
.. the 2 ABG is a rt. angle.
Similarly the 2 CBG is a rt. angle.
’, BG is perp. to the plane of AB, BC.
And, by construction, BG is perp. to the plane of ED, EF;
‘, these planes are par’, - Theor, 88,
Q.E.D.
 PARALLEL PLANES. 367

THEOREM 91. (Euclid XI. 17.]

Straight lines which are cut by parallel planes are cut pro-
portionally.

Let the st. lines AB, CD be cut by the three jee sis
GH, KL, MN at the pointsA, E, B, and©, F, D.
It is required to prove that AE: EB=CF : FD.
Join AC, BD, AD; and let AD meet the plane KL at the
point X : join EX, XF.
Proof. Because the two par' planes KL, MN are cut by the
plane ABD,.
", the lines of section EX, BD are par’. Theor. 89
And because the two par’ planes GH, KL are cut by the
plane DAC,
’, the lines of section XF, AC are par’.
Now since EX is par’ to BD, a side of the A ABD,
eRe EB AXCaXD: Theor. 60.
Again because XF is par' to AC, a side of the A DAC,
- AX: XD = CF #FD:
Hence AE : EB=CF : FD.
Q.E.D.
EXERCISES.
1. If two intersecting planes are cut by two parallel planes, the
lines of section of the first pair with each of the second pair contain
equal angles. Point out an exceptional case.
2. Shew how to determine in a given straight line the point which
is equidistant from two fixed points, When is this impossible?
368 GEOMETRY.

DEFINITION. The projection of a line on a plane is the |

locus of the feet of perpendiguaa’drawn from all points in~
the given line to the plane.

Thus in the adjoining figure the

projection of the line AB on the
plane PQ is the line ab. ,

THEOREM 92.
The projection of a straight line on a plune is itself a straight
line.

Let AB be the given st. line, and XY the giver. plane ;and
from P, any point in AB, let Pp be drawn perp. to the plane XY
and meeting it at p.
It is required to shew that the locus of p is a straight line.
Suppose Aa and Bb to be the perps. from A and B to the
plane XY.
Proof. Now A, Pp, Bb, being all perp. to the plane XY, are
par' to one another. Theor, 83, Converse.
And since these par" all cut AB, they are co-planar.
.. the point p is in the line of section of the planes Ad, XY;
that is, p is in the st. line ab.
But p is any point in the projection of AB,
.. the projection of AB is the st. line ab. QED
 PROJECTION OF A LINE ON A PLANE. 369

CoROLLARY 1. The angle which a straight line AB makes

with a plane XY may be measured by the angle between AB and its
projection ab on the plane ; for a straight line and its projection
are co-planar.

Thus, if AB and ab, produced if

necessary, meet at O, the angle
between AB and the plane XY is
measured by the angle BOb.

CoROLLARY 2. To find the length of the projection of a straight

line AB on a plane XY in terms of AB and the angle which wt makes
with the plune.
Let a be the angle which AB makes with the iis XY.
Draw Ad’ par’ to ab, cutting Bb at J’.
Then the 2 BAS’ =the corresponding Z BOb=a.
Now from the rt. angled A BAU’, = = COS @.;
hence ab=Ab’=AB cosa.
[See p. 272, Ex. (ii).]
Note. Since as a increases from 0 to 90°, cosa decreases, it follows
that as the inclination of AB to the plane increases, the projection ab
decreases.
EXERCISES.
1. Ifa straight line is parallel to a plane, shew that it is parallel to
its projection on 1 that plane.
2. Compare the length of AB with that of its projection on the
plane XY, when AB
(i) is parallel to the plane;
(ii) is perpendicular to the plane ;
(iii) is inclined to the plane at an angle of 60°.
3. Shew that equal obliques drawn toa plane from an external point
have equal projections on that plane.
4. Prove that parallel straight lines have parallel projections on a
plane. Is there any exception to this?
5. If A’B’, C’D’ are the projections of two parallel straight lines
AB, CD on any plane, shew that
ABsGD=A'B’ :C’D’;
H.8.G. 2A
370 GEOMETRY.
»*

THEOREM 93.
Tf a straight line outside a given plane is parallel to any straight
line drawn on the plane, it is also parallel to the plane itself.

Xx
Let AB be par' to CD drawn on the plane XY.
It is required to prove that AB is par" to the plane XY.
Proof. Suppose AD to be the plane of the par’ AB, CD; so
that CD is the line of section of the planes AD, XY.
Then AB, being in. the plane AD, must meet the plane XY, if
at all, at some point in CD.
But, by hypothesis, AB can never meet CD ;
‘*, AB can never meet the plane XY, and is consequently
par' to it. Q.E.D.

Conversely. Jf a straight line is parallel to a plane, then any

plane passing through the given line and intersecting the given plane,
will cut it in a line parallel to the given line.
In the above Figure let the st. line AB be par’ to the plane
XY, and let AD, any plane through AB, cut the plane XY in
the line CD.
It is required to prove that CD is par’ to AB.
Proof. Now AB, being par' to the plane XY, can never meet
CD which lies in that plane.
Moreover AB and CD are in the same plane AD ;
*, AB and CD are par’. Q.E.D,
 LINES IN SPACE. 371

CoroLuaRy. Through either of two skew straight lines a plane

may be made to pass to which the other line is parallel.

Let AB and CD be two skew

st. lines, that is, two lines which
are not co-planar.
Through any point O in AB let
cOd be drawn par! to CD. Then
AB and cd determine a plane XY,
to which CD is par!; for CD is
par! to the line cd which lies in
that plane.

DEFINITION. The angle between two skew straight lines

is measured by the angle contained by one of them and
the line drawn through any point in that line parallel to
the other.
Thus, in the above diagram, the angle between the skew lines AB,
CD is measured by the angle contained by AB and the line cd drawn
through any point O in AB parallel to CD.

EXERCISES.

1. Ifa straight line AB is parallel to a plane XY, then

(i) Every line parallel to AB is also parallel to the piane ;
(ii) Lvery line parallel to the plane 1s also parallel to AB.
Which of these statements is true, and which false?

.2. A straight line AB revolves about the point A, keeping always

parallel to a given plane XY. What surface does AB generate?

3. If two intersecting planes pass respectively through two parallel

lines AB, CD, shew that their line of section is parallel to AB, CD.

4. A straight line PQ is parallel to each of two intersecting planes ;

shew that PQ is also parallel to their line of section,

5. Shew that through a given point P a plane may be constructed

parallel to each of two skew lines AB, BC.

6. Shew that through any two given skew lines two parallel planes
may be passed, one through each line.
372 GEOMETRY

THEOREM 94.
If two straight lines neither intersect nor are parallel, then
(i) there is one straight line perpendicular to both of them;
(ii) this common perpendicular is the shortest distance between the
given lines.

Let AB and CD be the two given skew lines.

(i) Lo prove that there is one line perp. both to AB and CD.
Through E, any point in AB, let EF be drawn par' to CD,
and let XY be the plane of AB, EF.
Suppose the projection of CD on the plane XY to be QK,
cutting AB at Q: and let P be the point of which Q is the pro-
jection.
Then PQ will be perp. both to AB and CD.
Proof. Now CD, being par' to EF, is par' to the plane XY ;
', CD is also par' to its projection QK. Theor. 93.
And because PQ is perp. to the plane XY,
‘, each of the 2* PQB, PQK is a rt. angle ;
. the 2 QPD is a rt. angle ; Theor. 13.
that is, PQ is perp. both to AB and CD.

(ii) Zo prove that PQ is the shortest distance between CD and AB.

Let HE be any other st. line drawn from CD to AB; and let
HK be the perp. from H to the plane XY.
Then the perp. HK is less than the oblique HE. Theor. 85
.. PQ, which is equal and par' to HK, is also less than HE.
Q.E.D.
 DIHEORAL ANGLES. ale

DEFINITIONS.

1. When two planes meet along a line of section they are

said to form a dihedral angle.

2. A dihedral angle is measured by the plane angle con-

tained by two straight lines drawn from any point in the
line of section at right angles to it, one in each plane.

Thus in the adjoining Figure, AB is the

line of section of the two intersecting planes
BC, AD; and from Q, any point in AB, st.
lines QP, QR are drawn perpendicular to AB,
one in each plane.
Then the dihedral angle formed by the two
planes is measured by the angle PQR.

Notes. (i) This definition assumes that whatever point Q is taken

in AB, the angle PQR is of constant magnitude: the truth of which
follows readily from Theorem 87.
For let ML, MN be drawn perp. to AB, one in each plane, from any
other point M in AB.
Then clearly, LM, MN are respectively par! to PQ, QR;
‘. the LLMN=the ZL PQR.
(ii) Since AB is perp. to PQ and QR, it is perp. to the plane con-
taining them. Hence the dihedral angle between two given planes BC,
AD may be determined by cutting them with any plane perp. to AB
their line of section.

3. Two planes are perpendicular to

one another when the dihedral angle
formed by them is measured by a right
angle.
374 GEOMETRY.

THEOREM 95. [Euclid XI. 18.]

If a straight line is perpendicular to a plane, then any plane
passing through the perpendicular is also perpendicular to the given
plane.

Let the st. line PQ be perp. to the plane XY, and let CB be
any plane passing through PQ.
It is required to prove that the plane CB is perpendicular to the
plane XY.
Proof. Let QR be drawn in the plane XY perp. to AB the
line of section of the given planes.
Then PQ, being perp. to the plane XY, is perp. to QB and QR.
Thus the 2 PQR is art. angle ; moreover the 2 PQR measures
the dihedral angle, since PQ, QR are both perp. to the line of
section AB.
‘, the plane CB is perp. to the plane XY.
Q.E.D.

CoROLLARIES. With the same construction, it may be

shewn conversely that:
(i) If the two planes CB, XY are perpendicular to one another,
then any line PQ drawn in the plane CB perpendicular to the line of
section AB will also be perpendicular to the plane XY.
(ii) If the plane CB is perpendicular to the plane XY, and from
any point P in the first plane a perpendicular PQ is drawn to the
second, then PQ is contained in the plane CB.
 DIHEDRAL ANGLES. 375

THEOREM 96. [Euclid XI. 19.]

If two intersecting planes are each perpendicular to a third plane,
their line of section is also perpendicular to that plane.

: i4 i haane ‘
ir
a
ha
i Y
= |

S avoS
a
| 2 1

Let each of the planes AB CD, whose line of section is PQ

be perp. to the plane XY.
It is required to prove that PQ is perp. to the plane XY.
Proof. If from any point P, common to the planes AB, CD,
a perpendicular is drawn to the plane XY, then this per-
pendicular must lie in each of the planes AB, CD, for each
plane is perp. to XY. Theor. 95. Cor, (ii).
Hence the perpendicular must coincide with PQ the line of
section ; that is, PQ is perp. to the plane XY. Q.E.D.

EXERCISES.
1. Through any straight line a plane can be passed perpendicular
to a given plane.
2. Prove that a straight line makes equal angles with any parallel
planes which it cuts.
3. Ifa plane cuts two parallel planes, the corresponding dihedral,
angles are equal.
4. ABCD represents the floor of a room, and A’B’C’D’ its ceiling.
If the length AB=7°50 metres, the breadth AD=6-00 metres, the
height AA’=4°50 metres, find the cosine of the dihedral angles between
(i) the plane ABC’D’ and the floor;
(ii) the plane AB’C’D and the floor.
5. P is a point 2 feet vertically above the in-centre of a horizontal
square ABCD. If AB=1 ft. 2 in., find the cosine of the dihedral angle
. between the plane PAB and the plane of the square.
376 GEOMETRY.

SOLID ANGLES.

1. When three or more planes taken in order intersect,

each with the next, in lines which meet at a point, they are
said to form a solid angle. The point of concurrence is called
the vertex ; the lines of intersection of consecutive planes are
the edges of the solid angle; the angles between consecutive
planes are its dihedral angles ; and the plane angles formed by
consecutive edges are its face-angles.

Thus the planes ASB, BSC, ..., cutting consecutively in the con-
current edges SB, SC, ..., form a solid angle at the vertex S. The
solid angle is denoted by (S, ABCDE), or by the single letter S.

2. A solid angle formed by three concurrent planes is said

to be trihedral ; if formed by more than three, it is said to be
polyhedral. The face angles and dihedral angles of a trihedral
angle are known as its six parts.

3. ‘Two solid angles are identically equal when one can be

superposed upon the other, that is to say, exactly fitted into
it; in which case the face-angles of the first solid angle are
severally equal to the eceangioe of the other, and the dihedral
angles of the first to the dihedral angles of the other, these
parts being taken in order the same way round.
 SOLID ANGLES. Sit

Note 1. The necessity for the

last condition may be seen on
comparing a solid angle with that
determined by producing its edges
through the vertex.

Here the solid angles (S, ABCD), (S, A’B’C’D’) have the face-angles and
dihedral angles of one severally equal to the face-angles and dihedral
angles of the other in the order indicated by the letters. But to an
observer looking into each solid angle in turn the above sequence of
letters will appear clock-wise in the first case, and counter-clockwise in
the second. Hence the two solid angles, though equal as to their
several parts, cannot be fitted one into the other, and are therefore
not identically equal. Solid angles so related are said to be symmetrical.

Nore 2. Jf two trihedral angles (S, ABC), (S’, A’B’C’) have the three
face-angles ASB, BSC, CSA in one equal respectively to the three face-
angies A’S’B’, B’S’C’, C’S’A’ in the other, the dihedral angles of the first
will be equal to the corresponding dihedral angles of the other.
S s
Make SA, S’A’ of any
equal lengths.
In the planes ASB, ASC
draw AB, AC each perp. to
SA.
In the planes A’S’B’,
A’S’C’ draw A’B’, A’C’ each
perp. to S’A’.

Then the Zs BAC, B’/A’C’ measure corresponding dihedral angles.

Join BC, B’C’.

Outline of Proof. Prove the following pairs of triangles congruent:

(i) the As SAB, S’A’B’; (Theor. 17.)
(ii) the As SAC, S’A’C’; (Theor. 17.)
(iii) the A® BSC, B’S’C’; (Theor. 4.)
(iv) the As BAC, B’A’C’; .( Theor. 7.)
bence the 2BAC=the LB/A’C’.

Similarly the remaining dihedral angles may be proved equal; and if

the sequence of equal angles goes the same way round in each case, the
two trihedral angles are identically equal.
378 GEOMETRY.

THEOREM 97. [Euclid XI.- 20.]

In a trihedral angle the sum of any two of the face-angles is
greater than the third.
S

B Cc
agente, Mee wil

Let (S, ABC) be a trihedral angle, of which ASB, BSC, CSA are
the face-angles ; and of these let the 2 BSC be the greatest.
It is enough to prove that
the sum of the L* ASB, ASC is greater than the 2 BSC.
In the plane BSC make the 2 BSD equal to the 2 BSA; and
cut off SD, SA of any equal lengths.
In the plane BSC draw any st. line through D cutting SB, SC
at B and C; and join AB, AC.
Proof. Then in the A* BSA, BSD,
since BS, SA are equal to BS, SD respectively,
and the 2 BSA=the z BSD,
*, BA=BD. Theor. 4.
Now from the A BAC,
BA+AC is greater than BC,
that is, greater than BD + DC;
", AC is greater than DC.
Again, in the A* ASC, DSC,
since AS, SC=DS, SC respectively,
but AC is greater than DC,
*, the LASC is greater than the 2DSC. Theor. 19. Con.
’, the sum of the 2* ASB, ASC is greater than the sum of the
L’ BSD, DSC,
that is, greater than the 2 BSC.
Q.E.D.
 SOLID ANGLES 379

EXPERIMENTAL ILLUSTRATION OF THEOREM 93.

