A school geometry / by H.S. Hall, and F.H. Stevens.

Hall, H.S.
London : Macmillan and Co., Ltd., 1921.

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 & A SCHOOL- GEOMETRY.
x ** : *.
pn; théºyecom lendations of the Mathematical
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A SCHOOL GEOMETRY

II. AND III.

I.,

PARTS
 MACMILLAN AND CO., LIMITED
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SCHOOL GEOMETRY
II., AND III.

I.,
PARTS

Part I. Limes and Angles. Rectilineal Figwres

Part II. Rectilineal Figures

of
Areas
Part III. Cºrcles

III.

I.,
(Containing 1-34,

of
the substance Euclid Book Book
and part IV.)

of
Book

BY

H. HALL, M.A.
S.

AND

H. STEVENS, M.A.
F.

SECOND EDITION REVISED

MACMILLAN AND CO., LIMITED

ST, MARTIN'S STREET, LONDON
92
I
I
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2nd a
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COPYRIGHT.

First Edition July 1903.

Reprinted October 1903(twice),
Second Edition, Revised, March 1904.
Reprinted 1905(twice), 1906,1907(twice), 1908,1909,1910,
1911,1912,1913,1914,1915,1916,1917,1918(twice),
1919(twice), 1920,1921.

GLASGow; PRINTED AT THE 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:
pupil should gain his first geometrical ideas
(i)

We agree that
a

short preliminary course practical and experimental

of

from
a
a

the present book would

to
A

character. suitable introduction

Easy Exercises Drawing illustrate the subject
to
of of

in

consist
the Definitions; Lines and Angles;
of
matter Measurements
Compasses Protractor; Problems Bisection, Per on
of

Use and
pendiculars, and Parallels; Use Set Squares; The Construction
of

be
Triangles These problems should
of

and Quadrilaterals.
accompanied by informal explanation, and the results verified
by

to be

measurement. Concurrently, there should series ex of

a

Drawing and Measurement designed lead inductively

in

ercises
the more important Theorems Part 1-34]* While
of

[Euc.
to

I.
I.

strongly advocating some such introductory lessons, we may point

out that our book, goes, complete itself, and from
as

in
as

far
is
it

illustrated by numerical and graphical examples

of

the first the

is

easiest types. Thus, throughout the whole work, graphical

a

and experimental course provided side by side with the usual

is

.
.

deductive exercises.

(ii) Theorems and Problems are arranged separate but parallel.

in

studied pari passw.

be

This arrangement
to

courses, intended
is
by

made possible the use, now generally sanctioned, Hypothetical

of

These, before being employed the text, are care

in

Constructions.
by on

fully specified, and referred which they depend.

to

the Axioms
Experi.
an

in

Such introductory course now furnished our Lessons

is
*

mental and Practical Geometry.

Vi - PREFACE.

(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. * VT1

The examples are numerous and for the most párt 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. S. 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 I. 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. S. HALL.
F. H. STEVENS.
March, 1904.
 CONTENTS,

PART I.
PAGB
- as tº º tº
Axioms. Definitions. Postulates.
HYPOTHETICAL CONSTRUCTIONS tºº *º º º sº sº
INTRODUCTORY - sº- º ſº sº º wº º & *

SYMBOLS AND ABBREVIATIONS - º - * * & **

Lines and Angles.

THEOREM 1. [Euc. 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. 10
CoR. l. If two straight lines cut one another, the four
angles so formed are together equal to four right angles. 11

sº
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. 11
(i)

the same angle equal.

of

CoR. 3. are
ll
(ii)

Complements the same angle are equal.

of

If,

[Euc. point straight line,

in
at
2.

THEOREM 14.]
I.

of a
it,

two other straight lines,

on

opposite sides make the

two right angles, then
to

adjacent angles together equal

these two straight lines are straight line.
in

one and the same

[Euc. two straight lines cut
If
3.

THEOREM 15.] one

I.

another, the vertically opposite angles are equal. 14

Triangles.
DEFINITIONS gº *- º -> sº * 4 16
-
-

THE COMPARISON OF Two TRIANGLES sº wº tº- - tº 17

[Euc. two triangles have two sides

I,

4.]
If
4.

THEOREM
the other, each
to

the one equal each, and

of

two sides
to
of

the angles included by those sides equal, then the triangles

are equal all respects. 18
in
K CONTENTS.

PAGE
THEOREM 5. [Euc. I. 5.] The angles at the base of an isosceles
triangle are equal. *
20
CoR. 1. If
the equal sides of an isosceles triangle are pro
duced, the exterior angles at the base are equal. 21

CoR. 2. If a triangle is equilateral, it is also equiangular. 21

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.

THEOREM 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.

THEOREM 8. [Fuc. I. 16.1 If

one side of a triangle is pro
duced, then the exterior angle is greater than either of the
interior opposite angles. 28
CoR. l. Any two angles of a triangle are together less
.
than two right angles. 29
CoR. 2. Every triangle must have at least two acute
angles. 29
CoR. 3.Only one perpendicular can be drawn to a
straight line from a given point outside
it.

[Euc. triangle greater

If

of
9.

is
THEOREM 18.] one side
I.

to a

than another, then the angle opposite the greater side

is
greater than the angle opposite 30
to

the less.

10. [Euc. one angle triangle

If

of

is
THEOREM 19.]
I.

to a

greater than another, then the side opposite the greater

31
to

angle greater than the side opposite

is

the less.

THEOREM ll. [Euc. Any two sides triangle are

of

20.]
I.

together greater than the third side. 32

Of

all straight lines from given point

to

THEOREM 12.
a

given straight line the perpendicular 33

is

the least.
OC the shortest straight line from
If

to
is

CoR. the
O
l.

straight line AB, then OC perpendicular AB. 33

to
is

Two obliques OP, OQ, which cut AB

at

equal
2.

CoR.
the perpendicular, are equal. 33
of

distances from the foot

C
Of

two obliques OQ, OR,

at
3.

CoR. OR cuts AB the

if

the foot the perpendicular, -then

of

reater distance from

C

greater than OQ. 33

is
R

Parallels.
PLAYFAIR's Axiom tºº - * wº t- gº rº- gº 35
-

straight line cuts two

so 27

THEOREM 13. [Euc.

If

and 28.]
I.

a
.

other straight lines

as

angles
(i)
to

make the alternate

 CONTENTS.

PAGº
equal, or (ii) an 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. [Euc. I. 29.] If

a straight line cuts two parallel
lines, it makes the alternate angles equal one another;
(i)

to
to
(ii) the exterior angle equal the interior opposite angle on
the cutting line; (iii) the two interior angles
of
the same side
two right angles. 38

to
on the same side together equal
PARALLELS ILLUSTRATED BY ROTATION. HYPOTHETICAL CON
STRUCTION tºº * ºs tº ſº sº * sº º e 39

THEOREM 15. [Euc. 30.] Straight lines which are parallel

I.

the same straight line are parallel

to

to
one another.

Triangles continued.
The three angles triangle

of
THEOREM 16. [Euc. 32.]
I.

a
.

two right angles.

to

are together equal

All the interior angles any rectilineal figure,
of
1.

CoR.
together with four right angles, are equal

as
twice many to
right angles
as

the figure has sides.

rectilineal figure, which has no
If

of
2.

CoR. the sides

a

re-entrant angle, are produced order, then all the exterior

in

four right angles. 46

to
so

angles formed are together equal

THEOREM 17. [Euc. two triangles have two angles
of If

26.]
I.

one equal two angles the other, each each, and any
of

to

to
to

the first equal the corresponding the other,

of

of

side side
the triangles are equal all respects.
in

ON THE IDENTICAL EQUALITY OF TRIANGLES * º 4- * 50

THEOREM 18. Two right-angled triangles which have their

hypotenuses equal, and one side one equal
of
to
of

one side
other, are equal all respects. 5]
in

the
THEOREM 19. [Euc. two triangles have two sides
If

of

24.]
I.
to

to

the one equal two sides the other, each each, but the
of

angle included by the two sides one greater than the angle
of

included by the corresponding sides the other; then the

of

that which has the greater angle greater than the

of

base
is

base of the other. 52

* * sº * * *
OF

19

CoNVERSE THEOREM 53
-

Parallelograms.
DEFINITIONS º * * tº sº - tº sº * *
*

THEOREM [Euc.
20. 33.] The straight lines which join the
I.

two equal and parallel straight lines towards

of

extremities
the same parts are themselves equal and parallel. 57
xii CONTENTS.

tºº
PAGA
THEOREM 21. [Euc, I. 34.]The opposite sides and angles of a
are equal to one another, and each diagonal
isects the parallelogram. i 58
CoR. l. If
one angle of a parallelogram is a right angle, all
its angles are right angles. 59
CoR. 2. All the sides of a square are equal; and all 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. a triangle ABC, if a set of lines
In Po, 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 * tº ſº tº tº - 66

Practical Geometry. Problems.

INTRODUCTION. NECESSARY INSTRUMENTS sº * • - 69
PROBLEMs on LINES AND ANGLES.
PROBLEM To bisect a given angle.
1. 70
PROBLEM 2. To bisect a given straight line. 7]
PROBLEM 3. To draw a straight line perpendicular to a given
straight line at a given point in 72
it.

To draw straight line perpendicular given

to
4.

PROBLEM
a

straight line from given external point. 74

a

At given point given straight line

in

to
5.

PROBLEM make
to a

a
an

angle equal given angle. 76

a

Through given point straight line

to

draw
7. to 6.

PROBLEM
a

parallel given straight line. 77

a

PROBLEM To divide given straight line into any number

a

equal parts. 78
of

THE CONSTRUCTION OF TRIANGLES.

To draw triangle, having given the lengths
of
8.

PROBLEM
a

the three sides. 80

To construct triangle having given two sides
9.