S

To construct a solid angle, draw

the three face-angles ASB, BSC,
CSA’ in the same plane, placing the
greatest angle BSC between the
other two.

Suppose the diagram cut out, and folded about SB and SC, witha
view to bringing SA and SA’ into coincidence.
Now (i) if LBSA+ ZCSA’ is less than L BSC,
then SA and SA’ cannot be brought together, and therefore no solid
angle can be formed ;
(ii) if LBSA+ LCSA’= LBSC,
then SA and SA’ can be made to coincide, but in the plane of the £BSC,
so that no solid angle can be formed ;
(iii) if LBSA+ CSA’ is greater than 2 BSC,
then SA and SA’ would overlap when brought into the plane BSC, so
that they can be made to coincide outside that plane, and a solid angle
can consequently be formed.

EXERCISES.

1. Prove that in general three planes meet at a point. Indicate

three exceptional cases.
2. Shew that the sum of the angles of a skew quadrilateral is less
than 360°.
3. OA, OB, OC are three straight lines drawn from a given point
O not in the same plane, and OX is another straight line within the
solid angle determined by OA, OB, OC: shew that
(i) the sum of the angles AOX, BOX, COX is greater than half
the sum of the angles AOB, BOO, COA.
(ii) the sum of the angles AOX, COX is less than the sum of the
angles AOB, COB.
(iii) the sum of the angles AOX, BOX, COX is less than the sum
of the angles AOB, BOC, COA.
380 GEOMETRY.

THEoREM 98. [Euclid XI. 21.]

In a convex solid angle the sum of the face angles is less than feur
right angles.

Let (S, ABCDE) be a convex solid angle.

It is required to prove that the sum of the face-angles ASB, BSC,
CSD, DSE, ESA is less than four right angles.
Let a plane XY cut the planes of the face-angles in the lines
AB, BC, CD, DE, EA, which form the convex polygon ABCDE.
Within the polygon ABCDE take any point O, and join OA,
OB, OC, OD, OE.
Proof. In the trihedral angle A,
LSAB+ZSAE is greater than ZEAB, Theor. 97.
that is, greater than 2 OAE + 2 OAB.
Similarly for each of the angular points B, C, D, E.
Hence the sum of the base 2" of the A* with vertex $ is
greater than the sum of the base 2° of the A* with vertex O.
And since these two sets of A* are equal in number, the sum
of all the 2* of one set is equal to the sum of all the z* of the
other.
It follows that the sum of the angles at S is less than the
sum of the angles at O.
But the sum of the angles at O is 4 rt. 2";
’, the sum of the angles at S is less than 4 rt. 2".
 SOLID ANGLES. 381

EXERCISES.

(Miscellaneous )
1. Shew that the angle made by an oblique to a plane with its
projection ts less than that which it makes with any other line which meets
wt in that plane.
2. Through any point on an inclined plane shew how to draw on that
plane the LINE OF GREATEST SLOPE (that is, the line making the greatest
angle with the horizontal plane).
3. O is a fixed point ina plane, and Pa fixed point outsideit. Find
the locus of the feet of perpendiculars drawn from P to all lines in the
plane through O.
4. From a point A two normals AP, AQ are drawn one to each of
two intersecting planes: shew that
(i) the line of section is perpendicular to the plane of AP, AQ;
(ii) the dihedral angle between the two planes is equal or supple-
mentary to the angle between the normals.
5. -If AB and CD are two skew lines, shew that the joining lines
AC, BD are also skew.
On what planes will the projections of the skew lines AB, CD be
parallel?
6. Shew that through any point in space there is always one straight
line which cuts each of two given skew lines.
7. OA, OB, OC are three concurrent lines, each of which is perpen-
dicular to the other two; then
(i) if OX, OY, OZ are perpendicular to BC, CA, AB respectively,
shew that XYZ is the pedal triangle of the triangle ABC ;
(ii) if OP is perpendicular to the plane of ABC, shew that P is the
orthocentre of the triangle ABC.

8. ABCD is an inclined plane; AB, CD

horizontal lines in it; AD, BC lines of
greatest slope in_the plane ; AF, BE the
projections of AD, BC on the horizontal
plane.

If _CBE=a, LDAC=8, LCAE=0, LFAE=4, prove the following

relations :
(i) sin 0=sinacos; (ii) tan @=tan Bseca;
(iii) tan a=tan 6 sec ¢; (iv) sin 8=sin ¢ cos @.
382 GEOMETRY.

NOTE ON FIXING THE POSITION OF A POINT IN SPACE

BY MEANS OF AXES OF REFERENCE.

Ni
Let OX, OY be two fixed st. lines ne at right angles at the
origin O. Draw OZ perp. to the plane of OX, OY. Then each of the
lines OX, OY, OZ is perp. to the other two, and consequently to the
plane determined by the other two. The lines OX, OY, OZ are taken
as axes of reference, and the position of any point P with regard to them
is fixed in the following manner.
Let N be the projection of P on the plane XOY; let OM, MN be the
coordinates of N with reference to OX, OY; and let a, y, z denote the
lengths of OM, MN, NP. Then OM, MN, NP, taken together, are
said to be the coordinates of P; the point P is denoted by (a, y, z), and
its position is known if the numerical values of a, y, and z are given.
Exameue. Plot the point whose coordinates are 5, 3, and 4.
First, with reference to the axes OX, OY, plot the point whose
coordinates are 5, 8: call this point N, and draw NP perp. to the
plane XOY, making NP 4 units of length. This gives the position of P.
Tt will be seen that the coordinates of a point P are its distances from
the three planes of reference YOZ, ZOX, xoY. Now these planes divide
space into eight regions, and in each of these there is a point whose
distances from the planes are 5, 3, and 4. The coordinates of these
points are distinguished by the use of signs on principles analogous to
those explained on page 133.
Lines measured along or parallel to the axes OX, OY, OZ, in the
senses indicated by the letters, are positive. Lines-measured in these
directions but in senses opposite to the above are negative.

EXERCISES.
1. Draw diagrams to indicate the positions of the following points:
(i) (8, 5, 4); (ii) (-5, 4, 3); (iii) (-8, —4, 5); (iv) (4, 5, -3).
¥. The coordinates of P are 6, 8, 10; find the coordinates of Q, the
mid-point of OP.
3. If P is the point (x, y, z), shew that OP?=2
+4? +22,
Find OP when P is given by (3, 4, 12).
 383

SOLID FIGURES.
DEFINITIONS AND PRELIMINARY THEOREMS.
1. If a plane figure is
cut by a system of equi-
distant parallel lines, and
if the rectangleiscompleted
between each pair of con-
secutive parallels, as shewn
in the diagram, then the
area of the external recti-
lineal figure may be made
to differ from the area of .
the given figure by as little as we please by indefinitely
diminishing the width of the strips; that is, the area of the
given figure may be regarded as the limit of the sum of all such
rectangular strips, when theur breadth is diminished indefinitely.
2. If AB is any plane figure, and ab its projection on a
second plane making an angle 6 with the first, then
area of the projection ab = (area of the given fig. AB) x cos 0.
For the fig. AB may be
divided into narrow strips
by parallels drawn perpen-
dicular to the section LM
of the planes. Let PQRS
be any such strip, and pgqrs
its projection.
Then the strips pqrs,
PQRS, when taken very
small may be regarded as
rectangular and as having
the same breadth (viz. P'S’),
and pq the length of the
former, is PQ cos 0;
’, area of strip pars= (area of PQRS) x cos @.
And this is true for every pair of corresponding strips in
the limit when their breadth is diminished indefinitely ;
*, area of fig. ab=(area of fig. AB) x cos @.
384 GEOMETRY.

3. A solid figure, or solid, is any portion of space bounded

by one or more surfaces, plane or curved.
These surfaces are called the faces of the solid, and the intersections
of adjacent faces are called edges.

4. A polyhedron is a solid bounded by plane faces.

Norn. A plane rectilineal figure must at least have three sides; or
four, if two of the sides are parallel. A polyhedron must at least have
four faces ; or, if two faces are parallel, it must at least have five faces.

5. A parallelepiped is a solid bounded by three pairs of

parallel plane faces.

Fig.r.

Since in Fig. 2 the parallel planes ABCD, A’B’C’D’ are cut by the
plane ABA’B’, the edges AB, A’B’ are parallel [ 7'heor. 89].
Thus it may be shewn (i) that each of the six faces of a parallelepiped
is a parallelogram ; (ii) that opposite faces are congruent ; (iii) that the
twelve edges fall into three groups, each group of four edges being
equal and parallel to one another.

6. The four diagonals of a parallelepiped are concurrent and

bisect one another.
Let AC’, BD’, CA’, DB’ be the diagonals of the parallelepiped
(ABCD, A’B‘C'D’),
Join BD, B’D’.
Then since BB’, DD’ are equal and par’, the fig. BDD’B’ is a par™ ;
*, its diagonals BD’, DB’ bisect one another:
that is, BD’ passes through O the middle point of DB’.
Similarly, by joining DA’, BC, it may be shewn that A’C
passes through O the middle point of DB.
In the same way AC’ is also bisected at O.
 SOLID FIGURES. PRELIMINARY THEOREMS. 385

7. <A parallelepiped whose faces are rectangular is called a

euboid, or rectangular solid.
If the faces are all squares the parallelepiped becomes a cube

Since in the above Figures, the 2*COA, COB are rt. Z°,
.. OC is perp. to the face AB.
Similarly each edge is perp. to the two faces which it cuts;
and each face is perp. to the four faces which it cuts.

8. The square on a diagonal of a rectangular solid is equal to

the sum of the squares on three concurrent edges.
Let OP be a diagonal of a cuboid, in which three con:
current edges OA, OB, OC measure a, ), and ¢ units of length.
Join OQ.
Then PQ, being perp. to the face AB, is perp. to OQ.
OP? = 0Q? + PQ?=0Q? +0.
But OQ?=0OA?+AQ?=a?+4+0?, for OAQ is a rt. «4.
OP? = a? + 67 + 2,

CoROLLARY 1. The diagonals of a cuboid are equal.

Corotiary 2. Jf a denotes cach edge of a cube ; then

(diagonal)? =3a?; — .”.. diagonal = a,/3.

Nors. If (x, y, z) are the coordinates of a point P, then, O being

the origin,
OP?=27
+ y#+23,
H.8.G 2B
 ~~”

388 GEOMETRY.

13. (i) Any plune section of a pyramid taken parallel to the

base is similar to the base.
(ii) Zhe area of such a section varies as the square of tts
distance from the vertex.

In the pyramid (S, ABCD), let \

abed be a plane section par’ to the c
base ABCD. a

(i) Because the plane aJAB meets Eat

the parallel planes abcd, ABCD, the
iines of section wb, AB are parallel. Z
B

Similarly bc and BC, cd and CD, du and DA are parallel ;

‘, corresponding angles of the figs. abed, ABCD are equal.
ab be cd da,
And, from similar A‘, —— Se SS
AB BC CD DA

hence the figs. abed, ABCD are similar.

(ii) Let the perp. drawn from S to the base, meet the
sections abed, ABCD at 2 and X. Join az, AX.
Then _ fig. abed : fig. ABCD
=al” : AB? Theor. 73.
= Sa* : SA’, by similar A’,
we ORs GRA. svivevoncvs sieve

CoroLuary. If two pyramids stand on bases of equal area and

have equal altitudes, then plane sections taken in each pyramid
parallel to the base and at the same distance from the vertex are
equal in area.
 SOLID FIGURES. 389

es
EXERCISES.

1. A square sheet of galvanized iron, each side being 12 feet, rests

against a wall, and is inclined to the horizon at an angle of 60° ; what
area of ground will if, protect from a vertical rain ?

2. (i) In the rectangular solid re-

presented in the margin, if OA=12 cm.,
OB=9 cm., OC=8 cm., find the values
of OP, cos QOP, area of tig. OAPR.

(ii) If OP makes with OA, OB, OC the angles a, f, y respectively

shew that
cos*a + cos*B + cos*y =1 ;
and verify this result with the dimensions given above.
(iii) What plane passing through OP is parallel to BQ? If
OA=a, OB=b, OC=c, shew that the shortest distance between OP
and BQ is be/Vb? +c.
3. Ifa parallelepiped is cut by a plane which intersects two pairs of
opposite faces, shew that the lines of section form a parallelogram.
4. Shew that the polygons formed by cutting a prism by parallel
planes are identically equal. ;
5. If each edge of a tetrahedron is equal to the opposite edge, shew
that the sum of the plane angles in each corner is 180°.
6. Two planes cut at an angle of 45°. On one plane a circle of
radius 5 cm. is drawn, and this circle is projected on the other plane.
Find (i) the length of the greatest chord, (ii) the area of the pro-
jection.
7. If a tetrahedron is cut by any plane parallel to two opposite
edges, the section will be a parallelogram.
8. Prove that the shortest distance between two opposite edges of
a regular tetrahedron is one half of the diagonal of the square on au
edge.
9. Ina tetrahedron if two pairs of opposite edges are at right angles,
then the third pair will also be at right angles.
10. Ina tetrahedron whose opposite edges are at right angles, the
sum of the squares on each pair of opposite edges is the same.
390 GEOMETRY.

SURFACES AND VOLUMES.

14. The volume of a solid is the amount of space contained

within its bounding surfaces.
A cubic inch is the volume of a cube each of whose edges is one inch
in length. Similarly a cubic centimetre is the volume of a cube of
which each edge is one centimetre in length. Thus the unit of volume
is the volume of a cube on an edge of unit length.

15. To find the surface and volume of a rectangular solid.

Fig.2. mA

Surface. In the cuboid represented in Fig. 1, let the

length AB=a units, the breadth AC=0 units, and the height
AD=c units.
Then the whole surface is the sum of the three pairs of
opposite and equal rectangular faces.
Now the faces DE, DB, DC contain respectively ab, ac, be,
units of area ;
. whole surface of cuboid = 2ab+ 2ac + 2be units of area.
If a=b=c, the rectangular solid becomes a cube on an edge
of a units, and whole surface of cube= 6a? units of area.

Volume. Consider a cuboid whose length AB is 5 inches,

breadth AC 4 inches, height AD 3 inches. Fig. 1 shews that
the solid may be divided into 3 equal slices, each 1 inch thick.
And each slice may be subdivided (as in Fig. 2) into cubical
blocks whose edges are 1 inch, that is, into cubic inches.
Now the number of cubic inches in one slice is 5x4; so
that the number of cubic inches in the whole solid is 5 x 4 x 3,
or 60.
 RECTANGULAR SOLIDS. 391

Similarly, if the length=a linear units, the breadth =8

;
;
finear units, and the height=c linear units,
the rectangular solid contains abe units of volume.
And if each edge of a cube =a linear units,
the cube contains a® units of volume.
These statements may be thus abridged :
volume of cuboid = length x breadth x height ........... (i)
(NG: OF UCSC) 4 MCLGtn. weno saae (11)
MOU Of CRUG (COGE)". sec sitena ane vestita ckbvee tet (ii)

COROLLARY. The cuboid (ABCD, PQRS)

is divided by the diagonal plane BDSQ into
two right prisms whose bases are congruent
right angled triangles. These prisms are
identically equal, and the volume of each is
half that of the complete cuboid.

EXERCISES.

1. The length, breadth, and height of a room being respectively

a, b, and c units, shew that un four walls contain 2c(a+b) units of
area.
If the area of the four walls is equal to 86°70 sq. metres, and the
height is 3:40 metres, find the perimeter of the floor.
2.° Express in litres the capacity of a rectangular cistern whose
length, breadth, and depth are respectively 125 cm. , 80 cm., and
65 cm.
Find in kilograms the weight of water required to fill the cistern to
four-fifths of its capacity.
3. The annual rainfall at a certain place is 65 cm.; to how many
litres per hectare is this equivalent?
4. What is the weight of a rectangular block of granite whose
dimensions are 1‘20 metres, 0°75 metres, 0°50 metres, at the rate of
2°64 kg. per cubic decimetre?
[For further Exercises on Rectangular Solids see p. 393.]
392 GEOMETRY.

16. To find the lateral surface of a right prism.

In the given prism let the sides of

the base AB, BC, CD, ... contain a, 8,
¢,... units of length, and let the
height=h. Then, the prism being
vight, each side-edge=h, and each
side-face is a rectangle.

Then the area of the rect. ABQP=ah, and the areas

of the
remaining side faces are bh, ch, ....
.. lateral surface of prism=ah+bh+ch+...
=(a+b+c+...)h units of area
= (perimeter of base) x height.

17. To find the volume of a right prism.

(i) First consider a prism eee she eg
(ABC, PQR) on a friangular base
ABC; let its height be h.
Through AP draw the plane
APYX perp. to the face BCRQ,
thus dividing the given prism
into two prisms whose bases are
the right-angled A*AXB, AXC.
Through A draw LM par' to BC;
complete the rect. BLMC, and on
BLMC as base construct a cuboid
of height /.
Then prism on base AXB= $(cuboid on base AXBL);
and prism on base AXC = $(cuboid on base AXCM) ;
.. given prism on base ABC = }(cuboid on base LBCM)
=} rect. LBCM x height
= (area of base ABC) x height.
 RIGHT PRISMS. 393

(li) Now a prism on a_ polygonal

base may always be divided, as shewn
in the diagram, into a set of prisms on
triangular bases and having the same
height as the given prism;

.. volume of any right prism = (sum of triangular bases) x height

= (base of given prism) x height.

EXERCISES ON RECTANGULAR SOLIDS AND RIGHT PRISMS.

[1 litre 1s equivalent to 1 cubic decimetre.

1 cubic decimetre of water weighs 1 kilogram.
The specific gravity of a substance is the ratio of the weights of
equal volumes of the given substance and water.
Thus if the specific gravity of steel is 7°8, it follows that the weight
of 1 cubic decimetre of steel is 7°8 kilograms.]

1. The area of the section of a boring is 1325 square feet, and the
excavating machine is driven forward 4 feet a day. How many cubic
yards of earth are excavated in a day?

2. Find in kilograms the weight of water in a trench, whose length

and breadth are 21:25 metres and 1°50 metres, the depth of the water
being 64 cm.

3. Find the weight of a steel bar 1°28 metre long, 15 cm. wide, and
5 cm. thick, the specific gravity of steel being 7°8.

4, <A level seam of coal has an average thickness of 34 feet: find

approximately the yield in tons:per acre.
[1 cu. ft. of water weighs 1000 oz., and the specific gravity of
coal = 1°28. ]
394 GEOMETRY. —

EXERCISES. (Continued.)
(On Rectangular Solids.)

5. Find to the nearest penny the cost of painting the four inner
sides and bottom of a tank, which measures internally 2'50 metres long,
1:24 metres wide, and 1°50 metres deep, at 7d. per square metre.

6. Find the weight per square metre of sheet zinc 3 mm. thick, the
specific gravity of zinc being 7°14.
7. A chest whose external length, breadth, and height are re-
spectively 1°65 metres, 1°25 metres, and 0°55 metres, is made of oak
25 mm. thick. What are its inner dimensions? Find to the nearest
penny the cost of lining the sides and bottom with thin metal at 1s. 3d.
per square metre.

8. What is the length of the edge of a cube of which

(i) the surface is 2°5350 square metres ?
(ii) the volume is 274,625 cubic era.?

9. A closed box is built of wood of uniform thickness. Its ex:

ternal dimensions are 12 cm., 10 em., and 8 cm.; and the inner surface
is 376 sq. cm.; find the thickness of the wood.

10. The whole surface of a rectangular block is 1332 sq. em. Find
the length, breadth, and height; it being given that these dimensions
are proportional to the numbers 4, 5, 6.
11. The whole surface of a rectangular block is 214 sq. cm. The
hase contains 42 sq. cm., and one vertical face contains 35 sq. em.
Find the edges.
12. Find to the nearest millimetre the edge of a cube whose
diagonal is 10 cm. Find also the whole surface and volume of the cube,
13. <A rectangular block stands on a base of 48 sq. cm.; its height
is 3 cm., and its diagonal is 13 cm. Find the length and breadth.

14. The diagonal of a rectangular solid is 17 cm. and the whole

surface is 552 sq. cm. Find the sum of the three dimensions.
15. A rectangular tank stands on a base measuring 20 feet by 16
feet. If water flows in from a pipe which admits 40 gallons a minute,
tind at what rate (in inches per hour) the water will rise, reckoning
6} gallons as a rough equivalent of 1 cubic foot.
 RECTANGULAR SOEIDS AND RIGHT PRISMS. 395

On Right Prisms.)

16. The base of a right prism is a triangle ABC right-angled at C.

If AC=15 cm., CB=8 em., and the height of the prism=12 cm., find
the volume and lateral surface.

17. A right prism stands on a triangular base whose sides are

17 cm., 10 cm., 9 em.; and the height is 10cm. Find the volume and
whole surface.

18. The base of a right prism is a trapezium whose parallel sides

are 17 cm. and 13 cm., the distance between them being 8 cm. If the
height of the prism is 1 metre, find the volume in cubic centimetres.

19. Sand lies against a wall covering a strip of ground 4 feet wide,
and resting with its surface inclined at 30° to the horizon. Find, to
the nearest tenth of a cubic foot, how much sand may lie on this strip
per foot length of wall.

20. The vertical cross-section of a trench is a trapezium, measuring

15 feet across the top, 9 feet across the bottom; depth of trench
uniform 8 feet; length of trench 624 feet. Find roughly how many
gallons and how many tons of water the trench will hold.
[1 cubic foot of water is equivalent to 6} gallons nearly, and weighs
a little less than 1000 ounces.] ,

21. <A bed of coal 14 feet thick is inclined at 23° to the surface.
Calculate the number of tons of coal that lie under an acre of surface.
{The thickness is to be measured perpendicular to the coal-bed. One
ton of coal occupies 28 cubic feet. Cos 23°=0°9205. ]

22. Through a wcoden pipe, whose cross-section is a square on a

side of 8 cm., water flows uniformly at the rate of 40 metres a minute.
How long will it take to discharge a million litres?

23. Compare the lateral surfaces and the volumes of the following
right prisms :
(i) base a regular hexagon on side 8 em., height 6 cm.
(ii) base a regular octagon on side 6 cm., height 8 cm.

24, A railway cutting 850 metres in length is to have a uniform

depth of 4°50 metres, and its dimensions across the top and bottom are
to be respectively 31:20 metres and 16°80 metres. If 450 tons are
excavated on an average per day, and 1 cubic inetre weighs 2} tons,
how long will the work require?
396 GEOMETRY.

18. To find the volume of an oblique prism.

The diagram represents an

oblique prism (ABCDE, A’B’'C'D’E’),
and Abcde is a right plane section,
that is, a section perp. to all the
side-edges,

Now suppose the slice cut off between the base ABCDE and
the right section Abcde to be removed and placed on the other
end ABC'D'E’, so that A falls on A’, B on B, and so on;
then the given oblique prism is converted into the right prism
(Abcde, A’l’c'd’e’), whose edges are equal to those of the given
prism, and whose volume = (area of its end Abcde) x AA’.

Hence the volume of an oblique prism

= (area of right section) x Cdge. ....1.s.ereveeenes (i)

Now let 0 be the angle between the base ABCDE and the
right section Abcde (which is a projection of the base); then @
is also the angle between the perp. height h and the edge AA,
for the two lines are respectively normal to the two planes,
Thus we have
right section Alcde = base ABCDE x cos @ ; (Art. 2.)
also h=AA' cos 0.

Applying these values to (i),

Volume of oblique prism = base ABCDE x cos @ x AA’

= base ABCDE x h.
 OBLIQUE PRISMS, 397

Hence of oblique as of right prisms,

the volume = (area of base) x (perpendicular height).
We leave as an exercise to the student to prove that
the lateral surface of an oblique prism= (perimeter of right section) x edge.

19. Volume of oblique prism. [Alternative method.]

Cut the prism at equal in-

tervals by a series of planes
parallel to the base; and be-
tween each pair of consecutive
planes, and on the section made
by the lower of them, con-
struct a right prism. Then the
volume of such a_ prismatic
slice = (area of its base) x thick-
ness.

Now let the number of slices be increased indefinitely, and

consequently the thickness of each one diminished indefinitely;
then in the limit the whole prism is equivalent to the sum of
all the slices.
But the base of each slice=the base of the prism ; and the
sum of the thicknesses = the perp. height.
Hence
volume of prism = (area of base) x (perpendicular height).

CoroLuaRy. Prisms of the same perpendicular height and

standing on bases of equal area are equal in volume.

Nors.—The above proof applies also to parallelepipeds, which are a

special form of prism; the essence of the proof being the fact that
plane sections parallel to the base are identically equal.
398 GEOMETRY.

PYRAMIDS,

20. The slant surface of a pyramid (S, ABCDE) is the sum

of the triangular faces SAB, SBC, SCD, ..., of which the areas in
the general case must be separately found.
A convenient expression may however be obtained for the
slant surface when the pyramid is right and on a regular base.

21. To find the Slant surface of a right pyramid on a regular

base of n sides.
S

The pyramid being right, and the

base regular, it follows that the slant
edges SA, SB, SC, ... are all equal;
and the faces SAB, SBC, SCD, ... are
equal isosceles triangles.

SP, drawn perp. to a side of the base and therefore bisecting

it, is called the slant height of the pyramid, and is the same for
each slant face.
If SO is normal to the base, it follows (as in the Corollary to
Theorem 83) that OP is perp. to AB.
Let each side of the base=a, the perp. height SO=/, and
the slant height SP=/. Then
slant surface of pyramid = ASAB x n
=4AB.SPxn
=}.naxl units of area
=4(perimeter of base) x (slant-hetght),

The whole surface=the slant surface

+ the area of the base,
 PYRAMIDS. 399

22. To prove that two pyramids (S, ABC), (S’, ABC’) standing
on bases of equal area, and having equal perpendicular heights, are
equal in. volume.

Place the pyramids with their bases ABC, A’B'C’ in the sam#
plane, and cut them at equal intervals by a series of planes
parallel to the plane of the bases.
Between each pair of consecutive planes, and on the section
made by the lower of them, construct’ prisms with lateral
edges parallel to SA and S’A’ respectively.
Then any pair of sections, such as abc, a’b’c’, made by the
same plane, are equal in area (Art. 13. Cor.).
_ Hence the prismatic slices standing on these sections are of
equal volume, for they are of the same thickness.

Now let the number of parallel sections be increased in-

definitely, and consequently the thickness of the slices
diminished indefinitely ; then in the limit each pyramid is
equivalent to the sum of its prismatic slices ;and each slice in
one is equal to the corresponding slice in the other.
Hence the whole pyramids are equal in volume.

Nore. For the sake of simplicity triangular pyramids have been

considered, but the reasoning is general.
400 GEOMETRY.

23. To find the volume of a triangular pyramid.

Let (S, ABC) be a triangular pyramid of perp" height h.

Through A and C draw par™ to BS, and cut them by a plane

through S par' to the base ABC. This determines a triangular
prism whose volume = AABC x h,

Draw the diag. AQ of the par™ ACQP.

Now the prism may be broken up into the pyr* (8, ABC),
standing on the triangle base ABC, and the pyr* (S, ACQP), on
the par"-base ACQP.
But the latter may be broken up into the two pyr“ (S, APQ),
(S, ACQ), which have equal volumes since they stand on equal
bases, and have the same vertex.

Again, regarding the pyr* (8, APQ) as (A, PSQ), we see that
pyr’(A, PSQ)
= pyr'(S, ABC),
for they have equal bases PSQ, ABC, and the same height.

Thus the prism is divided into three pyr“ of equal volume.

“. pyramid
(S, ABC) =}(volume of prism)
== 3 (area of base) x (perp. heaght).
 PYRAMIDS. 40]

_ COROLLARY. A pyramid on a polygonal base may always

be divided into a vet of pyramids on triangular bases and
having the same height as the given pyramid ; hence
Volume of pyramid on any base
=i (area of base) x height.

EXERCISES.

1. Find (i) the slant surface, (ii) the volume of a right pyramid
15 cm. high, standing on a square base whose side is 16 cm.
Here @S=15 em
AB= AD =16 ecm.
OP=4AB=S8 cm.
From the 4:SOP, rt.-angled at O,
SP?= OS? + OP?= 15? + 8?=289 ;
SP=J/289=17 cm.
Area of face SDA=4AD x SP
=4(16 x 17) sq. cm.
= 136 sq. cm.
Henee (i) slant surface= ASDA x 4=544 sq. em. ;
(ii) volume = (area of base) x height
=4(16? x 15) cu. cm.
=1280'eu. en

2. Find (i) the slant surface (to the nearest hundredth of a square
- inch), (ii) the volume of a right pyramid 7” high, standing on a square
base whose side is 6”.

3. Find the volumes of the pyramids in which

(i) the base is a rectangle measuring 11 cm. by 7 cin., and the height
is 12 cm. ;
(ii) the base is a triangle whose sides are 15 cm., 14 cm., 13 em.,
and the height is 10 cm.

4. A right pyramid of height 9” stands on a square base on a side of

8”. Find to the nearest hundredth of an inch (i) the slant height, (ii) the
slant edge.
H.S.G. 2c
402 GEOMETRY.

5. Find (i) the slant surface, (ii) the volume of a pyramid having the —
same base and height as a cube on an edge of 10 cm.

6. A right pyramid stands on a rectangular base whose sides are

24 cm. and 18 cm. ; and each of the slant edges is 17 cm, Find the
height and volume of the pyramid.

7. Avright pyramid of height 2-4” stands on a square base of which

each side is ]‘4”. Find (i) the cosine of the dihedral angle between each
side-face and the base ; (ii) the area of the projection of each side-face
on the base.

8. The base of a right pyramid is an equilateral triangle on a side

of 10 cm., and the vertical height is5 cm. Find (i) the slant height,
(ii) the area of one side-face, (iii) the cosine of the dihedral angle between
the side-face and base.
Construct a plane angle having the same cosine, and measure it with
your protractor.

9. Find to the nearest millimetre the height of a pyramid of which

the volume is 270 cu. cm., and the base a regular hexagon on a side of
6 cm.

10. The solid shewn in the diagram is *

called a wedge. Its base is a rectangle of
length a and breadth b; the two ends are the
A* EAD, FBC; the remaining faces AEFB,
DEFC are trapeziums having a common side
EF which is thus par! to two sides of the base,
and is called the edye.
If the edge EF =e, and the height=h, prove that
hh
volume of wedge = 7 {2a+e}.

[Draw planes through E and F perp. to the base, thus dividing the
wedge into two pyramids and a prism. The two poo fit together
into a single pyramid on a rectangular base of length a —e and breadth b.
The ends of the prism are the vertical sections, each of which has an
area 4bh, the length of the prism being e. Hence the volume of the
wedge may be calculated.}

ll. If in the last exercise the triangular faces of the wedge are
equally inclined to the base, shew that the slant surface is given
by the formula
4{(a +e) /402 + 0 + bn 4h?+ (a — €)?}.
 EXERCISES ON POLYHEDRA. 403

12. The accompanying diagram represents an

oblique frustum, or slice, formed by cutting a right '
triangular prism by a plane A’B’C’ not parallel to J
'
the base ABC. i
‘
1
1
!
Yo)
'
NS

If a, 6, c are the lengths of the lateral edges AA’, BB’, CC’, prove
that
volume of oblique frustum=(area of base) x $(a+b+c).