PROBLEM
a

82
to

and an angle opposite

of

one them.
PROBLEM 10. To construct right-angled triangle having given
a

the hypotenuse and one side. 83

THE Construction of QUADRILATERALS.
PROBLEM 11. To construct given the lengths
a

jºinisterºl,
the four sides, and one angle 86
of
 CONTENTS. xiii
PAGE
PROBLEM 12. To construct a parallelogram having given two
adjacent sides and the included angle. 87
PROBLEM 13. To construct a square on a given side. 88

Loci.
PROBLEM 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. 91
PROBLEM 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. 92
INTERSECTION OF LOCI - º - - - - - * 93
THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.
I. The perpendiculars drawn to the sides of a triangle from
their middle points are concurrent. 96
II. The bisectors of the angles of a triangle are concurrent. 96
III. The medians of a triangle are concurrent. 97
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. 97

PART II.
Areas.
DEFINITIONS te - - - * - - º - - 99
THEOREM 23. AREA OF A RECTANGLE. 100
THEOREM 24. [Euc. I. 35.] Parallelograms on the same base
and between the same parallels are equal in area. 104
AREA OF A PARALLELOGRAM - * - º - - -- 105
THEOREM 25. AREA OF A TRIANGLE. 106
THEOREM 26. [Euc. 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,
it,

and stand on the same base and on the same side of they
are between the same parallels. 108
ll
(i)

THEOREM 28. AREA OF

A

TRAPEZIUM.
2

(ii) ANY QUADRILATERAL. 112

AREA of ANY RECTILINEAL FIGURE - - - - tº 114
-

THEOREM 29. [Euc.,

In
I,

47. PYTHAGORAS's THEOREM.]

a

right-angled triangle the square described

on

the hypotenuse
on
to

equal
of

squares the other two

is

the sum the described

sides. 118
xiv. CONTENTS.

PAGE
ExPERIMENTAL PROOFS OF PYTHAGORAS's THEOREM *. sº 120

THEOREM 30. [Euc. 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
16.

be
PROBLEM To draw squares whose areas shall respectively
twice, three-times, four-times, given square.

of
that 124

...

a
,
Problems on Areas.
17. To describe parallelogram equal given

to
PROBLEM

a
triangle, and having one its angles equal given angle. 126

to
of

a
PROBLEM 18. To draw triangle equal given

in

to
a area

a
quadrilateral. 128

to
To drawparallelogram equal given

to in
PROBLEM 19. area
a

a
rectilineal figure, and having an angle equal given angle. 129

a
Axes of Reference. Coordinates.
gº º, gº wº sº º 132
ExERCISES FOR SQUARED PAPER

PART III.

The Circle. Definitions and First Principles. 139

-

-
SYMMETRY. SYMMETRICAL PROPERTIES OF CIRCLES º º 141

Chords.
THEOREM 31. [Euo. III.
straight line drawn from
If

3..]
a

circle bisects chord which does not pass

of

the centre
a

through the centre, right angles.

at

cuts the chord

it

right angles, 144

it.

Conversely,
at

cuts the chord bisects

it

it
if

The straight line which bisects right

at
1.

CoR. chord
a

angles passes through the centre.

h

145
straight line cannot meet
at

circle
2.

CoR. more than

A

two points. 145

it,

circle lies wholly within 145

of
3.

CoR. chord
A

THEOREM 32. One circle, and only one, can pass through any
*

three points not the same straight line. 146

in

The size and position circle are fully deter

of
if 1.

CoR.
a

pass through three given points.

to

mined known
it
is

147
Two circles cannot cut one another in more than
2.

CoR.
two points without coinciding entirely. 147
HYPOTHETICAL CONSTRUCTION * se º gº fº tº 147
-
 CONTENTS. XV

PAGE:
THEOREM 33. [Euc. 9..] III. 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. [Euc. 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. [Euc. 7..] III. 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. [Euo. 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

Angles in a Circle.
38.

III.
at

The angle
of

THEOREM [Euc. 20.] the centre

on a

standing
an

angle
of

circle
at
is

double the circumference

the same arc. 158
III. Angles the same segment
of

THEOREM 39. [Euc.

in

21.]
a

circle are equal. 160

-

THEOREM 39. Equal angles standing

on
or

CoNVERSE the
it,

same base, and on the same side have their vertices on

of

circle, which the given base 16]

of
of

is

an arc the chord.

a

[Euo. III. 22.] The opposite angles any

to of

THEOREM 40.
quadrilateral inscribed circle are together equal two
in
a

right angles.
OF

pair opposite angles

If

of

CoNVERSE THEOREM 40.

a

quadrilateral are supplementary,

of

its vertices are con

a

163
-

cyclic.
[Euc. III. The angle
in

is

THEOREM 41. 31.] semi-circle

a

right angle. 164

The angle segment greater than

in

CoR. semi-circle
is is
in a

a a

acute; and the angle segment less than semi-circle

a

obtuse. 165
xvi CONTENTS.

PAGF;
THEOREM 42. [Euc. III. 26.] In equal circles, arcs which sub
tend Équal angles, either at the centres or at the circum
ferences, are equal. 166
CoR. In equal circles sectors which have equal angles are
-
equal. º 166

THEOREM 43. [Euc. III.

27.] In equal circles angles, either at
the centres or at the circumferences, which stand on equal
arcs are equal. - 167
THEOREM 44. [Euc. III.
28.] In equal circles, arcs which are
cut off by equal chords are equal, the major arc equal to the
major are, and the minor to the minor. 168

THEOREM 45. [Euc. III.

29.] In equal circles chords which
cut off equal arcs are equal. - 169

Tangency.
DEFINITIONS AND FIRST PRINCIPLES - sº es sº &= sº 172

THEOREM 46. The tangent at any point of a circle is perpendi

cular to the radius drawn to the point of contact. 174
CoR. 1. One and only one tangent can be drawn to a
circle at a given point on the circumference. 174
CoR. 2. The perpendicular to a tangent at its point of
contact passes through the centre. 174
CoR. 3. The fadius drawn perpendicular to the tangent
passes through the point of contact. º 174
THEOREM 47. Two tangents can be drawn to a circle from an
external point. as: 176
CoR. The two tangents to a circle from an external point
are equal, and subtend equal angles at the centre. 176
THEOREM 48. If two circles touch one another, the centres and
the point of contact are in one straight line, 178
CoR. l. If
two circles touch externally the distance be
tween their centres is equal to the sum of their radii. 178
CoR. 2. If
two circles touch internally, the distance be
tween their centres is equal to the difference of their radii. 178

ºly
THEOREM 49. [Euc. III.
32.] The angles made by a tangent
to a circle with a chord drawn from the point of contact are
equal to the angles in the alternate segments of
the circle. 180
g
Problems. -
GEOMETRICAL ANALYSIS - * wº- sº e- tº-3 e * 182
PROBLEM 20. Given a circle, or an aré of a circle, to find. its
centre. 183
PBOBLEM 21. To bisect a given arc. 183
 CONTENTS. xvii
PAGE
PROBLEM 22. To draw a tangent to a 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. On a 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
PROBLEM 25. To circumscribe a circle about a given triangle. 193
PROBLEM 26. To inscribe a circle in a given triangle. 194
PROBLEM 27. To draw an escribed circle of a given triangle. 195
PROBLEM 28. In a given circle to inscribe a triangle equi
angular to a given triangle. 196
BROBLEM 29. About a given circle to circumscribe a triangle
equiangular to a given triangle. 197
PROBLEM 30. To draw a regular polygon (ii) about
(i)
in

a
given circle.
-
200
To draw circle (ii) about regular polygon.
in
(i)

PROBLEM 31. 201

a

Circumference and Area of Circle - *- - 202

a

-
-

Theorems and Examples on Circles and Triangles.

THE ORTHoCENTRE OF TRIANGLE tº- - º tº- 207
A

LOCI * as - - tº- º -
•

210
-

•
-

-
-

SIMSON's LINE - & wº tº- º <- - 212

t
-

THE TRIANGLE AND ITs CIRCLES * - tº º 213

-

THE NINE-PoſNTs CIRCLE - *- -> - * cº 216

-

Appendix.
of

On the Form Some Solid Figures.

to

Answers Numerical Exercises,

 GEOMETRY.

PART I.

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. ~ wº...º

)
e
For, ,instance:
& *:
*
Å, rºº
--~~ *
# =
****
~ .
~£
.à . . . .
Things which are equal to the same thing are equal to one
another.
R →, sº sº, e º

The following axioms, corresponding to the first four Rules

of Arithmetic, are among those most commonly used in
geometrical reasoning.

Addition.
&)ºf . , ,
If equals are added to equals, the sums are equal.
Subtraction. If equals
~
are taken from equals, the remainders
5 ºn
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
all

be general, because they apply equally to magnitudes of

Certain special axioms relating geometrical magni
to to

kinds.
tudes only will
be

as

stated from time time they are

required.
Qº
S.

H. G.
A
2 GEQMETRy.
º
*~
0.5°

ov,
o-oo

º
an

,
,
DEFINITIONS AND FIRST PRINCIPLES.

by
Every beginner knows general way what

in
meant

is
a

a
point, line, and But geometry these terms are

in
surface.

a
in a

used strict sense which needs some explanation.

a

^\GY, -

point has position, but

of no
A

to
said have magnitude.
1.

to is
This means that we are point no idea

as
size either

to
attach

i.
a
think only where
to

length breadth, but

or

dot
to

is
situated.

it

A
roughly representing

be
made with sharp pencil may

as
taken
a

a
as

but small such dot may be, still has some length and

it
a

readth, and therefore not actually geometrical point. The smaller

is

a
the dot however, the more nearly represents point.
it

a
^^
sº
(

line has length, but

no
to
said
A

have
2.

is

breadth.
traced out by
line moving point. the point pencil

If
is
A

of

is
a

a
of

moved over sheet paper, the trace left represents line. But such
a

a
trace, however finely drawn, has some degree

of
breadth, and

is
a

therefore not itself true geometrical line. The finer the trace left by
a

the moving pencil-point, the more nearly will represent line.

it

Proceeding similar manner from the idea a

in

of
line
3.

a
to

surface, we say that

of

the idea
g
A

surface has length and breadth, but

no

thickness.

And finally,
{}^^-).
§

solid has length, breadth, and thickness.

A

Solids, surfaces, lines and points are thus related one another:
to
3-

-
fö

&
Y
r

solid bounded by surfaces.