[Cut the frustum by a plane parallel to the base through A’ the

extremity of the leastof the lateral edges, and thus divide the given
solid into a pyramid and a prism.]

13. (EuLER’s TuEorEM). In any polyhedron if F; E, and V denote

the number of faces, edges, and vertices respectively, then E+2=F+V.

[Suppose the polyhedron to be built up by fitting together x faces,

one by one.
Beginning with one face, which has as many vertices as edges, we
have E=V.
On adding the second face, there are two vertices and one edge in
common with the first, so that the number of new edges is one more
than the number of new vertices.
SE =Veri,
The third face has three vertices and tivo edges in common with the
former faces; and, as before, the number of new edges is one more
than the number of new vertices.
, b=V 42:

Proceeding in this way, step by step, when n-1 faces have been
fitted together, we shall find
SS
\N/ sj =
In fitting on the last face no new edges or vertices are added, and
finally n=F.
p B= Vea a2. %
or E+2=F+V.]
404 GEOMETRY.

24, There cannot be more than five regular polyhedra,

Three plane angles at least are required to form a solid
angle, and the sum of such plane angles must be less than
360° [Theor. 98]. Tt follows that each angle of the faces
forming a regular polyhedron must be less than 120°. That
is, the faces can only be equilateral triangles, squares, or pentagons; —
for the angle of a regular hexagon is 120°, and any regular
polygon of more than six sides has an angle greater than 120°, —
Let D represent the number of degrees in a face-angle.
When the faces are equilateral triangles, D = 60°.
Then (i) 3D = 180", (ii) 4D = 240”, (ili) 5D = 300°, [6D = 360°].
Thus three, four, or five equilateral triangles, and not more
than five, can be used to form a solid angle in a regular
polyhedron.
When the faces are squares, D = 90°.
Then (iv) 3D = 270°, [4D = 360°].
Thus three squares, and only three can be used.
When the faces are pentagons, D = 108".
Then (v) 83D =324°, [4D = 432°].
Thus three regular pentagons, and only three can be used.
Hence there can only be five regular polyhedra.

25. If the plane faces which make up the surface of any of

the regular polyhedra are supposed to be unfolded so that they
all lie in one plane, we obtain in each case a plane figure made
up of equilateral triangles, squares, or pentagons. Such a
plane figure is called the net of the corresponding polyhedron.
The regular polyhedra, and their nets (drawn on a reduced
scale) are exhibited on pages 405-407.
 THE REGULAR POLYHEDRA. 405

THE REGULAR POLYHEDRA.

(i) The polyhedron of which each solid angle is formed by three

equilateral triangles is called a regular tetrahedron.

4 Faces
4 Vertices
6 Edges

The net of the regular tetrahedron consists

of four equal equilateral triangles placed as in
_ the adjoining figure.

(ii) The polyhedron of which each solid angle is formed by four

equilateral triangles is called a regular octahedron.

8 Faces
6 Vertices
12 Edges

The net of the regular octa-

hedron consists of eight equal
equilateral triangles.
406 GEOMETRY.

(iii) The polyhedron of which each solid angle is formed by five

equilateral triangles is called a regular icosahedron.

20 Faces
12 Vertices
30 Edges

The net of the regular

icosahedron consists of
twenty equal equilateral
triangles.

(iv) The regular polyhedron of which each solid angle is formed by

three squares is called a cube.

6 Faces
8 Vertices
12 Eages

The net of the cube consists of

six equal squares.
 THE REGULAR POLYHEDRA. 407

(v) The polyhedron of which each solid angle is fermed by ¢threz

regular pentagons is called a regular dodecahedron.

12 Faces
20 Vertices
30 Edges

The net of the regular dodecahedron consists of twelve equal regular

pentagons.

“4
Models of the regular polyhedra may be made from their nets in the
following manner. Draw the net on cardboard and cut it out along the
outside lines, and cut partly through along the dotted lines. The faces
may then be folded over till the edges come together, and the edges
may be kept in position by using strips of gummed paper.
408 GEOMETRY.

___Exampie. Find (i) the length of a diagonal, (ii) the surface, —

(iii) the volume, (iv) the dihedral angle of a regular octagon whose
edge is 2m.

It is evident from the diagram that a regular octahedron consists of

two pyramids on opposite sides of a common square base ABCD. One
of these pyramids is shewn on a larger scale in the right-hand figure. *
PO is the perpendicular from P, O being the central point of the
square ABCD. PR bisects AB at right angles.
Now AB=2m; «. RB=m, and RO=m.
PR=RB
tan 60° =mN3.
From the rt.-angled A POR,
OP?=PR?-— OR?=32n? — m2=2m? ;
- OP=mv2.
Hence (i) Diagonal of octahedron =20P=2m\’2.
(ii) Surface=8 A PAB=8RP. RB
=8 x mv/3 x m=8m?2V3,
(iii) Volume =2(vol. of pyramid whose vertex is P)
| =2x JOP x (area of base)

+ =2xim,/2x (2m)=—
83/2 :

(iv) Dihedral angle=twice the 2 PRO.

EO ils = s/2=1°4142...,
Now tan PRO =
OR” m
Whence, by means of a Table of tangents, 2 PRO=54° 44’ approxi-
mately.
Thus the dihedral 2 = 109° 28’ approximately.
 POLYHEDRA. 409

MISCELLANEOUS EXERCISES ON POLYHEDRA.

1. The dimensions of a rectangular block being taken as 8°5 cm.,

7-4 cm., 6°0 cm., find the volume. Supposing that each of the above
measurements may be too great or too small by as much as 1 mm.,
find the greatest possible error in your result in excess and in defect;
and express each as a percentage of the volume assumed in each case.

2. A deal plank 15” wide is placed against the top of a wall 8 feet
high, while the other end rests on the ground 6 feet from the wall.
The thickness of the plank being 14”, tind its weight, supposing 1 cubie
foot of deal to weigh 56 lbs.

3. An equilateral triangle ABC, of which each side=10 cm., is

projected on a plane passing through AB; and the area of the pro-
jection is 34°64 sq. cm. What is the angle between the } lanes?
[Calculate the cosine of the dihedral angle: then either use Tables, or
construct and measure the corresponding plane angle. N3=1°732...].

4. A right pyramid of height 8 cm. stands on a regular hexagonal

base on a side of 4em. Find :
(i) the slant surface to the nearest tenth of 1 sq. em.
(ii) the volume to the nearest tenth of 1 cu. em.

5. The base of a right pyramid is a square on a side of 6 cm., and

the slant faces are equilateral triangles. Draw the net of the pyramid,
and find its approximate height and volume.

6. The corners of the base of a right pyramid are at the points

(9, 5, 0), (-9, 5, 0), (9, -—5, 0), (-—9, —5, 0); and its vertex is at the
point (0, 0, 12). Jind the slant surface. [See p. 382].

7. A right pyramid stands on a regular hexagon having a side of

5 cm., and its slant faces are inclined to the base at an angle of 60°.
Find the volume.

8. If a perpendicular is drawn from a vertex of a regular tetra-

hedron on its base, shew that the foot of the perpendicular will divide
each median of the base in the ratio 2: 1.

9. Prove that the perpendicular from the vertex of a regular tetra-

hedron upon the opposite face is three times that dropped from its foot
upon any of the other faces.
410 . GEOMETRY.

10. If p is the perpendicular drawn from a vertex of a regulat

tetrahedron to the opposite face, and if each edge=2m, shew that
3p?=8m?.

11. The length of each edge of a regular tetrahedron being 2m,

shew that }
(i) the whole surface =4n2n/3 ; (ii) the volume = 2 m*r/2. {

12. Find approximately the dihedral angle between any two ad-)
jacent faces of a regular tetrahedron.

13. Prove that (i) the sum of the squares on the four diagonals of a
parallelepiped is equal to the sum of the squares on the twelve edges.
(ii) The sum of the squares on the edges of any tetrahedron is four
times the sum of the squares on the straight lines which join the middle
points of opposite edges.
14. In any tetrahedron the plane which bisects a dihedral angle
divides the opposite edge into segments which are proportional to the
areas of the faces meeting at that edge.

15. OA, OB, OC are conterminous edges of « cube, each of length

“units. Prove that
(i) volume of pyramid (O, ABC)=". ad;

(ii) area of triangle ABC a ats

(iii) perp. from O on plane ABC ae, a.

16. OA, OB, OC are lines in space, each perpendicular to the other
two; and their lengths are a, b, c respectively. Prove that

(i) volume of pyramid (O, ABC) =r. abe;

(ii) area of triangle ABC =3Vath? + 0%? + a? ;

(iii) perp. from O on plane ABC =abe/a*b? + b2c?+c®a*.

17. Shew how to cut a cube by a plane so that the lines of section
may form a regular hexagon.

18. The greatest possible cube is cut from a right pyramid h inches
high, standing on a square base whose side is a inches, one face of the
nabe being in the plane of the base of the pyramid. Prove that the
edge of the cube =ah/(a +h).
 SOLIDS OF REVOLUTION, 411

SOLIDS OF REVOLUTION.

THE CYLINDER.

26. DEFINITION. A right circular cylinder is a solid

generated by the revolution ef a rectangle about one of its
sides as axis.

Thus if the rectangle ABCD revolves about AB

as axis, the opposite side CD generates the curved
surface of the cylinder represented in the diagram.
The side CD, which moves parallel to the axis, is
called the generating line of the surface; and
since the sides AD, BC are aiways pernendicular
to the axis, they move in paralle! planes and
describe circular ends or bases. The height of a
right cylinder is the length of its axis AB.

It is easily seen that every plane section of a right circular cylinder

parallel to the base is a circle. Also every section parallel to the axis
is a rectangle.

27. In general any surface described by a generating line

which moves parallel to itself and slides continually over a
_ fixed curve (not in the same plane with it), is said to be
cylindrical. The fixed curve is called the guide; and in the
particular case of a right circular cylinder the guide is a
circle in a plane perpendicular to the generating line.
In the present Section, unless otherwise stated, only right
circular cylinders will be considered.

we

a——
412 GEOMETRY.

28. To find the surface and volume of a cylinder.

Consider a right prism standing on a regular polygonal base.

If the number of sides in the base is increased without limit,
the polygon ultimately becomes a circle [Page 203], and the
prism takes the form of a right cylinder. Thus a cylinder
may be regarded as a prism in its limiting form ; and accord-
ingly expressions for the surface and volume of the cylinder
follow from those given for the prism on p. 392.
Hence we have

(i) Curved surface of cylinder = (circumference of base) x height

r= Dorr x h
=27rrh units of area.

(ii) Volume of cylinder = (area of base) x height

=r xh.
=77°h units of volume.

Note l. The whole surface=curved surface

+ area of ends
= Qrrh + Qrr?
=2rr(h+r).

Nore 2. The volume of an ob/ique cylinder on any base, as of an

oblique prism, is also given by the formula :—
Volume =(area of base) x (perp. height).
 THE CYLINDER. 413

Norse 3. The formula for the curved surface of a cylinder may be

illustrated thus.

R Q

S P

Suppose the surface of the cylinder to be cut along a generating

line PQ, and then unrolled on a plane; the surface will take the form
of a rectangle PQRS, of which the length PS and breadth PQ are
respectively the circumference and height of the cylinder.
Thus curved surface =PS x PQ
_ =circumference x height.

Nore 4. Surfaces which can be unrolled (without stretching or tear-

ing), and represented by plane figures are said to be developable.

EXERCISES.

[In working examples which involve the use of 1m, the substitution of
a numerical equivalent should be postponed as long as possible, and in
each case the value should then be selected which will give a result of the
required degree of accuracy. See page 202.]

1. Find the curved surface (to the nearest square centimetre), and
the volume (to the nearest cubic centimetre) of the cylinders whose
dimensions are as follows:
G)tr=3 tm, h=S:0'em. (ul) 740 cm. k=7 Dems

2.“as Find to the nearest square centimetre the total surface of a

. cylinder whose height is 15°83 cm., and the diameter of whose base
is 8°4 cm.
3. Find to the nearest cubic centimetre the volume of a cylinder,
of height 12 em., which just tits into a right prism standing on a square
base whose side is 3°6 cm.
4. Find the locus of points at a given perpendicular distance from
a given finite straight line.
If the given distance=3°5 cm., and the length of the given straight
line=5'6 cm., find the surface of the locus to the nearest square
centimetre.
414 GEOMETRY.

EXERCISES ON CYLINDERS. (Continued.)

5. Find to the nearest square centimetre the whole surface of a
hollow cylinder open at the ends, if the length is 12 cm., the external
diameter 8 cm., and the thickness 2 em.

6. The volume of a cylindrical column standing on a base whose

diameter is 4 metres is 128°2 cubic metres; find its height to the
nearest centimetre.

7. A cubic inch of gold is drawn into a wire 1000 yards long; find
the diameter of the wire to the nearest thousandth of an inch.

8. The curved surface of a cylinder is 1000 sq. cm., and the

diameter of its base is 20 em.; find the volume of the cylinder. Also
find its height to the nearest millimetre.

9. A column is formed by inserting a block of wood in a closely

fitting cylindrical case, and then filling up the surrounding space with
concrete. If the wood is in the shape of a right prism 8} ft. long ona
rectangular base whose sides are 1 ft. 4 in. and | ft., find the require ad
amount of concrete to the nearest cubic foot.

10. Find the weight to the nearest 100 grams of a cylindrical iron
pipe 18 metres long, the external diameter being 5:4 cm., and the thick-
ness 4 nm.; assuming that the specific gravity of iron is 7°79.

11. A copper wire 2 mm. in diameter is evenly wound about a

cylinder whose length is 12 cm., and diameter 10 cm., so as to cove
the whole surface. Find the length and weight of the wire assuming
the specific gravity of copper to be 8°88.

12, The annexed diagram represents an oblique frustum of a cylinder

i this solid cut by a plane through c parallel to the base AB,
and hence prove that

(i) Curved surface =2rr. Cc=2rr x hy thy

<4 a

(ii) Volume = m9, Cox mtx thy,

where h, and hg denote the greatest and least heights of the frustum.,
 THE CONE. Ala

THE CONE.

29. DEFINITION. A right circular cone is a solid generated

by the revolution of a right-angled triangle about one of the
sides containing the right angle as axis.

Thus if the right-angled triangle ABC

revolves about AB as axis, the hypotenuse AC
generates the curved surface of the cone repre-
sented in the diagram. The hypotenuse AC,
which in all its positions passes through the
fixed point A, is called the generating line of
the surface ; and the circle described by the
radius BC is the base of the cone.

The point A is called the vertex, and the angle CAD (which is twice
the angle A of the revolving triangle) is the vertical angle. The height
of the cone is the length of the axis AB, and the slant height is the
length of the hypotenuse AC.
Every plane section of a right circular cone parallel to the base is a
circle. Also every section of the cone through the vertex is a pair of
intersecting straight lines.

30. In general any surface described by a generating line

which passes through a fixed point and slides continually over
a fixed guiding curve (not in the same plane with it) is said to
be conical. In a right circular cone the guiding curve is a
circle, and the vertex any point in the line through the centre
of the circie normal to its plane.
In the present Section, unless otherwise stated, only righé
circular cones will be considered.
416 GEOMETRY.

31. To find the surface and volume of a cone.

>

- aa me me,

Consider a right pyramid standing on a regular polygonal

base. .
If the number of sides in the base is increased without
limit, the polygon becomes a circle, and the pyramid ultimately
takes the form of a right cone. Accordingly, expressions for
the curved surface and volume of a cone follow from those
given for the pyramid on pages 398, 401.
Thus, if 2 denotes the vertical height, / the slant height of
the cone, and 7 the radius of the base,
(i) Curved surface of cone = } (circumference of base) x (slant heaght)
= 4x 207 xl
=77l units of area.
Again, since the volume of a pyramid is one-third that of
the prism of the same base and height, it follows that the
volume of a cone is one-third that of the corresponding cylinder.
Hence
(ii) Volume of cone= (area of base) x (height)
=ixaexh
=}77°h units of volume.