(i)
A A A

is

-
(ii) bounded by lines; and surfaces meet
in
is

surface lines.
by

(iii) line points; and lines meet

in
is

bounded (or terminated)

points.
V
\\
\

- **
3
be

line may straight

or
A

curved.
4.

straight line has the same direction from point point

A

to
its

throughout whole length.

its

curved line changes continually from point

A

- direction
;

point. -
..
.
.
.
..
to
 DEFINITIONS. 3
-

AxIOM. There - be only one straight line joining

can two given

is,
points: that
Two straight lines cannot enclose Space.

a
Sºn

$
ſº
v %

plane flat surface, the test flatness being that

of
A
5.

is
any two points are takena the surface, the straight line

in
if

between them lies wholly that surface.

in
When two straight lines meet

at
6.

a
to

are said
point, they ºv-, form an angle.

B
2

The straight lines are called thé arms

of
the
angle; the point
at

which they meet its vertex.

is

be f\ 6.0
The magnitude the angle may
\
of

thus

A
explained:
Suppose that the arm OA fixed, and that OB turns about
is
by

the point (as shewn the arrow). Suppose also that OB

O

began its turning from the position OA. Then the size

of
the
angle AOB measured by the amount turning required
of

to
is

bring the revolving arm from its first position OA into its
subsequent position OB.
an

on

Observe that the size angle does not

in

any way depend the

of

length
of

its arms.
-
lie

Angles which
on

of

either side
B
C
be

ad
to

common arm are said

a

jacent. **
J:

Sea.
-

For example, the angles AOB, BOC,

which have the common arm OB, are
A
O

adjacent.

When two straight lines such

as

AB, CD
C
O,
at

cross one another the angles COA, BOD

vertically opposite.
be

The
to

are said A
5

angles AOD, COB are also vertically opposite

to one another.
4 GEOMETRY.

7. When one straight line stands on an

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.
B O A
(i) point straight line AB, then

If

in
AXIOMS. line

O is
O

a
a
OC, which turns about from the position OA the position OB,

to
must pass through position, and only one,

in
one which

is
it
perpendicular AB.
to

(ii) All right angles are equal.

right angle divided into 90 equal parts called degrees (°); each
is
A

equal parts called minutes ('); each minute into

60

60
degree into equal
parts called seconds (").

OC revolves about oftom the

In

the above figure,

if

position OA into the position OB, turns through two right

it
**
or

angles, 180°.

OC makescomplete revolution about O, starting from OA

If

to a

and returning its original position, turns through four

it
or

right angles, 360°.

An angle which less than one right

8.

is
be
to

angle
•

said acute.
is
is,

That an acute angle

is

less than 90°.

O

An angle which greater

9.

is

than one right angle, but less than

two right angles,
be
to

said obtuse.
is

That is, an obtuse angle lies between

90° and 180°.
O

-
an

one arm OB angle turns

of

10.
If
.

until makes straight line with the ---,

it

5-4-5- A
so

other arm QA, the angle formed

is

-
called straight angle.
a

straight angle=2 right angles= 180°.

A
 DEFINITIONS. 5

11. An
angle which is greater
than two right angles, but less than () , ,
four right angles, is said to be :--
reflex. T (ii)
A º
is,

That reflex angle lies between

B
a

180° and 360°.

NOTE. When two straight lines meet, two angles are formed, one
greater, and one less than two right angles. The first arises by
have revolved from the position OA the longer way
to

supposing OB
round, marked (i); the other by supposing OB

to
have revolved the
shorter way round, marked (ii). Unless the contrary stated, the

is
straight

be

be
angle between two lines will that which

to

is
considered
less than two right angles.

Any portion by one

or
plane surface
of

12. bounded
a

more lines called plane figure.

is

..
a

circle plane figure contained

/ P
A

13.
is
a
by

line traced out by point which moves

a
a

so that its distance from certain fixed

a

point always the same.

O
is

Here the point

so

moves that its distance

P

P
•

from the fixed point always the same.

is
O

The fixed point called the centre, and the bounding line
is

called the circumference.

is

.
.
.
.
.
.
.
.
.

.
of

radius straight line drawn from the

A

circle
is

14.
a

all

of

follows that radii

to

It

centre the circumference.

a

circle are equal.

15.

straight line drawn through

of

diameter
A

circle
is
a

the centre, and terminated both ways by the circumference.

16.

An any part
of

of

arc circle the circumference.

is
a
6 GEOMETRY.

17. A semi-circle is the figure bounded

by a diameter of a circle and the part of the

by
off
circumference cut the diameter.

|
To divide into two equal parts.

to
18. bisect means

(i)

If
AxIOM's. point moves

O
a
from along the straight line

B
to

A
B
in it A

AB, pass through posi-

*
must one
tion which divides AB into two equal parts.
it

That say:
to
is

O,
line OP, revolving about
If

(ii) turns
a

from OA OB, must pass through

to

one
it it

position divides the angle AOB

in

which
into two equal parts.
That say:
to
is

HYPOTHETICAL CONSTRUCTIONS.
to

From the Axioms attached Definitions and 18,

it
7

follows that we may suppose

•
(i)

straight line perpendicular

to in be
A

drawn given
to

to
a
.
straight line from any point
by be it.

(ii) finite straight line point.

at
A

bisected
a

(iii) An
be

angle line.
to

bisected
a

SUPERPOSITION AND EQUALITY.

Axiom. Magnitudes which can

be

with- one
to

made coincide
are equal.
*

another
This axiom implies that any line, angle, taken up
be

figure, may
or
or

from its position, and without change form, laid down upon
in

size
a

second line, angle, figure, for the purpose comparison, and

of
or

it

exactly
be

states that two such magnitudes are equal when one can
placed over the other without overlapping.
to

This process called superposition, and the first magnitude

is

is

said
be

applied
to

the other.
 POSTULATES. 7

POSTULATES.

In order to draw geometrical figures certain instruments

(i)
are required. These are, for the purposes of this book,

a
straight ruler, (ii) pair The following Postulates

of
compasses.
a
(or requests) claim the use these instruments, and assume

of

be
that with their help the processes mentioned below may
duly performed.
granted:
be

Let
it

straight line may drawn from any one point

be
That

to
1.

any other point.

FINITE (or terminated) straight line may

be
That
2.

is,

PRODUCED (that prolonged) any length that straight

in
to

line.

may drawn with any point

be

as
That and
3.

circle centre
a

any length.
of

with radius
a

stated above, im
(i)

3,
as

NoTES. Postulate
plies that we may adjust the compasses
to

the
length any straight line PQ, and with radius
of

point
a be as

length any
of

this draw circle with

O
a
to

That say, the compasses may

is

centre.
transfer distances from one part
to

of

used
to

diagram another.

(ii) Hence from AB, the greater

of

two -
8traight lines, we may cut off part equal
l
a

g |X

PQ the less.
to

B
A,

For with centre and radius equal

if

PQ, we draw circle cutting

an
to

of

arc

§§
a
X,

obvious that AX equal

Q
is

is
it

i*\cºs
O
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 line, 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.

preliminary statement,
(i)

The General Enunciation

is
a

general terms the purpose the proposition.

of

describing
in

(ii) The Particular Enunciation repeats special terms

in
to

the statement already made, and refers diagram, which

it

follow the reasoning more easily.

to

enables the reader

directs the drawing

of

(iii) The Construction then such

required
be

to

may
as

straight lines and circles effect the

to

problem, prove the truth

of

purpose theorem.
or
of

a
a

(iv) The Proof shews that the object proposed problem

in in
a a

that the property stated

or

has been accomplished, theorem

true.
is

to

The letters Q.E.D. are appended theorem, and stand

4.

for Quod erat Demonstrandum, which was

be

proved.
to
 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 Part I.
for therefore,
= ...

for angle,

a
A
to,
is,

triangle.

,
are, equal
,

or

After Part
I.

pt. for point, perp. for perpendicular,

straight line, par"
st.

parallelogram,

,,,
line
,, ,

rt. right angle,

A.

rectil. rectilineal,

-
.
.
.
par' (or parallel, G) circle,
||)

gº
,

-.
.
..
Sq. Square, O* ,, circumference;
,,
as all

commonly occurring words,

of

and obvious contractions

such opp., adj., diag, etc., for opposite, adjacent, diagonal,
etc.

oral work, and prevent the rather common

of

[For convenience
to

by beginners, the above code signs has been

of

of

abuse contractions
introduçed gradually, and first somewhat sparingly.]
at
In

numerical examples the following abbreviations will º

be used.
m.

for metre, cm, for centimetre,

,
,

mm. millimetre. km. kilometre.

by

().

Also inches are denoted the symbol

Thus
5”

means inches.
.
..
.

5
 10 GEOMETRY.

ON LINES AND
ANGLEs.

THEOREM 1. [Euclid I. 13.]

The adjacent angles which one/straight line makes with another

so

sº,
it,
straight line on one side of are together equal two right angles,

to
D;
s
O

A
B

Let the straight line CO make- with the straight line

-
AB the
adjacent "AOC, COB.
4

required prove that the 'AOC, COB are together equal

It

to

-to
is

.
two right angles.
right angles to BA.
at

Suppose OD
is

Proof. "AOC, COB together

Then the
4

the three "AOC, COD, DOB.

=

.*

Also the "AOD, DOB together

4

. the three z*AOC, COD, DOB.

= = =
-

the "AOC, COB together the "AOD, DOB

...

4
.

two right angles.

Q.E.D.

PROOF BY ROTATION.

Suppose straight line revolving about turns from the position OA

O
a

into the position OC, and thence into the position OB; that
is,

let the
to 4."

revolving line turn succession through the AOC, COB.

in

Now passing from its first position OA its final position OB,
in

the revolving line turns through two right angles, for AOB straight
is
a

line, -
4."

Hence the AOC, COB together= two right angles,

 LINES AND ANGLES. 11

surn of * * *
D
If
2^
CoROLLARY straight lines
1. . two
* , º,
cut one another, the four angles so formed ' ". . .
are together equgl to four right angles. A O B
2, 3}{}>. X= >r2 -73, 3 º' -, *
For example, C
4. BOD + 4. DOA+ 4 AOC + 4-COB=4 right angles.

* &
*- - ~}^2,224
CoRoºſ/ARY 2. When any number of
the

straight lines meet at a point,

of
sum
the consecittà dhgles
so

formed equal
is
,

four right angles.

to

ºr,
Jº,
-

§§§
O,

For straight line and turning

in
about
a

succession
Z."

through the AOB, BOC, COD, DOE, EOA, will have made one
complete revolution, and therefore turned through four right angles.

e-
º
*
*

ºn tº
Or, ºr/ACN §N,
}

*
v

definitions. J-,
-

* .”.
*,
ºn
J.

>
is is of

>\*
*
(i) g

v.

>
5
v

.
3

.
:

.
.

to ..
Two angles whose sum two right angles, are said
be

supplementary; and each called the supplement

of

wº d's *** **,

the
other. **
Y.
>

Q
&

**

*
*
**
.
.

Thus the Fig. Theor. the angles AOC, COB are supplementary.
of
in

Again the angle 123° the supplement

tº
the angle 57°.
is

of

*.
** tº
.

Tºa tº
•
tº
}

oº ev
*

v
,
o

s
*•

(ii) Two angles whose sum one right angle are said
to
is
be

complementary; and each called the complement

, of
* is

the
other. S".
sº

***
,
~

*>
*
º

,
.
.
.
.
.
.
of .
.

of ,

,
:

.
.
.
.
.
.
.
.

Thus the Fig. ****

in

of 1,

Theor. the angle DOC the complement

is

the angle AOC. Again angles 34* and 56° are complementary, ºn
a
* a
A,

x^^
the the St

no
tº
y
.

2
of of o

~'s
°
5
o

*
.

,
.

:
(i)

CoRoLLARY Supplements same angle are equal.

3.

(ii) Complements same angle are equal.

 -
12 GEOMETRY.

THEOREM 2. [Euclid I. 14.] *

If,

point straight line, two other straight lines,

on
in
at
a

a
it,
of
opposite the adjacent angles together equal

to
sides make two
these two straight lines are

in
ºright angles, then one and the same
straight lime.

...Rºzz woº “ºver

"w

C
.
~q
→

g
n
~

Q
X

A
O
B

At the straight line CO let the two straight lines OA,

in
on O

OB, opposite sides CO, make the adjacent AOC, COB

of

4
"
two right angles: (that is,
together equal let the adjacent
to
be

AOC, COB supplementary).

**

required OB and OA the same straight

It

prove that
in
to
is

- are
•.
lime.
X;

Produce AO beyond any point

be
to

will shewn that

it
O

OX and OB are the same line.

’,

Proof. Since by construction AOX straight line,

of is
a

COX the supplement

...

the the COA.

is
4:

1.

Theor.
4
by

But, hypothesis, -
s

the supplement COA,

of

the COB the

is
A.

A.

COX the COB;

...

the
A.
=
4

.*. OX and OB are the same line.

by

But, construction, Ox same straight line

in

the
is

-
with OA;
the same straight line with OA.
in

hence OB also
is
-

Q.E.D.
 LINES AND ANGLES. I3

EXERCISES.

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

thirds of a fight 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.
Qº’
* , of
roº, isſon, uzov A.
j
4,311n the triangle ABC the angles ABC, ACB are given equal. If
prod;
º
the side BC is both ways, shew that the exºriº; angles so
t
23% are equal.
2.
vºteo

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.

Dºrismos. The lines which bisect an angle and the

its

adjacent angle made by producing one of arms are called

the internal and external bisectors the given angle.
of

Thus the diagram, OX and OY are the

in

and external bisectors the angle

ºnal
of

Prove that the bisectors the adjacent angles which one

of
6.

straight line makes with another right angle. That

to

contain
is
a
of

say, the internal angle right angles

at

and extermal bisectors an are

to

one another.

Shew that the angles AOX COY the above diagram

7,

in

and are
complementary.

Shew that the angles BOX and COX are supplementary; and
8.

also that the angles AOY and BOY are supplementary.

the angle AOB 35°, find the angle COY.

If

is
9.
 14 GEOMETRY.

THEOREM 3. [Euclid I. 15.]

If two

the
cut
straight lines one another, vertically opposite angles

6,
-}2
*

2.
Ó
are equal, *

>J
f1

*
D
A

let the straight lines AB, CD cut one another the point

O.
at
required
It

prove that
to
is

DOB;
(i)

4,
the AOC = = the
4

(ii) the COB the Z.AOD.

4

Proof. Because AO meets the straight line CD,

the adjacent AOC, AOD together two right angles;
=
‘..

4- 4
"
is,

that the supplement

of
the AOC the AOD.
is

4
Again, because DO meets the straight line AB,
the adjacent "DOB, AOD together two right angles;
...

=
4, 4
is,

the supplement
of

that the DOB the 4-AOD.

is

AOC, DOB the supplement the 4-AOD,

of

Thus each
of

the
is
4
"

... the AOC the DOB.

=
4

Similarly, the 4-COB the Z.AOD.

=

* Q.E.D.

PROOF BY ROTATION.

Suppose the line COD until OC turns into the

to

revolve about
O

OA. Then the same moment OD must reach the position

at

osition
(for AOB and COD are straight).
B

turning
to

as

required
of

Thus the same amount close the Z.AOC

to
is

close the 4-DOB.

the Z.AOC= the DOB.
A.
...
 15
LINES AND ANGLES.

EXERCISES ON ANGLES.
.
(Numerical.)
1. Through what angles does the minute-hand of a clock turn in

2.1

14

10
minutes, (ii) minutes, (iii) 43% minutes, (iv)
(i)

min. sec.
5

2
And how long will

to
take turn through (v) 66°, (vi) 222°?
it
at noon through what angles will the hour
2.

is

clock started
A

:
by

10
minutes past 52 And what will
(i)
3.45, (ii)
hand have turned
be

the time when has turned through 1723°

it

2
The earth makes complete revolution about its axis

in
3.

24 hours.
a

Through what angle will turn hrs. 20 min., and how long will in
it

it
turn through 130°2 3
to

take
If In

the diagram
of
(i) 4.

Theorem
3

the 4-AOC=35°, write down (without measurement) the value

A_*

COB, BOD, DOA. -

of

of

each the
(ii) *COB, AOD together make up 250°, find each
If

of
the the
Z

COA, BOD.
A
s

4–5

(iii) AOC, COB, BOD together make up 274°, find each

If

of
the -
O.
at

the four angles

(Theoretical.)

point AB two straight lines OC, OD are drawn

on
If

as in
5.

from
of O
a

to

make the angle COB equal

so

opposite sides AB the angle

to

AOD; shew that OC and OD are the same straight line.

in

Two straight lines AB, CD cross

O.
at

OX
If

of
6.

is

the bisector
the angle BOD, prove that XO produced bisects the angle AOC. ~.

Two straight lines AB, CD cross

O.
at

the angle BOD

If
7.

is

by OX, and AOC by OY, prove that OX, OY are

in

bisected the same

straight line. -
*
*:

angle AOB, shew that, by folding the diagram

an

OX bisects
If
8.

be

about the bisector, OA may coincide with OB.

to

made
-
How would OA fall with regard OB,
to

if

XOB;
(i)

AOX were greater

4- 4-

the than the

4

(ii) the AOX were less than the A-XOB?

AB and CD are straight lines intersecting O;
9.

right angles
at

to at
by

folding the figure about AB, that OC may

be

shew made fall

along OD.
&

10. straight line AOB

on
O, A

drawn paper, which

is

then folded
is

make OA fall along OB; shew that the crease

to
is so
as

left
in

about -
the paper perpendicular AB.
to
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 s 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

it all all

equilateral, when its sides are equal;

equiangular, when its angles are equal
;

regular, when both equilateral and equiangular.

is

Triangles are thus classified with regard their sides:

to
7.

be

triangle
to
A

said
is

of its

equilateral, when all sides are equal;

•
2

its

,,,,, isosceles, when two sides are equal;

º

all

scalene, when its sides are unequal.

Equilateral Triangle. IsoscelesTriangle, ScaleneTriangle.

B,
A,

triangle ABC, the letters

In

often denote
C
a

in

the magnitude the several angles (as measured

6.
of

&
b,
in a,

degrees); the lengths

of

and the letters the

c

or

opposite sides (as measured inches, centimetres,

length).
of

some other unit

&
B

C
 TRIANGLES. 17

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

its
vertex; and the opposite side then called the base.

is
an isosceles triangle the term vertea, usually applied
In

the point

is to
is
which the equal sides intersect; and the vertical angle
at

the angle
included by them.

Triangles are thus classified with regard

to
their angles:
8.

triangle

be
to
A

said
is

of its its
***!?ight-angled, angles right angle;

of of
when one

is is
a
42.2×obtuse-angled, when one angles obtuse;

its
3.30%acute-angled, all angles are acute.
when three
3.

~/
be

[It will Cor. that every triangle mw8t

1)
8.
seen hereafter (Theorem
least two acute angles.]
at

have

~1
Right-angled Triangle. Obtuse-angledTriangle.
2\
Acute-angled Triangle.

right-angled triangle the side opposite the right angle

In

to

is
a

called the hypotenuse.

any triangle the straight line joining

In

to
vertex the
9.

middle point the opposite side

of

called median.
is

THE COMPARISON OF TWO TRIANGLES.

triangle are called

(i)

The three sides and three angles

of
a
its

triangle may also

be

six parts. considered with regard

A

to its area.
be

(ii) Two triangles are said equal all respects,

in
to
in be
so

as

when one may placed upon the other exactly

to
it;

coincide with which case each part the first triangle

of

equal the corresponding part (namely that with which

to
it is

the other; and the triangles are equal

in
of

coincides) area.
In

two such triangles corresponding sides are opposite equal

to

angles, and corresponding angles are opposite equal sides.

to

Triangles which may thus coincide by super

be

or to

made
identically equal
be
to

position are said congruent.

S.

H. G. E
 18 GEOMETRY.

If two
** ‘.3
THEOREM
× 13%, sº
4... [Euclid I. 4. J
.# a 2-4
2’

the
triggles have two sides of

of
one equal

to
two sides the

the

by
other, éâch'fö’each, and angles included those sides equal,
gº
then the triangles are equal all respects. 6)G),

in
equ
9. tºgº.
º'º ano
. i

º' Tºº.
.3,
> 3,

r*.
2,

ºr
y
,
,
.
.
.