Nore 1. The whole surface=curved surface+ area of base

=mrl + rr*
=mrr(t+7).

Notr 2. The volume of an ob/ique cone on any base, as of an oblique

pyramid, is also given by the formula:
Volume=} (area of base) x (vertical height).
 THE CONE. 417

Nore 3. The formula for the curved surface of a cone may be illus-
trated thus
A

Suppose the surface of the cone to be cut along a generating line AC,
and then unrolled on a plane; the surface will take the form of a
sector ACD, of which the radius AC and the are CD are respectively the
siant height and the circumference of the base of the cone.
Thus curved surface=i.arcCD x radius AC
Ss) [Page 204]
tole
b
x Qrr
x 1.

Hence we see that the surface of a right circular cone is developable.

EXERCISES.

1. If S is the surface, V the volume, and a the semi-vertical angle

of a cone of height 4, on a base of radius 7, prove the following
formule :
2
(i) ene Veritas
cos a 3
a rr 1 re
2) Sr ee: Wisa? gave
Hence prove that the volumes of cones with the same vertical angle
are to one another as the cubes of their heights.
2. Find the surface to the nearest square centimetre, and the
volume to the nearest cubic centimetre of the cones in which
(ives Oren, ¢— LO em. (i) 72 em.) ==3-5 cn,

3. Find to the nearest square centimetre the whole surface of a

cone whose height is 40 cm., and the diameter of whose base is 18 cm.
4, Find to the nearest cubic centimetre the volume of a cone whose
slant height and vertical height are 5:1 cm. and 4°5 cm. respectively.
[For further Exercises on Cones see p. 422.]
H.S.@. 2D
418 GEOMETRY.

THE FRUSTUM OF A PYRAMID AND CONE.

32. DEFINITION. A frustum (that is to say, a slice) of a

pyramid or cone is the part cut off between the base and a
plane parallel to the base. 1
oO
A
ay4
/ ' \
( \
1
t
é OM
'
1
t

Fig.r. Fig.z.

Thus in Fig. 1 the frustum of the pyramid (O, ABCD) is the part cut
off between the base ABCD and the parallel section abed.
In Fig. 2 the frustum of the cone (O, AB) is contained between the
base AB and the parallel section ab.
In Fig. 1 the figures ABCD, abcd, and in Fig. 2 the figures AB, ab
are called the ends of the frustum. The ends of the frustum of a
pyramid are similar figures (Art. 13, p. 388). The ends of the frustum
of a cone are circles.
The slant surface of the frustum of a pyramid is made up of trapeziums.
If the base ABCD is regu/ar, and the pyramid is right, these trapeziums
are all equal.

33. Let the ends of the frustum contain E, and E, units

of area, and let the perpendicular from the vertex O cut the
ends at the points P and p respectively; then it has been
proved (Art. 13, p. 388) that
E, : E,= OP? : Op’.
 FRUSTA OF PYRAMIDS AND CONES. 419

34. To find the slant surface and volume of the frustum of a

right pyramid on a regular base of n sides.

In Fig. 1 of the preceding page, let & denote the thickness

Pp of the frustum, and let a,, a, denote the lengths of any pair
of corresponding sides BC, bc of the ends, and / the perpen-
dicular distance between these sides, namely the slant thickness
of the frustum. Also let the ends ABCD, abcd contain E, and
E, units of area. Then

(i) Slant surface of frustwm

=n times trapezium BCcb
=}(a,+a,)lxn Theor. 28,
= }(na, +na,yl units of area
=i(sum of perimeters of ends)
x (slant thickness).

: @ Pate the heights OP, Op by h, and h,, so that

peril 1:
- = =
Now 2 = i =m, where m is some constant;
1

» B=mi7, and E,=—mit,’.

Hence
Volume of frustwm = pyr* (O, ABCD) — pyr® (0, abed)
= 4 E hy — +4Eh,

= 3(h? —h)m
=43(hy — hy) (hy? + yg + ho?)m

=1k(mh.2 +S mh? mh? + mh,2)

=1i[eE, +VE,E,+E,}-
420 GEOMETRY.

35. To find the curved surface and volume of the frustum

ef a right cone.
Considering the cone as the limiting form of a right
pyramid on a regular polygonal base, we derive expressions
for the curved surface and volume of a frustum of a cone from
those given on the preceding page for the frustum of a
pyramid.
In Fig. 2 of page 418 let 7, and, denote the radii of the
ends AB and ab; let the thickness Pp be denoted by 4, and the
slant thickness Aa by 1.
Then £,<ar?, and, Ej=—a9,%
(i) Curved surface of frustum of cone
=4(sum of circumferences of ends) x (slant thickness)
=}(2rr, + 2rr,)l
=7(r,+7,)1 units of area.

(ii) Volume of frustum = *(E,+/E,E,+E,]

= fart + Vane ar tory

oe 5
ne,s(n? +77) +7,2] units of volume.

Note 1. Since 7,+7,=twice the radius of a circular section equi-

distant from the two ends, we have
Curved surface of /rustum=7(7,+7,)1=2r. att |
=(circumference of mid-section) x (slant thickness).
Nore 2. If E,, E, denote the areas of the ends, and M the area of
the mid-section, then
Volume of frustum= men? +1 Po +197) = . (27,2 + Qry774 + 2r,?)
le I 2
= mer? + (7+ 19)?+ 722) = 7 (rye+4. (As
:*) + rf)

=KE, +4M + E,).

This last result is called the prismoidal formula. It is applicable to

any solid whose ends are parallel (but not necessarily similar) figures of
the same number of sides, each eal of corresponding sides in the two
ends being parallel. Such a solid is called a prismoid ; and frusta cf a
pyramid and cone are special cases of it.
 FRUSTA OF PYRAMIDS AND CONES. 42]

EXERCISES.

On Frusta oF PyRAMIDS AND CONES.

1. The ends of the frustum of a pyramid are squares on sides of

20 cm. and 4cm.; if the frustum is-15 cm. in thickness, find its slant
surface.

2. Find the curved surface of the frustum of a cone whose slant

thickness is 5cm., and whose circular ends are 8 cm. and 6 cm. in
diameter.

5. Find the volume of the frustum of a pyramid, the ends being

squares on sides of 8 cm. and 6 cm., and the thickness being 3 em.

4. The slant thickness of a frustum of a cone is 5 cm., and the

radii of its ends are 4cm. and lcm. respectively, find its curved
surface to the nearest sq. cm., and its volume to the nearest cubic cm.

5. The ends of the frustum of a pyramid are squares on sides of

8:0 cm. and 1°4cm.; if the thickness of the frustum is 5'6 cm. find
its slant surface to the nearest sq. cm.

6. The ends of the frustum of a pyramid are triangles, the sides of

the base measuring 13 cm., 12 cm., 5cm., and the sides of the top
65 cm., 6 cm., 2°5 cm. ; if the thickness of the frustum is 8 cm., find
the volume.

7. If 7, 72 are the radii of the ends of a frustum of a cone, whose

height is h, shew that its volume is equal to the sum of the volumes of
a cylinder and cone, each of height /, on bases of radii $(7,+7,) and
4(r, — 7.) respectively.

8. If the height of a frustum of a cone is equal to twice the mean

proportional between the radii of its bases, shew that the slant side is
equal to the sum of the radii.

9. A cone, whose height is n cm., is cut through by a plane parallel

to the base and lcm. distant from it: express the volume of the
frustum so formed as a fraction of the volume of the whole cone.

10. The frustum of a square pyramid is 6 cm. thick, and the area
of one end is four times the area of the other. If the volume is 350
cubic centimetres, find the dimensions of the ends.
422 GEOMETRY.

EXERCISES.

(Miscellaneous Examples on Cones.)

1. ABC is a triangle in which a=6°5 cm., b=2:0cm., and the

perpendicular from C on AB=1'6cm. Find to the nearest cubic
centimetre the volume of the double cone formed by a complete
revolution of the triangle about the side AB.
2. Find to the nearest foot the length of canvas, 1 yard wide,
which will be required to make a conical tent 30 ft. in perpendicular
height, and covering 1386 sq. ft. of ground.
3. Find to the nearest tenth cf a metre the height of a conical tent
which stands on a circular base of diameter 8°0 cm. and which contains
90-478 cubic metres of air.

4. Find to the nearest 3q. em. the whole surface of the greatest cone
that can be cut from a solid cube whose edge is 20 cm., the base of the
cone being in the same plane as the base of the cube.
5. Water flows at the rate of 10 m. per minute from a cylindrical
pipe 5 mm. in diameter. How long would it take to fill a conical
vessel whose diameter at the surface is 40 cm. and depth 24 em.?
6. A cone is cut by a plane parallel to the base and the upper
portion is removed. If the remainder is $ of the surface of the whole
cone, find the ratio of the segments into which the cone’s altitude is
divided by the plane.
7. From a solid cylinder whose height is 2-4 em. and diameter
1:4 cm., a conical cavity of the same height and base is hollowed out.
Find the whole surface of the remaining solid to the nearest square
centimetre. :
8. <A conical vessel 7°5 cm. deep and 20°0 em. across the top is
completely filled with water. If sufficient water is now drawn off to
lower its level by 6°0 cm. find the surface of the vessel thus exposed, to
the nearest square millimetre.
9. Two conical vessels with a common vertex and axis are placed
with the vertex downwards and the axis vertical. The inner cone is
filled with oil, and the remainder of the outer with water. If the
diameters at the surfaces of oil and water are 7°0 em. and 11°2 em,
respectively, find the ratio of the weights of the oil and water, assuming
the specific gravity of the oil to be 0-92.
10. Into each end of a solid cylinder, whose length is 10 cm. and
diameter 8 cm., a conical cavity is bored; if the diameter of each
cavity is 6 em. and its depth 4 cm., find the whole surface of the
remaining solid to the nearest square centimetre.
 THE SPHERE. 433

THE SPHERE.

36. DEFINITION. A sphere is a solid generated by the

revolution of a semi-circle about its diameter as axis.
A

Thus if the semi-circle APB revolves Q

about the diameter AB, the semi-circum- i. \
ference APB describes the surface of a Pp
sphere. And since, as the semi-circum-
ference revolves, all points in it remain
at a constant distance from its centre O,
it follows that the surface of a sphere is the
locus in space of all points whose distance
from a fixed point is constant.

S)
The fixed point is called the centre, and the constant distance the
radius of the sphere. A diameter is any straight line through the
centre terminated both ways by the surface. Thus all diameters are
equal.

37. Every plane section of a sphere is a circle.

In the above Figure let the plane QPR cut the sphere whose
centre is O, and whose radius is 7; let P be any point on the
line of section.
Draw ON perp. to the cutting plane, and let p be the length
of ON. Join OP, PN.
Then since ON is perp. to NP in the plane QPR,
PN? = OP? — ON?
=7? — pr;
PN =,/7? — p? =a constant.
.. the locus of P is a circle whose centre is the fixed point N.

DEFINITION. The diameter AB perpendicular to a plane

section QPR is said to be the axis of the section, and the
extremities A, B are called its poles.
424 GEOMETRY.

38. If the plane section passes a SA a

through the centre, N coincides
with O, and the radius of the circle
Cc D
QPR assumes its greatest length,
being then equal to the radius of
the sphere.

B
The line of section of a sphere by a central plane is called a
great circle ; all other plane sections are called small circles.

39. In a sphere of radius 7, if 7, is the radius of any

plane section whose distance from the centre is p, it has
been shewn that 7,=\/7*—p*. Hence if the cutting plane
moves parallel to itself away from the centre O, then as p
increases, 7, will decrease. Thus the radius of a plane section of
a sphere continually diminishes as the distance of the plane from
the centre increases; and ultimately when N coincides with A,
r, vanishes, so that the. plane meets the sphere at A only, and
is then said to be a tangent plane at that point. It follows
that at any point on a sphere there is ONE tangent plane, namely
the plane perpendicular to the radius through that point.

40. Any straight line drawn on a tangent plane through

its point of contact will meet the surface at one point only, and
is said to touch the sphere at that point. Thus at any point on
a sphere there may be an infinite number of tangent lines, cach
being perpendicular to the radius through that point; and a
tangent plane may be generated by rotating a tangent line
about the radius through its point of contact.

41. If AB is a diameter of a sphere, there is ONE great circle

having AB as axis, and an infinite number of great circles passing
through the poles A and B.
 THE SPHERE. 425

42. Through any TWO given points on a sphere (not the extremi-
ties of a diameter) one, and only one, GREAT circle can be drawn;
for the two given points and the centre of the sphere determine
one, and only one, central cutting plane.

Note. The minor are of a great circle through two given points on
a sphere is called their spherical distance. It will be shewn (p. 437)
that this are is the shortest line that can be drawn on the surface
between the two points. Now since all great circles of a sphere are
equal, an are of any great circle is measured by the angle which it
subtends at the centre. [Theorem 69].
Thus in the diagram the spherical distance between the points Q
and C is measured by the angle QOC and estimated in degrees. The
spherical distance of any point on the great circle CD from its pole A
is 90°.

43. Through any THREE given points on a sphere one, and only
one, circle (not necessarily great) can be drawn on the surface ; for
the three points determine a plane which cuts the sphere in
a circle passing through those points. ;

44, An infinite number of spheres can pass through TWO given

points, and thew centres lie in @ FIXED PLANE.
For if Q and R are the given points, it is clear that the
locus of points equidistant from Q and R is the plane which
bisects the join QR at right angles. Hence with any point in
this plane as centre a sphere may be drawn through Q and R.

45. An infinite number of spheres can pass through THREE given

points, and their centres lie on a fixed STRAIGHT LINE.
For in the Figure of the preceding page, let P, Q, R be any
three points, and N the centre of the circle through them.
Let AB be the perpendicular to the plane of P, Q, R through N.
Then if O is any point in AB, it is easily proved that the
A* OPN, OQN, ORN are congruent; so that OP=OQ=OR.
Thus with any point in AB as centre a sphere may be drawn
through P, Q, R. In other words, an infinite number of
spheres can pass through P, Q, R; and the locus of their
centres is the straight line AB, namely the perpendicular to
the plane of the APQR through its circum-centre.
496 GEOMETRY.

46. One sphere, and only one, can pass through any FOUR points
not in the same plane.

Let A, B, C, D be four points not in the same plane, and let

F and G be the circum-centres of the A* ABC, ADC.
Let FH, GK be the perps. to the planes ABC, ADC through F
and G respectively.

Then every point in FH is equidistant from A, B, C; and

every point in GK is equidistant from A, D, C; hence every
point in FH and in GK is equidistant from A and C.
But the locus of points equidistant from A and C is the
plane which bisects AC at right angles ;
*, FH and GK both lie in this plane; and since they ecan-
not be par' (being perp* to intersecting planes) they must
meet at some point O.
Then O, the only point common to FH and GK, is equi-
distant from A, B, C, and D.
‘, a sphere having its centre at O, and radius OA, will
through A, B, C, and D; and this is the only sphere that can
pass through the four given points.
 THE SPHERE. 492 |a

EXERCISES ON THE SPHERE.

(Theoretical. )

1, Shew that any tangent plane to the inner of two concentric

spheres will cut the outer in a circle of constant radius.

2. Find the locus of points on a sphere at a given distance from a

given point P. Illustrate by diagrams the several cases when P is inside,
on, or outside the sphere.

3. Two spheres of radius 7 and 7’ have their centres a cm. apart.

What are the conditions that the spheres (i) touch (ii) cut one another?
If the spheres cut, prove that their line of section is a circle.

4. How many tangent lines may be drawn to a sphere from an

external point? What surface do such tangents generate? Prove that
they are all equal ; and find the locus of their points of contact.

5. If a fixed sphere is cut by planes which pass through a giver

point, find the locus of the centres of the sections. Distinguish between
the cases which arise when the given point is (i) within, (ii) on,
(iii) outside the fixed sphere.

6. A fixed point O is joined to any point P in a given plane not con-

taining O; and on OP a point Q is taken such that OP. OQ=constant.
Find the locus of Q.

7. Shew how a sphere may be inscribed in aay tetrahedron, so as to

be touched by each of its faces. Shew also that four spheres may be
. escribed to a tetrahedron.

8. If Rand r denote the radii of the spheres circumscribed about,

and inscribed in, a regular tetrahedron each edge of which measures 2a,
shew that
hon 2/6.

9. Find the locus of points in a given plane at which a straight line

of given length and fixed position in space subtends a right angle.

10. If a sphere can be placed in a wire tetrahedron so as to touch

all the edges, the sum of each pair of opposite edges is the same.
428 GEOMETRY.

47. To find the surface of a sphere.

Let O be the centre of a sphere of radius 7, generated by the
revolution of a semi-circle APB about the diameter AB.
Let PQ be a side of half a
regular polygon (of an even Q
number of sides) inscribed in the
semi-circle.
Draw OM perp. to PQ and
therefore bisecting it.
A B
Draw Pp, Mm, Qq perp. to AB.
Then as the semi-circle revolves about AB, the side PQ will
generate the curved surface of a conical frustum ;
", curved surface of frustum = 27.Mm x PQ. (Art.35, Note 1.)
Now, if @ denotes the angle between pg and PQ,
Mm
py=PQ
= cos 0 = PQ —"0"

for Mm and MO are respectively perp. to pq and PQ;

*. Mm, PQ=OM. pq.
‘, eurved surface of frustum = 27 .OM x pq.
And when the number of sides is indefinitely increased, and
PQ in consequence indefinitely diminished, the surface of this
frustum ultimately becomes the belt of the sphere generated
by the arc PQ. Also in the limit, OM=r.
’, the area of belt = 27r x (projection of PQ on AB).
But the surface of the sphere is the sum of all the belts corre-
sponding to the successive sides ;
and the sum of the projections of all the sides = AB= 2r,
.. the surface of sphere = 2rr x 2r
= 47rr?,

Notr. The surface of a sphere is thus four times the area of a great circle,
 SURFACE OF A SPHERE. 499

48. DEFINITIONS. A frustum of a sphere is the part cut

off between two parallel planes. The curved surface of a
frustum of a sphere is called a zone.

Either part of a Athan cut off by a

single plane is called a segment. The
curved surface of a spherical segment is
sometimes called a cap.

If in a frustum of a sphere one of the planes of section PP’ moves

parallel to itself until it becomes a tangent plane (Art. 39.) the circular
end will vanish, and the frustum becomes a segment.

49. From Art. 47

the area of a zone=2rr x (distance between planes)
= Dark,
where 7 is the radius of the sphere, and & the thickness of the
frustum.
This was proved when the thickness was indefinitely small; but, by
addition of narrow zones, the formula holds good for all values of &.

Similarly the curved surface of a segment =27rh, where r is

the radius of the sphere, and / the height of the segment.

Note 1. Since the area of a zone depends only on the radius of the
sphere and the thickness of the frustum, it follows that @ zone of given
thickness has the same area from whatever part of the sphere it as cut.

Norte 2. Ifa cylinder is circum-

scribed about a sphere, and a frus-
tum cut by two planes perpendicular
to the axis of the cylinder, then the
area of the zone is equal to the area
of the corresponding cylindrical belt,
since each area=27rk.

Hence the whole surface of the sphere is equal to the curved surface of
the circumscribed cylinder.
430 GEOMETRY.

50. To find the volume of a sphere of radius r.

The surface of the sphere may
be divided into minute portions
or elements of area, as indicated
in the diagram. If these elements
of area are diminished inde-
finitely, each one tends to become
ultimately plane, and may be
considered the base of a pyramid,
whose vertex is at the centre,
and whose height is the radius
of the sphere.
Now the volume of any such pyramid
=1(the element of surface) x r.

But the sum of all the elements of surface is the whole

surface of the sphere;
and the sum of all the corresponding pyramids is the whole
volume of the sphere.
", volume of sphere =1(surface of sphere) x r
=ixdarxr
= 577,

51. Derrition. A sector of a sphere is the solid bounded

by a segmental cap and the conical surface traced out by a
radius of the sphere moving round the edge of the cap. (See
Fig. p. 431.)

52. By the method of Art. 50 it may be shewn that the

volume ofa sector of a sphere is 3Sr, where S denotes the
surface of the segmental cap.
 VOLUME OF A SPHERE AND SEGMENT. 43]

53. The volume of the segment

PAQ may be found by taking the
difference of the solid sector (O, PAQ)
and the cone (O, PQ).
Let r denote the radius of the
sphere, 7, the radius PN of circular
base, and / the height AN of the seg-
ment.

Then volume of segment = - Qrrh — senitr —h)

x ka eax al (he 1) eae a eee (i)

But by Theorem 57, B20 hea Ae he ORCe eRe (ii)

To obtain the required volume in terms of 7 and h, substi
tute in (i) the value of 7,? obtained from (ii) ;
and to obtain the volume in terms of 7, and h, substitute in
(i) the value of 7 from (ii).
Simplifying in each case, it will be found that

volume of segment=nhe(r ~ 5)ee ee (iii)

et ne(Si eternctent 8(iv)

54. The volume of a frustum of a sphere of thickness

& may be found by taking the difference between two
segments of heights h, and h, respectively, and remembering
that h, -—h,=h.
Use the result marked (iii), and on reduction by means of
(ii), it will be found that
volume of frustum of sphere = ae(37,2 + 37.2 + 2),

where 7, and r, are the radii of the circular ends.

 432 GEOMETRY.

EXERCISES ON THE SPHERE.

(Numerical.)

1. Find to the nearest square centimetre the surfaces of the spheres

whose radii are (i) 2°4 cm., (ii) 10°5 em. Also find their volumes to the
nearest cubic centimetre.

2. Find to the nearest penny the cost of gilding a hemispherical

dome 12 ni. in diameter at ls. 6d. per square metre.

3. Find the radius of a sphere whose surface is equal to that of a

circle 2‘8 cm. in diameter.

4. How many solid spheres 6 cm. in diameter could be moulded

from a solid metal cylinder whose length is 45 cm. and diameter 4 cm. ?

5. Find to the nearest cubic centimetre the volume of a spherical

shell whose internal and external radii are 6 cm. and 5 em. respectively.

6. Find to the nearest cubic centimetre the whole surface of a

hemispherical bowl, 1 em. in thickness, and 10 cm, in external diameter,

7. A solid metal sphere, 6 cm. in diameter, is formed into a tube

10 cm. in external diameter and 4 cm. in length; find the thickness of
the tube.

8. Find the whole surface and weight of a hemispherical copper

bowl 12 cm. in external diameter and 1 em. in thickness; assuming
that 1 cu. cm. of copper weighs 8°88 grams.

9. A sphere whose radius is 3°5 cm. is enclosed in a hollow cylinder

of the same radius, whose length is equal to its cireumference : how
many cubic centimetres are there in the remaining part of the cylinder?

10. Find to the nearest millimetre the radius of a sphere whose

surface is equal to the whole surface of a cylinder of height 16 cm, and
diameter 4 cm.

11. Assuming a drop of water to be spherical, and one tenth of an

inch in diameter, to what depth will 500 drops fill a conical wine glass,
the cone of which has a height equal to the diameter of its rim?

12. A sphere of diameter 6 om. is wae into a cylindrical vessel

. partly filled with water, The diameter of the vessel is 12cm. If the
sphere is completely submerged, by how much will the surface of the
water be raised ?
 EXERCISES ON THE SPHERE. 433

13. The weights of two spheres are in the ratio of 8 to 17, and the
weights of a cubic foot of the material in the two spheres are in the
ratio of 289 to 64: compare their radii.

14. Find the approximate weight of 50,000 spherical lead lullets,

each 8 mm. in diameter, the specific gravity of lead being 11°35.

15. Assuming the specific gravity of copper to be 8°88, find the

weight of a spherical copper shell, 12 cm. in external diameter, and
2 cm. thick.

16. How many metres of wire 0-4 mm. in diameter may be drawn
from the amount of copper required to mould a solid sphere of diameter
18 cin.?
If the diameter of the wire is decreased by 5 per cent., by how much
per cent. will its length be increased?

17. Find the whole surface and volume of a spherical segment

greater than a hemisphere, its height being 18 cm., and the radius of
the sphere being 13 cm.

18. Find the whole surface and volume of a spherical frustum, the
diameter of the sphere being 20 cm., and the distance of the plane ends
from the centre (on the same side) being 6 cm. and 8 cm.

19. Find the area of a spherical zone, the radii of the two ends
being 12 cm. and 5 cm., and the thickness 7 cin.

20. A sphere of diameter 24 feet is placed so that its centre is 37

feet from an observer’s eye. Find the area of that part of the sphere’s
surface that is visible to the observer.

21. Considering the Earth as a sphere of diameter 8000 miles, find

roughly in feet at what height above the ground would one-millionth of
the Earth’s surface be visible.

92. AB is anarc of a circle whose centre is O. If the sector OAB

rotates about the radius OA, shew that the curved surface of the solid
generated is expressed by 7(chord AB)’.

23. A cylinder is circumscribed about a hemisphere, and a cone is

inscribed in the cylinder so as to have its vertex at the centre of one
end, and the other end as base ; shew that
vol. of cylinder vol. of hemisphere
_vol. cf cone
3 2 ih

9)
i.S. G. ~E
434 GEOMETRY.

LINES OF REFERENCE ON A SPHERE. LATITUDE AND

LONGITUDE.

55. In the sphere whose centre

is O a diameter NS is taken as axis.
The great circle XQx’, of which NS
is the axis, is then called the equator,
and the points N and § are respec-
tively its North and South poles.

Then NS will be the axis, and N, S the poles, of all small

circles in planes parallel to the equator. Let BPC be any such
small circle, and P any point on it.
Through the poles N and § an infinite number of great
circles may be drawn: let NPS be that which passes through
P, and let it eut the equator at Q.
Now it may readily be proved (as in Art. 37) that the 2NOP
is constant for all points on the small circle BC ; hence the are
NP, called the North polar distance of P, is constant for all
such points. And since the LNOQ=90", the 2 POQ is also
constant.
The 2POQ, or the corresponding are PQ is called the
latitude of the point P, and may be defined as its spherical
distance from the equator.
The small circle BPC is called a parallel of latitude, for the
latitude of all points on it is the same. This latitude is denoted
in the diagram by 0.
Of the system of great circles which pass through the poles
N and §, semi circumferences, such as NPQS, are called
meridians.
The position of a meridian relative to some fixed meridian
of reference is determined by the angle between their planes.
Thus if NXS is the fixed meridian of reference, the position of
the meridian NPQS is determined by the .XOQ, for OX and OQ —
are perpendicular to the line of section NS.
 LATITUDE AND LONGITUDE. 435

The 2 X0OQ, denoted in the diagram by 4, is called the

longitude of the meridian NPQ§, or the longitude of all points
on that meridian.
It is now clear that the position of a point P on the
surface of a sphere is fixed if we know on what parallel of
latitude and on what meridian of longitude it lies: briefly, if we
know its latitude and longitude. These angular data corre-
spond to the linear coordinates which fix the position of a point
on a plane.

EXERCISES.

1. A plane section of a sphere is taken in latitude 6. If the radius

of the sphere is 7, and the radius of the circular section 7,, shew that
T,=r cos @.
2. Taking the equatorial radius of the Earth as 3960 miles, find
approximately, using Four Figure Tables
(i) the length of the equator ;
(ii) the length of a knoé (1 minute of equatorial arc) ;
(iii) the length of the parallel of latitude 55°;
(iv) through how many miles an hour London moves in consequence
of the rotation of the Earth [Latitude of London=51° 30’).
eco 8 segment is cut off from a sphere of radius r by a plane in
latitude @ ; deduce from the formule of Arts. 49 and 53.
(i) Surface of segmental cap=27r?(1 — sin-6).
(ii) Volume of segment = }173(1 —sin @)(2—sin 6 - sin? 4).

4,= Shew that the area of a spherical zone included between the
latitudes @, and 0, is given by the formula
27rr*(sin 6, — sin 4).
5. Taking the mean diameter of the Earth as 7922 miles, Aad as
nearly as you can with Four-Figure Tables
(i) the whole surface of the Earth;
(ii) the surface of the Arctic Cap [lat. of Arctic Circle=664°] ;
(iii) the surface of the Tropical Zone [lat. of Tropics
= 233° N. and §.}.
6. If@and ¢are the latitude and longitude of a point (a, y, z) on
the surface of a sphere of radius 7, shew that
x=reosécos¢?, y=rcos@sng, z=rsin$.
436 GEOMETRY.

LUNES OF A SPHERICAL SURFACE.

56. Any two planes cutting a

sphere centrally must intersect in a
diameter; hence any two great circles
must cut at the extremities of a dia-
meter.
The angle at which two great circles
cut one another is called a spherical
angle, and is measured by the angle
between the tangents to the circles
at either point of intersection.
Thus the angle between the great circles ABA’, ACA’ is that between
the tangents AP, AQ. Now these tangents are respectively in the
planes of the circles, and they are perpendicular to the line of section
AA’ ; therefore the angle between them is the measure of the dihedral
angle between the planes.

Hence a spherical angle is measured by the dihedral angle

between the planes of the intersecting great circles.

57. Derinirion. <A Lune of a sphere is a part of the sur-.

face cut off by two planes passing through a diameter. The
boundaries of a lune are therefore two great semi-circles ; and
the spherical angle betiveen them is called the angle of the
lune.

58. To find the area of a lune.

It will easily be seen that the areas of lunes are proportional
to their spherical angles; moreover the whole surface of a
sphere may be regarded as a lune whose angle is 360°.
Hence, if the angle of a lune is D degrees,
Area of lune =(whole surface of sphere) x a
D
=r? , 50
 SPHERICAL TRIANGLES. 437

SPHERICAL TRIANGLES.

59. DEFINITION. A triangle drawn on the surface of a

sphere, the sides being the arcs of great circles, is called a
spherical triangle.

60. If the vertices of a spherical

triangle ABC are joined to the centre
O of the sphere, the three planes AOB,
BOC, COA form at the point O a
trihedral angle closely related to the
triangle ABC.

For instance: the sedes of the spherical triangle ABC may be

considered either as the arcs AB, BC, CA, or as the fuce-angles
AOB, BOC, COA. Thus the sides a, , ¢ cf a spherical triangle
may be measured in degrees.
Again, the spherical angles A, B, C have the same measure
as the dihedral angles of the solid angle (O, ABC.
From this connection between a spherical triangle and a tri-
hedral angle we conclude that:
(i) The sum of any two sides of a spherical triangle is greater
than the third.
For the sum of the face-angles AOB, AOC, which measure the sides
AB, AC, is greater than the third face-angle BOC, which measures the
side BC. (ZVheor. 97.)

(ii) The sum of the sides of a spherical triangle is less than

the circumference of a great circle.
For the sum of the face-angles AOB, BOC, COA is less than four
right angles (Theor. 98); hence the sum of the correspending arcs is less
than four quadrants.

Norr. From the first of these results we see that any side of a spher-
ical polygon is less than the sum of the remaming sides. And hence: The
shortest line on the surface of a sphere between two points is the minor arc
of the great circle which passes through them ; for any other line may be
regarded as the sum of minute arcs of great circles, in the limit when
each arc is diminished indefinitely.
438 GEOMETRY.