A..

D
s
. C

F
!
!

E
be

Let ABC, DEF two triangles which

in

AB DE,
= = =

AC DF,
and the included angle BAC the included angle EDF.

all
required
It

prove AABC DEF

to

in
A
that the the
is

Tespects.

Proof. Apply the AABC DEF,

to

the
A

the point
on
so

that the point.

D,

falls
A

and the side AB along the side DE.

Then because AB DE,
=

the point must coincide with the point

...

E.
B

And because AB falls along DE,

and the BAC the EDF,
A.

4.

AC must fall along DF.

...

And because AC DF,

=

the point must coincide with the point

...

F, F.
C

E,

Then since coincides with and with

B

... the side BC must coincide with the side EF.

Hence the ABC coincides with the DEF,

A
A

all

therefore equal respects.

in
to

and
is

it

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 Z. BAC = the 4. EDF.

From these data we prove that the triangles coincide on

superposition.
BC = EF,
Hence we conclude that the 4 ABC = the 2. DEF,
and the 4. ACB = the 4 DFE ;
also that the triangles are equal in area.
the

angles which are proved equal the two triangles

in
Notice that
are opposite sides which were given equal.
to

D
NoTE. The adjoining diagram shews A
make two congruent
to
in

that order
be

triangles coincide, may necessary

it

of

reverse, that is, turn over one

to

them C
before superposition. F
º

Jºus;* E
º;
*... XXERCISES.

*ś %
l.

of

Shew that the bisector angle an isosceles triangle

%the
perpendicular
#!".
(i)

to

bisects th9, base, (ii) the

e
}

is
i)

be
let

line AB, and OC

of

be the hiddle pbint straight

O
2.

Let
a

any point OC, prove that PA=PB.

in
if

perpendicular
to
it.

Then
is
P

~
Y
*
f

its

square are equal, and that

of

Assuming that the four sides

3.

the square' ABCD, the

in

angles are all right angles, prove that

diagonals AC, BD are equal. te
M, and are the middle points AB,
of
L,

square, and
N

ABCD
is
4.

BC, and CD prove that

:

LM MN. (ii) AM DM.

(i)

=
= =

(iii) AN AM. (iv) BN =DM.

[Draw separate figure
in

each case.]
a

isosceles triangle: from the equal sides

AB, AC two
an

§" gº
ABC
is
5.

AX, AY are cut off, and BY and CX are joined. Prove that
 20 GEOMETRY.

5.]
THEOREM 5. [Euclid I.
ºr

f
o-f

an
isosceles triangle are

of
at
The angles rºo

#
the base gual.
ºjºs, ºf

24

54

J
at
*4 wº

&
%24

,
Lº
2

A
5–5–3
be
an

Let ABC isosceles triangle, which the side AB

in
the

=
side AC.

required ACB.
It

prove that the ABC

to

=
4.
is

the

4
Suppose that AD the line which bisects the BAC, and

4.
is
in
D.

let meet BC
it

A"

Then BAD, CAD,

in

1st Proof. the

BA CA,
to =

because AD common both triangles,

is

and the included BAD the included CAD

Z.

Z.
=

3.
;
4- all

the triangles are equal respects;

...

in

4.
Théor.
so that the ABD the ACD.
Z.

-
Q.E.D.
be

Suppose the AABC folded about AD.

= to

2nd Proof.
Then since the A-BAD the CAD,
4.

AB must fall along AC.

...

And since AB AC,

=
on

is on
C,

must fall and consequently DB DC.

B
‘.

ACD, and
...

the ABD will coincide with the therefore

4.

4.

equal
it.
to

Q.E.D.
 ISOSCELES TRIANGLES. 21

COROLLARY 1. If the
equal sides AB, AC of
an isosceles triangle
are produced, the exterior
angles EBC, FCB are equal ; for they are the
B C
supplements of the equal angles at the base.

E F
CoROLLARY 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

of
its

the bisector VERTICAL angle.

An equilateral triangle symmetrical about the bisector

of
ANY
is
of
its

ONE angles.

EXERCISES.

ABCD figure whose sides are all equal, and the

1.

is is

four-sided
a

diagonal BD drawn; shew that

the angle ABD the angle ADB
(i)

= = =

; ;

(ii) the angle CBD the angle CDB

(iii) the angle ABC the angle ADC.
ABC, DBC are two isosceles triangles drawn
on
2.

the same base

BC, but it: prove (by means Theorem
on

opposite sides that

5)
of

of

the angle ABD the angle ACD.

=

ABC, DBC are two isosceles triangles drawn on the same base
3.

it; employ Theorem

on

to

prove that
of

BC and the same side

5

the angle ABD the angle ACD.

of =

AB, AC are the equal sides isosceles triangle ABC

an
M,4.

and
;
L.,

are the middle points AB, BC, and CA respectively: prove

of
N

that
LM NM. (ii) BN CL.
(i)

(iii) the angle ALM the angle ANM.

=
 22 GEOMETRY.

6.]
THEOREM 6. [Euclid I.
If
triangle are equal

of
two angles one another,

to
then the

a
are

one
the

are
sides which opposite equal angles equal

to

to
yºurº* 22/
another.
•

**-
,
f
# { ...

sſe 9v
*Y

*
º'
a
9

*
§,
p

**** *

A
fºr 3~ **** syave
y

* >

a
*

es.” * -->
º*
$3*

,-\,

D
>
25

g
,
,

C
B

|
triangle
be

Let ABC which

in
a

the ABC the ACB.

Z.
=
4

the

required
If It

prove that side AC=the side AB.

to
is

AC and AB are not equal, suppose that AB the greater.

is
to

From BA cut off BD equal AC.

Join DC.
Proof,
A"

Then DBC, ACB,

in

the
DB AC,
is =

to

because both,
ſ

BC
common
land the included DBC
the included ACB;
=

4.
/

y
4.

the AACB area,

in

Theor.
...

the DBC
A

the part equal the whole; which

is to

absurd.
is

not unequal
..'.

to

AB AC
;
is,

that AB AC.
=

Q.E.D.

triangle equiangular
if

COROLLARY.
is

is

Hence also
it
a

equilateral.
 A THEOREM AND ITS CONVERSE. 23

NOTE ON THEOREMS 5 AND 6.

A
Theorems 5 and 6 may be verified ex- * \
perimentally by cutting out the given * ,
AABC, and, after turning it over, fitting , ,
\
it thus reversed into the vacant space left /
in the paper.
f *
C’ C 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
fall on A'.
In either case the given triangle reversed will coincide with its own
“trace,” so that the side and angle on the left 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 assume, and is called the hypothesis; the
second tells us what it 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 = the 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 “ . S : AB = AC ;
it is required to prove that the angle ABC= the angle Ace.}
Now in Theorem 6
assumed that the angle ABC = the angle ACB;
it is
it is
required to prove that AB = AC.
S 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
intrue; 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
p.

necessarily true.
is

verse [See 25.]

24 GEOMETRY.

2 THEOREM 7. [Euclid I. 8.]

z; 2 ºf £72. re-ſki
If two

the

the

the
triangles have

of
one equal

to
. three sides three

the

all
of other, each each, they are equal respects.

in
to
sides

be

Let ABC, DEF two triangles which

in
AB DE,
== =

AC DF,
BC EF.
the

all
required triangles are equal
It

in
prove that respects.
to
is

Proof. Apply the AABC DEF,

to

the
A
,

and BC along EF, and

on
E,
so so

that falls
A B

on

EF opposite F. to
C of

D.
that the side
is

Then because BC EF, must fall on

=

the new position

be

Let GEF the AABC.

of

Join DG.
Because ED EG,
=

... the EDG the EGD.

5.
Z.

A.
=

Theor.

Again, because FD FG,

=
...

the FDG the FGD.

=
/.

Honce the whole EDF the whole EGF,

= =
/

/
is,

that the EDF the BAC.

A.

A.

EDF;
A"

Then BAC,
in

the
BA ED,
= =
|-

because AC DF,
the included BAC the included EDF;
A.

4.
all

the triangles are equal respects.

...

in

4.

Theor.
Q.E.D.
 CONGRUENT TRIANGLES. 25

Obs. In this Theorem

it BC = EF,
is given that AB = DE, CA= FD;
and we prove that 4 C = 4-F, 4-A = 4. D, A-B = A. E. ii
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.

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

As EDF, EGF.
Two other cases might arise:
DG might fall outside the EDF, EGF [as 4.5

1].
Fig.
(i)

2). in
2
(ii) DG might coincide with DF, FG [as Fig.

in
A

21
D
A

~!
F
C
B

B
E

Fig.1. Fig.2.

G
G

These cases will arise only when the given triangles are obtuse
angled right-angled; and (as will
be
or

seen hereafter) not even then,

we begin by choosing for superposition the greatest side the AABC,
of
as if

diagram page 24.

in

the
of

Two triangles are said be equiangular

to
2.

NOTE
to

one another
when the angles one are respectively equal the angles
of

of
to

the other.
two triangles have the three
of

one severally equal

if

Hence sides
to

the
of

three sides the other, the triangles are equiangular

to

one another.
by

The student should state the converse theorem, and shew

a
.

diagram that the converse not necessarily true.

is

*** At this stage Problems 1-5 and [see page 70] may
8

taken, the proofs affording good illustrations

be

conveniently
of

the
Identical Equality
of

Two Triangles.
26 GEOMETRY.

EXERCISES.

ON THE 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,
bisè& the vertical angle
(i)

:
(ii) to the base.
is

\?Zºnſ,

Aſ
tº

ºf
3.
`.

9
2

is,
a
jść "...º.
ABCD rhombus, figure;
If

º
is

an eqüilateral
2.

foursided
a

shew, by drawing the diagonal

A
that
(i)

the
of

(ii) AC bisèëts'éâch the angles BAD, BCD.

quadrilateral ABCD the opposite sides are equal, namely

If
in
= 3.