61. The diameters AA’, BB’, CC’

through the vertices of a spherical tri-
angle ABC meet the surface at the
vertices of a spherical triangle A’B'C’,
which is said to be the opposite or
symmetric of the first.

62. Now just as the trihedral angles (0, ABC), (O, A’B’C’)
are equal in their several parts, but yet are not superposable
[see page 377], so the sides and angles of the spherical triangle
ABC are severally equal to the sides and angles of its opposite
ABC’; but owing to the curvature of their faces the twe
triangles are not in general superposable.
For if we look on the convex face of each triangle, the
sequence of the vertices A, B, C is clockwise, while that of the
corresponding vertices A’, B’, C’ is counter-clockwise.
Now this disparity, if the triangles were plane, might be met
by turning one of them over before superposition {see Part L
p- 19]; but such a reversal in the case of spherical triangles
would bring the two convex faces into contact, so that coin-
cidence would be impossible.

Norr. An isosceles spherical triangle and its opposite are super-

posable. For, A and A’ being the vertices, we shoul ate
AB=AC=A’B’=A(C’;
so that by placing A’ on A, B’ might be made to fall on C, and C’ on B.

63. Though a spherical triangle and its opposite are not

generally superposable, they are always equal in area.
For the triangle ABC may be divided into minute triangles,
each of which will have its opposite in the triangle A’B'C’.
Now each such pair of opposites, if taken small enough, may
be regarded as plane and therefore superposable. Thus the
triangle ABC is made up of elements of area each of which has
its equal in the triangle ABC.
 SPHERICAL TRIANGLES. 439

Norr. Just as a plane triangle may be divided into three isosceles

triangles by joining the vertices to the circumcentre, so a spherical
triangle ABC may be divided into three isosceles spherical triangles by
joining the vertices to the pole of the plane ABC. Now each isosceles
triangle has for its opposite an equal and superposable isosceles triangle;
and the three opposites makeup the whole opposite triangle A’B’O’.

64. To find the area of a spherical triangle.

Let A denote the area of the

spherical triangle ABC, formed by arcs
of the great circles ABA’, BCB’, CAC’ ;
the points A’, B’, C’ being respectively _
opposite to A, B, C.
n

Now :
lune of ZA =A+ AA'BC;
lune of 2B =A+ AABC;
lune of LC=A+ AABC’=A+ AABC; (Art. 63.)
.., by addition,
lune A + lune B + lune C= 2A + {A+ AA’BC+AABC+ AABC}
“2, .
or, ae {A+B+C} =2A+surface of hemisphere ;

rr
or, Pe a eee
* wre °
: A= Tap iAtB+C — 180 as aeaiadecnies (i)
|

Norsr. Since A must be a positive quantity, we see from (i) that

A+B+C must be greater than 180°; that is, the sum of the angles
af the spherical triangle is greater than two right angles. The angle
A+B-+C- 180° is called the spherical excess, and is denoted by E.
2
Hence io xE, if E is measured in degrees ;
or A=r"xE, if E is given in radian measure.
440 GEOMETRY.

M{SCELLANEOUS EXERCISES.

(In some of the following examples numerical work may be shortened by

the use of Four-Figure Logarithms. )

1. A cube and a sphere being of equal volume, find the ratio of the
radius of the sphere to the side of the cube.
2. The diagonal of a cube is 584 cm.; find the radius of a sphere
whose surface is equal to that of the cube.
3. Find the surface and volume of a cone whose base is 97°6 sq. em.,
and such that the height is to the radius of the base as 11 to 6.
4. A solid, consisting of a right cone standing on a hemisphere, is
placed upright in a right cylinder full of water and touches the bottom,
Find to the nearest cubic foot the volume of the water remaining in the
cylinder, having given that the radius of the cylinder is 3 ft. and its
height 6 ft., the radius of the hemisphere 2 ft., and the height of the
cone 4 ft.
5. The diameter of a sphere is 32°4 cm. and its volume is 5°6 times
as great as that of a cone whose height is 702 cm. Find the radius of
the base of the cone to the nearest millimetre.
6. Find the slant surface and volume of the frustum of a pyramid
whose thickness is 15 em., and whose ends are squares on sides of 40 cm.
and 24 cm. respectively.
7. How many cubic feet of air are contained in a tent in the form
of a circular cylinder surmounted by a cone, the radius of the base being
118 inches, the vertical sides 124°3 inches, and the extreme height to
the vertex of the cone 217°9 inches ?
8. If a cone of lead 24°6 cm. in height can be hammered into a
solid sphere 15:0 cm. in diameter, find to the nearest millimetre the
radius of the base of the cone.
9. Find the surface and volume of the solid generated by the
revolution of an equilateral triangle about one of its sides, each side
being 7°4 em. in length,
10. A cireular room, surmounted by a hemispherical vaulted roof,
contains 5236 cubie feet of air, and the internal diameter of the buildin
is equal to the height of the crown of the vault above the floor. Fin
the height.
11. Prove that the volume of a spherical shell is equal to that of the
frustum of a cone whose height is four times the thickness of the shell,
and the radii of whose bases are the outer and inner radii of the shell.
 MISCELLANEOUS EXERCISES. 44)

12. If V and S are the volume and whole surface of a cone, and
V’, S’ the volume and surface of an inscribed sphere, prove that
V:W=S:S’.
13. To the ends of a cylinder, whose length is equal to its diameter,
are applied hemispheres of the same diameter as the cylinder. If the
volume of the whole solid thus formed is 464 cubic centimetres, find its
surface.
14. Water passes into a reservoir from a cylindrical pipe 30 cm. in
diameter, flowing through the pipe at the rate of 1°25 metre a second.
Find how many thousand litres enter the reservoir in 24 hours.
15, Find in kilograms, as nearly as possible with Four-Figure Tables,
the weight of sufficient mercury to fill a cylindrical vessel of depth 64 cm.,
and internal diameter 8 cm. The specific gravity of mercury being
taken as 13°6.
16. If in measuring the diameter of a sphere the error on either
side may be as much as one per cent. of the true diameter, by how
much per cent. may the calculated value of the volume exceed the real
volume.
17. Find, as nearly as possible with Four-Figure Tables; how many
metres of wire 0°4 mm. in diameter can be drawn from 593 kg. of coppey
of which the specific gravity is 8 88.
18. Find roughly, using Four-Figure Tables, how many spherical
shot, each 2°6 mm. in diameter, can be made from 10°45 kg. of lead,
the specific gravity of lead being 11°35.
19. The volume of a frustum of a cone is given by the formula

V=5(A+VAB+B),
h denoting the height of the frustum, and A and B the areas of the two
ends. Calculate the volume to the nearest cubic inch, when
h=45", A=28"5 sq: in.; B=78"6 sq; in.
What does the formula become (i) when A=B, (ii) when A=0? Give
the geometrical interpretation in each case.
20. A water-tube boiler has 350 tubes of 2°5” internal diameter,
and the length of each tube is 8 feet. Find the total heating surface
(i.e., the interior curved surface of the tubes) in square feet to the
nearest integer.
21. A right circular cone was measured in such a way that the
diameter of the base is known to be between 16°2” and 16°3”, and the
height between 27°5” and 27°6’. Taking the value of 7 as 31416, find
the volume of the cone using (i) the lesser (ii) the greater dimensions.
If you give your answer in cubic inches to seyen significant figures how
many are useless?
449 GEOMETRY.
—
is
22. <A cubical block of metal, each edge of which is 36°4 cm. is
melted down into a sphere. Find the diameter of the sphere as
correctly as possible with Four-Figure Tables.
23. The weights of two spheres are in the ratio 8:11, and their
specific gravities are respectively 1-21 and 0°64. If the diameter of the
first is 5°6 cm. find the diameter of the other.
24. A right circular cone is cut by two planes parallel to the base
and trisecting the height. Compare the volumes of the three parts inte
which the cone is divided.
25. Supposing the Earth to be a sphere of diameter 7926 miles, find
the length of the Arctic Circle [lat. 66° 30’] as correctly as possible with
four-Figure Tables. Also the area of the zone between latitudes 60°
and 65°.

26. Find, as correctly as possible with Four-Figure Tables, the

volume of the greatest cube that could be cut from a sphere of diameter
3762 inches.
27. The water contained in a cubical cistern, of which each internal
edge is 6 feet, is found to lose by evaporation 0°04 of its volume in a
day. Assuming the loss to arise from evaporation only, find how many
yunces of water will be left in the cistern after 10 days,
28. A Sere tetrahedron weighs 10°70 kg., its substance bein
jead of which the specific gravity is 11°35. Find the length of each
edge as nearly as possible with Four-Figure Tables.
29. Find the volumes of the following solids of revolution, con-
sidering them as the sum or difference of cylinders, cones, or frusta of
cones. The solids are generated by the revolution of
(i) an equilateral triangle (side=a) about one side ;
(ii) an equilateral triangle (side=a) about a line through a vertex
parallel to the opposite side ;
(iii) a square (side=a) about a line through one vertex parallel to —
the diagonal which does not pass through that vertex.
(iv) a regular hexagon (side=a) about one side.
In each case shew that the volume is equal to that of a prism whose
base is the a uLa A figure, and whose height is the circumference of
the circle described by the centre of the figure.
30. Assuming the principle indicated at the end ot the last example,
find the volume of a solid ring generated by the revolution of « circle
of radius 1?” about a line 7” from its centre,
 ANSWERS TO NUMERICAL EXERCISES.
Since the utmost care cannot erasure absolute accuracy in graphical work, vesulte %
obtained are likely to be only approximate. The answers here given are those found by
calculation, and being true so far as they go, furnish a standard by which the studend
may test the correctness of his drawing and measurement. Results within one per cont
of those given in the Answers may usually be considered satisfactory.

Exercises, Page 15.

30° s 126°; 261°s 85°. 11 mim; 37 min.
1124°; 155°; 5 hrs. 45 min. 8. 50°; 8 hrs. 40 min.
@) 145°, 35°, 145°, (iii) 55°, 66°. 86°, 94°.

Exercises. Page 27.

1. 68°, 37°, 75° v. nearly. 2. 60cm. 4 2:2’, 50°, 73° nearly.
5. 37 ft. 6. l0l metres. 7. Q7ft. 8. 424 yds.,nearly; N.W.
9. 281 yds., 155 yds., 153 yds. 10. 214 yds,
Exercises. Page 41.
E, 1957555", 125°. 12. 15 secs., 30 secs,

Exercises. Page 43. :

a SI’. 4 Q7°. 5: 92°, 46°. Go NGimn Ge.

Exercises. Page 45.

4. 30°, 60°, 90°. 2) Misbs teers: se LO SOm SO
3. 40°. 4 siete, iss 9. (i) 34-3 (ai) 107°.
6. 68°. te WAS 8. 36°, 72°, 108°, 144%
9. -165:. Li eon lo:

Exercises. Page 47.

2. (i) 45°; (ii) 36°. Se G)) 125; (eld,

Exercises. Page 54,

4, (i) 81°; ¢ (ii) 55°.

Degrees |15° | 30° | 45° 75°

10.
Om. |41 |46 |57 |8-0 |156
Degrees |oO” | 30° | 60° | 90” | 120° |150° |180
180° |
11.
Cin |1-0 |2-0 |3°6 |50 |6:1 |6:3 | 7-0
12, 37. ft. 13. 112 fb. 14, 346 yds. 693 yds.
 GEOMETRY.

Exercises. Page 61.

14. 54°, ‘72°, 54°. 15. 36°. 1B.
18. (i) 16; (ii) 45°; (iii) 113° per sec.

Exercises. Page 68.

6°80 cm. oon oes 4. 0:39. 5. 2°54. 8. 10°6 em
Biatsh 10. 20 miles; 12°6 km.
1 Ie 147 miles; 235 km. 1 cm. represents 22 km. .
12. 1” represents 15 mi. ; 1” represents 20 mi.

Exercises. Page 79.

0°53 in. 4 1:3 em. 6. 24",

Exercises. Page 84.

4°3 cm., 5°2 cm., 6°1 cm. 2, Viga0 8. 200 yards,
Go, 77 m., 61 m., 66m. 5. 6°04 knots. S, 15° E, nearly.
Results equal. 9 cm. 7. 4°38 cm.; 9°8 cm., 60°; 120°.
(i) One solution ; (ii) two; (iii) one, right-angled ; (iv) impossible.
380 yds. 10. 6:5 cm. 11. 6'9'om,
Two solutions; 10°4 cm. or 45 cm. 16. 2°8cm.,4°5cem., 53cm.
-_Se
=
SSS
~—_ 5°8 cm., 4°2 cm. 19. 7 cm., 8 cm.

Exercises. Page 89.

60°, 120°. 2 3:54". Bo Sis", 4. 44cm.
16°4 cm., 3°4%. 6. 90°. %. Gyae2s’s. (nn) B= D=90",

Exercises. Page 102.

6 sq. in. 2. 6 sq. in. 3. 2°80 sq. in. 4 3°50 sq. in,
3°30 sq. in. 6. 3°36 sq. in. 7%. 1988q.m. 8. 428q. ft.
10,000 sq. m. 10. 110sq. ft. 11. 5 em, 12. 2°6 in.
900 sq. yds. ; 48 yds.; 4°8”. 15. 11700 eq. m.
1 em. =10 yds. is, aaa 18, 600 sq. ft. 19 1152 sq. f&
100 sq. ft. 21. 156 sq. ft. 22. 110 sq. ft.
288 sq. ft. 24, 72 sq. ft. 25. 75 sq. ft.

Exercises. Page 105.

(i) 22 em.; (ii) 3 6’. 2. 3:4 8q. in. 8. 574°5 sy. in.
18%, B, 108", 75:
 ANSWERS. ii

Exercises. Page 107.

(i) 180 sq. ft. ; (ii) 84 sq. in. ; 1 hectare.
(i) 13°44 sq. cm. ; (ii) 15°40 sq. em. ; (iii) 20°50 sq. cm.
15 sq. cm. 4. 6:3 sq. in.
ke (i) 873 (i) 13 emi
PESO 6. 3°36 sq. in.

Exercises. Page 110.

11400 sq. yds. 2. 6312 sq. m.
24 eme > 5:1 em: Ay 250455 2:22

Angle |0° |30° | 60° |90° |120° |150° |180°

|

Area in gq. om. | Oma: (180 |15-0 150 15 | 0

i pa llig

Exercises. Page 111.

66 sq. ft. 2. 84 sq. yds. 8. 126 sq. m.
132 sq. cm. 5. 180 sq. ft. 6. 306 sq. m.

Exercises. Page 113.

6 sq. in. 2. LO sq. tt. 8. 615 sq. m. 4. 8:4 sq. in.
31°2 sq. em. 6. 5°20 sq. in. 7. 24 sq. cm.

Exercises. Page 115.

(i) 25°5 sq. em. ; (ii) 15°6 sq. em.
(i) 8°95 sq. in.; (ii) 9°5 sq. in. 8. 12500 sq. m.

Exercises. Page 116.

3°3 sq. in. 5. 7°5 cm. 6. 3°6 sq. in.

Exercises. Page 121.

(i) oem. 3 Gi) Goreme; (in)'3°7". 2p (wi e6s s(n) 2*Siem:
41 ft. 4. 65 miles. oO. Oolikimns 6. 16 ft.
48 m. 8. 25 miles. De a te 10. 62 ft.

Exercises. Page 123.

(i) and (iii). ll. 2°83". 12. 4:24 cm.; 18 sq. cm,
70-7 Wsq., Tot 14, »y=6°93 cm.
(i) 20 cm.; 15cm. (ii) 40 cm.; 39 cm.

5*1 cm. nearly.

iv GEOMETRY.

Exercises. Page 127.

71 cm. 4. 40cm. 5. 16% 6 3:lem.; 15°6 sq. om.

Exercises. Page 130.

23°90 sq. cm. 2. 8°40 sq. in.
27°52 sq. cm. 4, 129800 sq. m.

Exercises. Page 134.

(i) (8, 5)5 (ii) (10, 10).
(i) (4, 5) 5 (ii) (4, 5)5 (iii) (-4, -—5) ; (iv) (-4, -5).
(6, 5), (12, 10). 6. (5, 8).
(i) 17 ; (ii) 173 (iti) 2°65"; 2°75”.
(i) and (ii) 5; (iii) and (iv) 17 ; (v) and (vi) 37. 9% 10
(0, 0). (7, 5). 15. 13; (9, 6).
A straight line passing through the points (4, 0), (0, - 4).
117 units of area in each case. 18, Asquare. 2sq.in. 1sq. in
Each=70 units of area, 20. 9 units of area. 31°, 71°, 78°.
(i) 96 ;(ii) 80; (iii) 120; (iv) 104.
(i) 503 (ii) 605 (iii) 120; (iv) 132.
Sides 5, 13; area 63. 24. (i) 27; (ii) 21; (iii) 305 (iv) 27°5.
(i) 50; (ii) 65°5 ; (iii)21 5 (iv) 83-5.
Each side 13; area 120. 27. 13, 10, 15, 8-24, 42, 30.
AB=10, BC=9, CD=17, DA=12°7, Area=130°5.
10, 13, 5, 5, 3. Area=60. 30. 160,000sq. yds. 1000 yds. 320 yds,
Side =15'23; area=232 units of area,

Exercises. Page 145.

5 cm. 9, 24", 3. 06", 0:8”. 4, /7=26 om.
1 ft. 6. O6sq.in. 7% 0°8%
. Exercises. Page 149.
147” 2. 3/2=4-2 cm. 3. 2N3=3'5 om,
ts 6. 5 cm.

Exercises. Page 151.

4 cm. 7; ee

Exercises. Page 153.

1:85”, o T62. 6. OBB" y (G1", 21")'s SOT",
 ANSWERS. : bad

Exercises. Page 155.

81” 6. 1:6" 1-57, 06°.

Exercises. Page 157.

(8, 11). 5. 17; 10; (0, —8).

Exercises. Page 161.

74°, 148°, 16°. 2 ee 2305, Sh Gay, iy Gaye

Exercises. Page 177.

8:0 cm. 2s °O'6". 3. 8-7 cm. AO Gia by Has.

Exercises. Page 179.

3 cm. and 17 cm.

Exercises. Page 181.

72°, 108°, 108°.

Exercises. Page 187.

1:6”. Se ey 4, 1:98”, 16".

Exercises. Page 198.

2°3 cm., 4°6 cm., 6°9 cm. ae. So
6°9 cm.; 20°78 sq. cm. 7. 32cm.

Exercises. Page 199.

2°12”; 4°50 sq. in. 4 85cm. 6. 2a,

Exercises. Page 200.

1287; 1°73”.
Exercises. Page 201.
3 46”; 4:00". 2. 259°8 sq. cm.
(i) 41°57 sq. cm.; (ii) 77°25 sq. cm.

Exercises. Page 205.

(i) 28°3 om.; (ii) 628-3em. 2 (i) 16°62 sq. in.; (ii) 352-99 so, Im
11°31 cm.; 10°18 sq. cm. 4, 56 sq. cm. 5. 43°98 sq. in,
30°5 sq. cm. Se O'S. aie: shale 10. 12°57 sq. in.
Circumferences, 4°4”, 6°3”.
oo Areas, 1°54 sq. in., 3°14 sq. in.
 GEOMETRY.

Exercises. Page 225.

6°4 sq. cm, 4. 3:7’. 5. 10 em. 6 xX,

Exercises. Page 228.

630 sy. cm. 15 cin.

Exercises. Page 231.

85cm. 90°. 8. A circle of rad. 6 cm.
NS 5:20”, 6. 0°25".

Exercises. Page 235.

(i) 16 sq.cm. (ii) 16 sq. cm.
2. (i) 16sq.em. (ii) 16 sq. cm,
os. & (i)12. @ij12@s5em. 6.° (167,417. (i) 8h an.
Two concentric circles, radii 2 em, and 6 cm.

Exercises. Page 237.

26". 2. 48 ft.; 8 ft. 8. 2cm.; 32 cm,
3°6". 5. 8100 miles; 10 miles.

Exercises. Page 239.

4 cm. a, 2: 3. 1°94”, 4. 1:97".
6°6 cm. G (616; 24, 1,. po2, 48, 8, 3:5 cm.
oa11Maier 10. 9°6, 2°6.

Exercises. Page 241.

2°47". 2. -3'24",

Exercises. Page 245.

8, 2. Mm” Ty ts
9, -4. 6. 11°32, - 4°32.

Exercises. Page 246.

8. 2. 36, 45. E 8. 16,12
(10, 124); 124. 5. (17, 18); 12/2=16-97.
PP
2 6. .&. 40 9. Four. (26, 15). 10, 1284,
 ANSWERS. Vii

Exercises. Page 253.

(i) 35; (ii) 8; (ili) a
rus 4°0”, 5°6”. 4. 16:5 cm., 12:0 cm.
4:0 cm., 2°4 cm. ; 16°0 cm., 9°6 cm.

Exercises, Page 258.

(i) each=3 :2; (ii) each=5:33 (iii) each=5: 2.
(i) 1-4”; (di) 08”; (iii) 6-4 cm., 2°4 cm.
(i) 56cm. (ii) 7°7 cm., 2°8 cm.

Exercises. Page 259.

0:9", 0:67 s.4:5", 3°07 3"3'¥ 2:
+N
we
a 2:0 cm., 1°55 cm. ; 14:0 cm., 10°5 cm.

Exercises. Page 262.

(i), 1-2" = Gi),2:0’>.(in) 7-7 om: (i): 2"1'"5. (Ti), Gedcom,
QB=35 > Bh=2.. 3:2 cm., 4°2 cm.
DSfatien(isne 5 ft., 123 ft., 94 ft.
Ba sn 2395". PP
2m 58 cm.
ee 0°8 cm., 1°4 cm., 2°1 cm.
ee

Exercises. Page ao
17, gry 4, fe % 87. sine= 22, cosine=38, tangent=32,
hid. eee res 6. 37°
13)? 129 85> 36° B :
3D, 26°, 45°. 8. A=58°, cosA=0°53.
AC=7'8, A=39°, sin A=0°63, cos A=0°78. Il. 3:4, 35°.

Exercises. Page 278.

(i) 1:0”; (ii) 0°9" ; (iii) 6-0 cm.
1:4”, 0°6"; 3-5”, 1:5”. 8. (i) 2-0; (ii) 2°8; (iii) 20.
1°6 cm., 2°4 cm., 3°2 cm. Oy re, 079". 6. O75
(i) 1°73 ; (ii) 3°16 ; (iii) 1°67. 8. (i) 35 (i) 3°21; (iii) 2-26.
PN (i) 1:2”, 1:6”, 2:0" ; (11) 3°0:em., 3°6 cm, 4*5 em.
eS
oS
(iii) 2°5 em., 4°3 cm., 5:0 cm. (iv) b=3 “a! c=2°1", nearly.

Exercises. Page 279.

140 m., 160 m. ; 125 m. 2. 124 yds.
424 miles. 4. 30 ft., 4 ft.
24 ft., 2 ft. 4 in. 6. 60 ft. he ©G2 ibe
2S
ae 106 ft.
H.8.G QF
 GEOMETRY.

Exercises. Page 284,

0°52, 5. 31:28, nearly.

Exercises. Page 287.

lo 10°5 sq. in, 2. 3:0 cm. 3. 64 sq. cm.
I-07 5. 33°9 acres.

Exercises. Page 289.

4°9 cm. 7. 80cm.

Exercises. Page 291.

> 2. 20 sq. ft. 8. 10 sq. cm. 4 706 5. 5-6".

Exercises. Page 294.

3°46", 4°33”, 5°54”. 8. 9 ft. 3 in. 4. 3°78 sq. cm.
4°5 cm. 6. 15°48 sq. in. 7 36m. 15m.
90 acres. 9. 512 acres. 10. 1 cm. represents 15 metres,

Exercises. Page 295.

1: V2. 6. x: 7. 256: 81,

Exercises. Page 297.

2°5 sq. cm., 6°4 sq. em. 4 4:1,
ia, 8. 6:2 cm., 3°8 cm.

Exercises. Page 299.

1: 2, 8. 46cm. 4, 69 cm.

Exercises. Page 301.

11°56 sq. in. 2. (i) 100, (ii) 576 units of area ; (iii) 18°5.
(1°3, 24), (1°, -2°4); 4:8; 2-4, nearly.
Exercises. Page 305.
10. (i) 10%” (ii) 158 ft.

Exercises. Page 313,

15; (8, 6). 2. 15, 3°75; (0, 20), (0, 5). 8. 0°83",
al 17"
os 6. 0°55". 7. 0°98".
 ANSWERS: ix

Exercises. Page 317.

%. 15°3 sq. cm. 2. 8°55”. 3. 43°30 sq. cm.
4 90°. S) 6:83" 1-17" 10. 20°78 sq. cm., 20°8 cm,
12. 56°. 14, 1:2: 227°. 15. 63 ft.
16. 18 sq. cm. 17. 604 sq. cm.

Exercises. Page 322.

a | 1s’ | 30° | 45° | 60° | 75°

17°6 cm. ; 25°.
PR | 8:3 | g-2 | 11-3 |16-0 |30:9
2. (i) 8°9 sq. cm. ; (ii) 72° or 108° ; (iii) at right angles.
8. When the rod is equally inclined to the rulers.
4 When P isthe mid-point of AB (i) is a maximum, (ii) is a minimum.
5. Minimum when a=3.
6. 45°. ye Se as 8; 29,

Exercises. Page 326.

2. (i) 6-0”; (ii) 1°2 cm.

Exercises. Page 329.

L 40cm., 8°8 cm. Xe (iO Gi) abs

Exercises. Page 355.

4. 70cm. 6. S07 3 2157; 6. (ii) 42°4cm.; (iii) 56°6 cm.

Exercises. Page 361.

OF 1s 6. 0°324.

Exercises. Page 362.

4. 10'4cem. 35cm. 102 cm. 0°385, 0°339.

Exercises. Page 375.

4, (i) 0°8000; (ii) 0°8571. §. 0:2800.

 GEOMETRY.

Exercises. Page 382,

3, 4,5. a 1S:

Exercises. Page 389,

72 sq. ft. 2. 170m.; +73; 144°48 sq. cm.
10cm. ; 55°54 sq. cm.

Exercises. Page 391.

25°50 m. 2. 650 litres; 520 kg. * 8. 6y millions,

1188 kg.

Exercises. Page 393.

196°3. 2. 20400 kg. 8. 74°88 kg.
5445. 5. 8s. 4d. 6. 21°42 kg.
5s. 1ld. 8. 65cm.; 65cm. 9. lem.
12 cm., 15 cm., 18 cm. ll. 7 cm., 6 cm., 5 cm.
5°8 cm., 200 sq. cm., 192°4 cu. cm.
12 cm., 4 cm. 14. 29 cm. 15. 14:4”,
720 cu. cm., 480 sq. cm. 17. 360 cu. cm., 432 sq. em.
12,000 cu. em. 19. 4°6 cu. ft.
375,000 gall., 167 tons. 21. 23661.
65 hrs. 61 mins. 28. (i) 3:43 (ii) 717: 1000.
459 working days.

Exercises. Page 401.

(i) 91°38 sq. in. ; (ii) 84 cu. in.
(i) 808 cu. cm. ; (ii) 280 cu. cm.
(i) 9°85”; (ii) 10°63”.
(i) 223°6 sq. em. ; (ii) 333°3 cu. cm.
8 cm., 1152 cu. cm. 7. (i) 0°28; (ii) 0°49 sq. in.
(i) 58cm. ; (ii) 28°87 sq. om. ; (iii) #. 60°.
8°7 cm,
 ANSWERS. xf

Exercises, Page 409,

377°400 cu. cm. 15°612, 16:050; 4:31%, 4-08%.
871 lbs. 3, 36" 52%
(i) 104°6 sq. em. ; (ii) 110°8 cu. cm.
4°2 em., 50°9 cu. cm. 6. 384 units of area,
Pp
ae
PS 162°4 cu. cm. 12, 70? 32".

Exercises. Page 413,

(i) 151 sq. em.; 226 cu. cm. (ii) 204 sq. em. ; 458 cu. cm.
528 sq. cm. $. 122 cu. cm, 4. 123 sq. cm.
528 sq. cm. 6. 10:20 m. 7. 0:006”.
Doe
Shake
5000 cu. cm. ; 15:9 cm. 9: “7 cur ft.
10. 88'1 kg. 11. 18°85m.; 525°9 gr.

Exercises. Page 417.

(i) 188 sq. cm.; 302cu. cm. (ii) 14sq.cm.; 5 cu. cm.
1414 sq. cm. 4." 27 cu.;cm.

Exercises. Page 421.

816 sq. cm. 2. 110 sq. cm. 3. 148 cu. cm.
os
ol
| 79 sq. cm. ; 88 cu. cm. 5. 122°2 sq. cm.
n3—(n—-1)8
140 cu. cm. 9, ——,—n> _.. 10. 10cm., 5 cm.

Exercises. Page 422,

20 cu. cm. 2. 268 yds. 1 ft. 3. 5°4m,
1017 sq. cm. 5. 5l min. 12secs. 6 1:2.
18 sq. cm. 8. 376°99 sq. cm. 9. 23:39.
ee 390 sq. cm.
oa
od
xii ANSWERS.

Exercises. Page 432.

1. (i) 72sq.cm.; 58cu. cm. (ii) 1385 sq. cm. ; 1849 cu. cm.
2. £16. 19s. 4d. 8. 0°7 cm. 4. 15.
5. 381 cu. cm. 6. 286 cu. cm. 7. Jom.
8. 418 sq.cm.; 1°694 kg. 9. 667 cu. cm.
10. 4:2 cm. pO ee a 12. lcm.
1S, eSeiki. 14, 152kg. 15. 5654 kg.
16. 24300 m.; 10°8%. 17. 1922°66 sq.:em. ; 7125°15 cu, cm
18, 439°82 sq. cm. ; 318°35 cu. cm.
19. 571°77 sq. cm. 20. 611°3 sq. ft. 21. 42 ft.

Exercises. Page 435.

2. (i) 24880 mi. ; (ii) 1°152 mi. ; (iii) 14270 mi. ; (iv) 645 m..
6. (i) 197,100,000 sq. mi. ; (ii) 8,172,000 sq. mi.;
(iii) 78,590,000 sq. mi,

Exercises. Page 440.

1, 31:50. 2, 23 cm.
8. 204 sq. cm. ; 332 cu. cm. 4. 136 cu. ft.
5; 6°6 cm. 6. 2803 sq. cm.; 15680 cu. cm.
7. 3937 cu. ft. 8. 8°3 cm,
9, 298sq.cm.; 318cu.cm. 10. 20 ft.
13. 315 sq. cm. 14, 7634000. 15. 43°75 kg.
16. 3:0% nearly.- 17. 531,500 m. 18. 100,000.
19. 232 cu. in. 20. 1833 sq. ft.
21. 1889 cu. in. 1920 cu. in. Data suffice for 4 significant digits only.
22. 45°16 cm. 23. 7°7 cm. MB 127319.
25. 9926 mi.; 3,976,000 sq. mi.
26. 10250 cu. in. 27. 143700 oz. 28. 20:00 cm.

29 (i) ae (ii) $ra*; (iii) ra*/2; (iv) rat, $0. 423°16 cu. in,

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