AB CD and AD CB; prove that the angle ADC the angle ABC.
=

on
ABC and DBC are two isosceles triangles drawn
If
4.

the same
base BC, prove (by means that the angle ABD the
of

7)
Theorem

=
angle ACD, taking triangles

on
(i)

the case where the are the same side

BC, (ii) the case where they are on opposite sides

§
of
BC.
of

ABC, DBC are two isosceles triangles drawn on opposite

If
5.

the same base BG, and joined, prove that each

be
of

of
sides AD the
if

angles BAC, BDC will

be

into two equal parts.

()
()
Il

r
J
Y

Shew that the straight lines which join the extremities

of
6.

the
an isosceles triangle the middle points the opposite sides,
of

to

of

base
are equal
to

one another.

Two given points an isosceles triangle are equi

of
in
7.

the base
distant from the extremities shew that they are also
of

the base
:

equidistant from the vertex.

that the triangle formed by joining the middle points

8.

Shew
equilateral triangle
; C an an

also equilateral.
of

is of

is

the sides
ABC triangle having AB equal AC; and the
9.

to

isosceles
angles by BO and CO: shew that
at

and are bisected

B
(i)

BO CO
=

(ii) AO bisects the angle BAC,

º

10. Shew that the diagonals [see Ex.

2)
of

rhombus bisect one

a

right angles.
at

another

11. The equal sides BA, CA isosceles triangle BAC are pro
an
of

the points
F,

that AE
so

duced beyond the vertex equal

to

is

and
A

AF; and FB, EC are joined: shew that FB equal

to

to

EC.
is
 TRIANGLES. 27

EXERCISES ON TRIANGLES.

(Numerical and Graphical.)

1. Draw a triangle ABC, having given a =2.0", b=2'1", c=l'3".

Measure the angles, and find their sum.

2. In the triangle ABC, a = 7-5 cm., 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 cm., 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 b, c, and A. What conclusion do you draw Ż
5. A ladder, whose foot is placed 12 feet from the base of a house,
reaches to a window 35 feet above the ground. Draw a plan in which
l" represents 10 ft.; and find by measurement the length of the ladder.
6. I
go due North 99 metres, then due East 20 metres. Plot my
course (scale l 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
ft.);
10
on

1"

diagram (scale
ft.

to

shadow 30 long. Represent this

a

and find by measurement the approximate height the pole.

of

From point surveyor goes 150 yards due East

D. to

then
B
8.

;
to a
a

finally 450 yards due West

to

300 yards due North Plot his

C
;

100 yards); and find roughly how far

A.

from
1"
to

is
D

course (scale
Measure the angle DAB, and say what direction
A.
in

bears from
D

260 yards apart,

on
be
to
9.

and are two points, known

a
C
B

straight shore. The angles CBA, BCA are

at

vessel anchor.
A
is
a

Find graphically the approxi.

be

33° and 81* respectively.

to

observed
C,

the vessel from the points and and from the

of

mate distance

ºgº º:
nearest point on shore.
10.

required find the distance between

In

to
is
as it

two points and B; but lake intervenes, direct measurement

A

a a
a

C,

The surveyor therefore takes third point from

be

cannot made.
he

Which both and are accessible, and finds CA=245 yards,

B
A

CB=320 yards, and the angle ACB =42°. Ascertain from plan the
a
B.

approximate distance between and

A
28 GEOMETRY.

THEOREM 8. [Euclid I. 16.]

/
If one side of a triangle is produced, then the exterior angle is
jºr than either of the interior opposite angles.

f B
\Z^
a. *
& Let ABC be a triangle, and let BC be produced to D.

It is required to prove that the exterior 4. ACD is greater than

either of the interior opposite 4 "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,

N AE = CE,
because EB = EF,
and the A. AEB = the vertically opposite / CEF;
all

the triangles are equal respects; Theor,

...

in

4.

so that the BAE the ECF.

Z.

4.

But the ECD greater than the ECF;

4. 4. 4,

is is is

A. 4. 4.

greater than the BAE;

is, ...

the ECD
that the ACD greater than the BAC.
by
AC
In

it to

the same way, produced supposing

G,

to
is
if

joined the middle point

be

BC, may proved that

be

is to

of

the BCG greater than the ABC.

4

But the BCG the vertically opposite ACD.

=
4

ACD greater than the ABC.

4.

4.
is
N.

Q.E.D.
*
\
 EXTERIOR AND INTERIOR TRIANGLES. 29

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

than two right angles.
* A
For the A. ABC is less than the A-ACD : Proved.
to each add the 4-ACB.
4-5

Then the ABC, ACB are less than the 4–5ACD, ACB,
therefore, less than two right angles.

D
C
B
Every triangle

at
COROLLARY
2.

must have least two acute

angles.

right angle, then by Cor. 1.each

or

one angle

of
For
beis

obtuse the
if

a a

other angles must less than right angle.

Only one perpendicular

be
COROLLARY drawn

to
3.

can

a
g
straight line from given point outside
it.
a

P
be

two perpendiculars could drawn AB from

If

to

triangle PQR
P,

which each
in

is of
we should have
a

PQR, PRQ would right angle, which

be

the
Z.

impossible.

Q
A

R
B
EXERCISES.

Prove Corollary by joining the vertex any point

to

in
1.

the
A
1

base BC.

any point within

If

ABC triangle and BD and CD are

it.
2.

D
is
a

joined, the angle BDC greater than the angle BAC. Prove this
is

by producing BD
(i)

meet AC.
to

(ii) by joining AD, and producing towards the base.

it

any side triangle produced both ways, the exterior

of
If
3.

is
a

formed are together greater than two right angles.

so

angles

given straight line there cannot point

be

4... To drawn from out

a

more than two straight lines the same given length.

of

side
it

triangle are produced,

an
5,

equal
of
If

the sides isosceles the

.

be

exterior angles must obtuse.

30 g GEOMETRY.

THEOREM 9. [Euclid I. 18.]

Ifone 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.

B C

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

the side AB.
It is required to prove that the 4. ABC is greater than the LACB.

From AC cut off AD equal to AB.

Join BD.

Proof. Because AB = AD,

... the 4. ABD = the 4. ADB. Theor. 5.

But the exterior 4. ADB of the A BDC is greater than the

is,

interior opposite / DCB, that greater than the ACB.

4-

greater than the ACB.

...

the ABD
4.

4.
is

Still more then the Z.ABC greater than the ACB.

is

4.
$

&

Q.E.D.

Obs. the following Theorem

of

The mode
in

demonstration
used
to
as

known the Proof by Exhaustion. applicable which

in
It
is

is

cases
true; and
be

certain suppositions must necessarily

of

in

one consists
it

shewing that each these suppositions false with one exception:

of

is

hence the truth the remaining supposition inferred.

of

is
 INEQUALITIES. zºº 31

THEOREM 10. [Euclid 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.

B’ C

Let ABC be a triangle, in which the 4. ABC is greater than

the 4. 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 4. ABC would be equal to the 4 ACB; Theor. 5.
but, by hypothesis, it is not.

Again, if AC were less than AB,

then the Z.ABC would be less than the 4 ACB ; Theor. 9.
but, by hypothesis, it is not.
to,
is,

That AC neither equal nor less than AB.

is

greater than AB. Q.E.D.

...

AC
is

10
on

[For Exercises Theorems and see page 34.]

9
32 GEOME1RY.

THEOREM 11. [Euclid I. 20.]

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

B C

Let ABC be a triangle.

It

its
is required to prove that any two of sides are together
greater than the third side.

enough shew that the greatest side, then

to
It

BC
is
if
is

BA, AC are together greater than BC.

making AD equal
D,
to

to
Produce BA AC.
Join DC.

Proof. Because AD AC,

=
...

5.
the ACD the ADC.
4-

Theor.
4.

But the BCD greater than the ACD;

A. A.

4.
is is

greater than the A-ADC,

...

the BCD
is,

that than the BDC.

4.

*
Hence from the ABDC,
BD greater than BC. Theor. 10.
is

But BD BA and AC together;

=
..".

BA and AC are together greater than BC.

Q.E.D.

This proof may serve

as

an exercise, but the truth

of

NoTE. the
to go

to

really self-evident. For along the

to
is

Theorem from
to C
B
go

straight line BC clearly shorter than

is

from and then

B

from to C. In other words

A

The shortest distance between two points the straight line which joine
is

them.
 INEQUALITIES. 33

THEOREM 12.

Of all straight lines drawn from a given point to a given straight

line the perpendicular is the least.

A R Q C P B s

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 is less than OP.

Proof. In the AOCP, since the 4. OCP is a right angle,

OPC right angle

8.
...

the less than Cor.

A.

is

Theor.
a

;
is,

the OPC less than the OCP.

A.

is

that
z
...

OC less than OP. 10.

is

Theor.
Q.E.D.

only
be

COROLLARY Hence conversely, since there can

1.

one perpendicular and one shortest line from AB,

to
O
If

OC the shortest straight lime from AB, then OC

to
is

is
O

perpendicular AB.
to

Two obliques OP, OQ, which cut AB

at

equal
2.

COROLLARY
the perpendicular, are equal.
of

distances from the foot

C

yſe

The A• OCP, by Theorem

be

be

OCQ may congruent

to

shewn
4

-
hence OP=OQ.
C Of

OQ, OR,
at
if

COROLLARY two obliques OR cuts AB

3.

the
the perpendicular, then
of

greater from the foot OR

is

distance
*
greater than OQ.

The Z.OQC acute, the Z.OQR obtuse;

...

is
is

the 4-OQR greater than the Z.ORQ

‘..

is is

OR greater than OQ.

...
"

H. S. G.
C
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

sºiâû
point D ; shew that AD is greater than either of the equal sides.

the
• -
If in àuadrīſātehā the greatest and least sides are opposite
5.

is
one another, then each the angles adjac to the least side
to

of

-
greater than its opposite angle. sºo

#"
triangle ABC, not greater than AB, shew that
In

AC
is
6.

if
a

by
any straight line drawn through the vertex and terminated the

Gl A
base BC, less than AB. 214ſtrº) fi/03]?()))
is

ABC triangle, which OB, OC bisect the angles ABC,

in
7.

is
a

ACB respectively: shew that, AB greater than AC, then OB

is
is
if

greater than OC.

triangle
of

The difference any is loss than the

of

two sides
8.

third side. * -

any point from the three angular

of

The sum
of

the distances
9.

points triangle greater than half its perimeter.

of

is
a
10.

quadrilateral
of
greater
of

The perimeter than the sum

is
a

its diagonals.
3/3/0/06)
11.

ABC triangle, and the vettical angle BAC bisected by

is
is
a

line which meets BC X; shew that BA greater than BX, and

in

is
a

proof
of

CA greater than CX. Hence obtain Theorem 11.

a

of Yay.'íſ
;

ºn
12.

any point within triangle from

is of

The sum
of

the distances
...

the triangle.
of

its angular points less than the perimeter

13. The sum the diagonals quadrilateral less than the

of
of

is
a
of .

the four straight lines drawn from the angular points any
to

sum
given point. Prove this, and point out the exceptional case.
triangle any two sides are together greater than twice the
In

14.
a

median which bisects the remaining side.

[Produce the median, and complete the construction after the
-
of

manner Theorem 8.]

any triangle the 8wm less than the perimeter.

of
In

15. the medians

is
 PARALLELS. 35

PARALLELS.

flºº• 9 º' {{{}A&YYYi \

;
3.

§§
twu. . . -
EFINITION. Parallel straight lines are such as, being in \
the same plane, do not meet however far they are produced Wyn Gio
its n aph Y
gº
.###"
w
OTE. §araj
vºy")
must be in the 8ame 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

inº,
- -
e.,

necessarily parallel.
Óºb
A®”Tºo

s
straight lines cannot parallel

be
both
-
third straight line.
to
a
In

other words: -

Through given point there can only one straight line parallel
be
a

aś
given Straight line.
to
a

This Playfair's Aaciom.

as

known
is

ion

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

third straight line EF, eight angles are formed,
to
which for
a

distinction particular names are given.

in of

the sake

the adjoining figure E

}{

Thus Øſ.
A
so 4 3, 1,
4, 2,
5, 7,

are called 1/2

6 8

B
are called interior angles, - 4/3
alternate angles;
be
to

and are said

6

also the angles and are alternate

3

* 5/6
as Of to

one another. 8/7

C

*
D

the angles
6,

and referred
is
2

2
to

the exterior angle, and

as

the
6

F
on

interior opposite angle the same side x.

as

EF. Such angles are also known corresponding angles.

of

Similarly are pairs

4,

corresponding
of
3,

and and and

8

5
7

angles.
36 GEOMETRY.

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

If a straight
(i) line cuts two other straight lines so as to make
the alternate angles equal,
an
or

on
(ii) angle equal the interior opposite angle

to
eaterior the
the cutting line,

of
Same side
or

right

on
(iii) interior angles equal

to
the the same side two
r
angles;
each case the two straight lines are parallel.
in

them

Let the straight line EGHF cut the two straight lines
(i)

to
so
as

AB, CD "AGH, GHD

at

and make the alternate

G

4
:

equal
to

one another.

parallel.
It

required prove that AB and CD are

to
is

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

If

if
-

produced, either towards

D,
or

C.

and towards and

B

possible, let AB and CD, when produced,

If

- meet towards
B

-
the point
-

.
D,
at

K.

and
KG

A;

triangle, produced
of

Then KGH
to

which one side

is

is
a

the exterior greater than the interior opposite

...

AGH
by4.

is

-
4-GHK; but, hypothesis, not greater.
it
is

.
AB

CD
...;

cannot meet when produced towards

D.

and and
B

Similarly shewn that they cannot meet

be

may towards
it

and C:
A

..".

AB and CD are parallel.

 PARALLELS. 37

(ii) Let the exterior / EGB = the interior opposite 4 GHD.

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

Proof. Because the A. EGB = the 4 GHD,

and the Z.EGB = the vertically opposite LAGH;
... the 4- AGH = the 4 GHD :

and these are alternate angles;

AB and CD are parallel.
...

be

(iii) Let the two interior BGH, GHD together equai

to
Z"

two right angles.

AB

required and CD are parallel.

It

prove that
to
is

Zºº

Proof. Because the BGH, GHD together two right

=

angles;
and because the adjacent "BGH, AGH together two right
=
/

angles;
-

"BGH, AGH together "BGH, GHD.

...

the the
=
4

From these equals take the BGH

Z.

then the remaining AGH the remaining GHD

=

4
/

and these are alternate angles;

..".

AB and CD are parallel.

Q.E.D.
&
*

,
'

DEFINITION. straight line drawn across given

of

set
A

lines called transversal.

is

For instance, the above diagram the line EGHF, which

in

crosses
the given lines AB, CD
is

transversal.
a
38 GEOMETRY.

14.
THEOREM [Euclid 29.]

I.
If
straight line cuts two parallel lines, makes

it
(i) a
-
the alternate angles equal one another;

to
(ii) the exterior angle equal

on
the interior opposite angle

to
the
the cutting line
of
same side

;
(iii) the two interior angles

on
the same side together equal

to
two
Tight angles.

Let the straight lines AB, CD

be

parallel, and let the

straight line EGHF cut them.
required
It

prove that
to
(i) is

AGH the alternate GHD;

A. =

the
4

(ii) EGB the interior opposite GHD;

=

the eaterior
4

(iii) the two interior BGH, GHD together two right angles.
=
A
*
(i)

Proof not equal GHD,

to

the
If

AGH the
4.

is

suppose the LPGH equal it;

to

GHD, and alternate

to

the
4

then PG and CD are parallel. Theor. 13.

by

But, hypothesis, AB and CD are parallel;

the two intersecting straight lines AG, PG are both parallel
...

CD: which impossible. Playfair's Aa.iom.

to

is

GHD;
to

not unequal
...

the 4-AGH the

is

4
is,

that the alternate "AGH, GHD are equal.

4
(ii)

Again, because the EGB the vertically opposite

A.

AGH
4

and the AGH the alternate GHD Proved.

= =
4. 4.

the exterior the interior opposite

...

EGB GHD.
4
 PARALLELS. 39

(iii) Lastly, the EGB = the 4 GHD;

4. Proved.
add to each the Z. BGH ;
then the 4 "EGB, BGH together = the angles BGH, GHD.
But the adjacent / " EGB, BGH together = two right angles;
the two interior "BGH, GHD together two right angles.

=
...

a
Q.E.D.

PARALLELS ILLUSTRATED BY ROTATION.

The direction straight line by the angle which

of

is
determined
a

makes with some given line

of
reference.
it

AB, relatively the given line YX, given by

of

Thus the direction

to

is
the angle APX.
Now suppose that AB and CD
in

the adjoining diagram are parallel; *

A
then we have learned that
the ext. Z-APX the int. opp. CQX;
=

/
is,

that AB and CD make equal angles

&
*
with the line of reference YX.

X
Q
Y

P
This brings
to
us

the leading idea

connected with parallels:

-
J’arallel straight lines have the 8ame
D

DIRECTION, but differ Position.

in
be

The same idea may illustrated

thus
h
:

through the 4-APX,

to
so
as

Suppose AB
to

rotate about take the

P

position XY. Thence let the opposite way through

Q

rotate about
it

be
the equal XQC will now take the position CD. Thus AB may
it
/

brought into the position CD by two rotations which, being equal

of
no

and opposite, involve final change

of

direction.
In

HYPOTHETICAL CONSTRUCTION. the above diagram let

fixed straight line, straight
be

AB fixed point, CD
Q
a

a
a
Q,

line turning about and YGPX, any transversal through

Q.
be

Then CD rotates, there must one position which the

in
==as

CQX the fixed APX.

4.

z.

we

Hence through any given point may line pass

to

assume
a

parallel any given straight line.

to

straight line, movements from towards

If

AB
is

Obs.
A
a

be

opposite
in
of B,

to

and from towards are said senses

B

the line AB,

40 geometry.

15.
THEOREM [Euclid 30.]

I.
the
Straight lines which are parallel straight line are

to
same
- -
parallel
to
one another.

E
a/

B
A
H/

D
C
P

Q
Z K
CD
Let the straight lines AB, be each . parallel the straight

to
line PQ.

are
AB

CD
required parallel
It

prove that and

to

to
is

one
another. .
Draw straight line EF cutting AB, CD, and PQ

in
the
a

-
H,

points
G,

K.

and
PQ

EF
AB

Proof. Then because and are parallel, and meets

:
them,
...

the AGK the alternate GKQ.

Z.
=
4.
CD

PQ

EF

And because and are parallel, and meets them,

the interior opposite
...

the exterior GHD GKQ.

Z.
=
4
...

the AGH the GHD

=
4

and these are alternate angles;

AB and CD are parallel. -
..'.

-
Q.E.D.

between AB and CD, the Proposition needs

no

NoTE. PQ lies
it If

proof; for inconceivable that two straight lines, which do not

is

meet an intermediate straight line, should meet one another.

this Proposition may readily deduced from Playfair's
be

The truth
of

-
"Axiom, which -
of

the converse.
it
is

For AB and CD were not parallel, they would meet when produced.
if

straight -lines both parallel

be

to

Then there would two intersecting

a

-
third straight line: which impossible.
is

Therefore AB and CD never meet; that they are parallel.

is,
 PARALLELS. 41

- EXERCISES ON PARALLELS.
cºw Aniº")
1. In the diºm of the previous page, if the angle EGB is 55°,
-
4.63express in degrees each of the angles GH C, HKQ, QKF.

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

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

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

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

the straight lines joining AC and BD are parallel.
N.V.O3-( .
– 6. Any straight line drawn parallel to the base of an isosceles
- tri
angle 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: show 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

an
of

the bisector
in

angle parallel the angle, and terminated by them, are

to

the arms
of

equal; and the resulting figure

is

rhombus.
a

"
'

X D,

11. AB and CD are two straight lines intersecting

at

and the
through any point
so

adjacent angles DC
in

formed are bisected

if

in a
:

straight line YXZ drawn parallel AB and meeting the bisectors

to
is
Z,

shew that XY equal XZ.

to
is

and
Y

Q,

12. Two straight rods PA, revolve about pivots

QB Pand PA
at

10.

minute, and QB making they

12

making complete revolutions

If
a

start parallel and pointing the same way, how long will before they
be
it

are again parallel, pointing opposite ways, (ii) pointing the same way?
(i)
42 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 4 "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 meets them,

... the 4. ACE = the alternate A. CAB.

Again, because BA and CE are parallel, and BD meets them,

the exterior ECD the interior opposite ABC,
=
4.

4.
...

of
...

ACD the two interior opposite

=
4.

the whole exterior the sum

Z_*

CAB, ABC.

To these equals add the BCA;

of

each
A.

then the "BCA, ACD together the three "BCA, CAB, ABC,
=
4

But the adjacent "BCA, ACD together two right angles.

=
/

"BCA, CAB, ABC together two right angles.

...

the
=
4

Q.E.D.

this proof the following most im

In

of

Obs. the course

portant property has been established.
If

triangle
of

produced the 60terior angle equal

to
is

is

side
a
a

the two interi Osite amales.

|

Namely, the ext. ACD the ACAB+ the ABC.

A.

4.
 THE ANGLES OF A TRIANGLE. 43

INFERENCES FROM THEOREM 16.

If A,

the
B, and C denote number

of

of
in
1. degrees the angles
triangle,
a

A+ B+ 180°.

=
then

C
If

two triangles have two angles the one respectively cqual

of
2.

equal
of

of
two angles the other, then the third angle
to

to
is
the one
of

the third angle the other.

In

any right-angled triangle the two acute angles are comple

3.

mentary.

triangle
If

of

of
one angle equal

to
4.

is the sum the other

a

two, the triangle right-angled.

is

any quadrilateral figure

of

of

the angles equal

to
5.

is
The Sum
four right angles.

EXERCISES ON THEOREMI 16.

triangle right
an

Each angle equilateral

of

two-thirds of
1.

is

a
In or

angle, 60°.

right-angled isosceles triangle each equal angles

of
2.

the
a

45°.
is

Two angles triangle are 36° and 123° respectively: deduce

of
3.

the third angle; and verify your result by measurement.

triangle ABC, the the 4-C=42°; deduce the 4-A,

In

B=111°,
A.
4.

and verify by measurement.

triangle ABC
to

produced
D.

One side BC the exterior

If
of
5.

is is
a

angle ACD 134°; and the angle BAC 42°; find each the remaining
of
is

interior angles.
In

figure 16, the Z.ACD=118°,

of
6.

the Theorem and the

if

B=51°, find the A-3A and and check your results by measurement.
4.

C
;

of

Prove that the three angles triangle are together equal

to to
7.

two
a

night angles by supposing line drawn through the vertea parallel the
a

base.

two straight lines are perpendicular two other straight lines,

If

to
8.

each, the acute angle between the first pair equal

to

to

each the acute

is

angle between the 8econd pair.

 whº,
44 GEOMETRY.
cººl jihnºlºgy
COROLLARY 1. All theft, angles of any
together with four right angles, are equal to twice as many right
angles as the figure has sides.

A B

Let ABCDE be a rectilineal figure of n sides.

all
It

rt.
is required to prove that the interior angles

+
4

A
"
2m rt.
=

".
A

Take any point within the figure, and join

to

of
each
O

O
its vertices.
Then the figure divided into triangles.
of is

rt.
And the three each together
A* A

= =

A ".
rt. A
4

2
of "

Hence all the all the together 2n

".
4
"
all

all

A"

But make up all the interior angles

of

the the
4
"

O,

rt.
the figure together with the angles
at
of

which
=

".
A
4
rt.

rt.
2n

all the int. the figure +4

of

"=
...

".
A

A
*

Q.E.D.
its
all

DEFINITION. regular polygon

all A

one which has

is

sides equal and its angles equal.

Thus degrees each angle
in
of

of

denotes the number

if
D

be

regular polygon sides, the abové result may

of

stated
n
a

thus:
360°
=

mD m. 180°.
+

EXAMPLE.

degrees each angle

of

Find the number

in
(i) of

regular hexagon sides);

(8 (6
s’

(ii) regular octagon sides);

a

*(iii) regular decagon (10 sides).

a
 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

the vertical angle;

of
Each base angle
(i)

double
is is

(ii) Each base angle four times the vertical angle.

The base triangle produced both ways, and the exterior
of
3.

is
X

to a

Con
be

angles are found 94° and 126°; deduce the vertical angle.
struct such triangle, and check your result by measurement.
a

triangle 162°, and their

of
the angles
of

The sum
at

is
the base
4.

a
difference 60°: find all the angles.
is

triangle are 84° and 62°; deduce

at

The angles
of
5.

the base
a
(i)

of
the vertical angle, (ii) the angle between the bisectors the base
angles. Check your results by construction and measurement.

,
In triangle ABC, the angles AB
at

and are 74° and 62°:

6.

of if
C
B
a
.

and AC are produced, deduce the angle between the bisectors the
exterior angles. Check your result graphically.
quadrilateral are respectively 114%.”,50°, and
of

Three angles
7.

75%’; find the fourth angle.

C,

re
B,

quadrilateral ABCD, the angles

In

at

and are
8.

D
a

to

spectively equal 2A, 3A, and 4A find all the angles.

;

an irregular pentagon sides) are 40°, 78°, 122°,

of

Four angles
(5
9.

and 135°; find the fifth angle.

2(n-2)
In

10. any regular polygon sides, each angle contains

of
n

7?,
right angles.
(i) Deduce this result from the Enumciation Corollary
of

1.

by joining one vertex

of

(ii) Prove independently

to

each the
it

A), thus dividing the

to

others (except the two immediately adjacent

polygon into n-2 triangles.

ll.
of

How many sides have the regular polygons each whose

(i)

angles 108°, (ii) 156°2

is

Shew that the only regular figures which may

be

12. fitted together

So

equilateral triangles, (ii) squares,

to

plane surface are

as

(i)

form
a

(iii) regular heavagons.

46 GEOMETRY.

If

the
rectilineal figure, which has

of
COROLLARY 2. sides

in a
To

ſº
re-entrant angle, are produced order, them all the eaterior
angles. J|}|JPlötſ
so
angles age together equal four|right

to
14

"rh
jejlſ

25
>
$)

tº
6)

1
T

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

n
and consequently vertices.
n

Now at each vertex

";
rt.
the interior 4-4 the exterior

=
2
Z

A
and there are vertices,
n

the int. '+ the sum "- 2n rt.

of

of
...

the sum the ext.

".
A

A
by

Dut Corollary
1,

Z."
the int.
... of

the sum rº.

+

";
2n rt.
Z

=
of *

A
the sum the ext. rt.
=

".
4
A

A
*

Q.I.D.
2nd Proof.
C

©
O,

Take any point and suppose Oa, Ob, Oc, Od, and Oe, are
C,
B,
A,

lines parallel the sides marked,

D,

(and drawn
to

which those sides were produced).

in

/ in

from the sense

O

Then the exterior between the sides and the Zaob.

=
B
A

And the other exterior the "bCC, cod, dOe, eOa,

=
A

4
*

respectively.
the ext. "a the sum
of
...

at
of

the sum the

O
A

4
"

rt.
=

".
4

A
 THE ANGLES OF RECTILINEAL FIGURES. 47

EXERCISES.

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

exterior angle is cqual to the interior angle of an equilateral triangle.

2. Express in degrees the magnitude of each eacterior angle of

regular octagon, (ii) regular decagon.
(i)
a

How many sides has a regular polygon each exterior angle

3.

is
if
a
30°, (ii) 24°3
(i)

straight line meets two parallel straight lines, and the two
If
4.

on

interior angles the same side are bisectod, shew that the bisectors
right angles.
at

meet
any triangle produced both ways, shew that the
If

of
5.

the base
is

the two exterior angles minus the vertical angle equal two

to
of

sum

is
right angles.
the triangle ABC the base angles are bisected by
In

at
6.

and

C
B
BO and CO respectively. Shew that the angle Boc-904.
the triangle ABC, the sides AB, AC are produced,
In
7.

a
exterior angles are bisected by BO and CO. Shew that
2 ''. A

BOC 90°
0
–
–

by

The angle contained two adjacent angles

of
of
8.

the bisectors
quadrilateral equal half the sum the remaining angles.
to

of
is
a

isosceles triangle ABC, and BA

an

produced
of
9.

is

the vertex
so A
is

BA;
D,

that AD equal DC drawn, shew that BCD

to

to

is
if
is

is

a
right angle.
10.

The straight line joining the middle point the hypotenuse

of

the right angle

to

right-angled triangle equal

ºf

half the
to

is
a

hypotenuse.

ExpºRIMENTAL proof THEOREM -C=180°.]

(A

16.
+
B

ABC, AD BC the
In

is

the
A

greatest side. AD right

º”
bi
is

ZY; and YP,

by

rp", on
If

is

now the folded

A

three
zºº

B,

coin
A,

dotted lines, the

ll

cide with the ºz.D.Y., ZDQ,

4

their sum 180°.

is
,
 -
48 GEOMETRY.

THEOREM 17. [Euclid I. 26.] . -

If two
triangles have two angles of one equal to two angles of the

the

the
other, each to each, and any side of first equal corresponding

to
are
the

the

all
triangles respects.Göyüşo 3'ſ
of

in
side other, equal

ºf a
wºº
*::. &
Fº, !.

..
me

p
#
***A&

*
\
TX; -
.
B

F
C

E
be

which

in
Let ABC, DEF two triangles

E, D,
the 4-A the
= = =

4 4.
the the
A.
BC B
let

also the side the corresponding side EF.

A'
ABC, DEF are equal all
It

required prove that the

in
to
is

Tespects.
ºr
2 B,

Proof. The sum "A,

of

the and
C
4

rt.
= =

Theor. 16.
".
A

F;
E,
the "D, and
of

the sum 4
F. "D
"A

and the and the and respectively,

==

E
CB
4

4. 4

-
... the the
4

Apply the DEF,

on
AABC E,
to

so

the that falls, and

A

BC along EF. -
º

Then because BC EF,

=

... must coincide with

F.
C

And because the the Z.E,

A.

=
B

BA must fall along ED.

...

And because the 4.6–the A.F,

CA must fall along FD.
...

the point
on

on

which falls both ED and FD, must coin

A,
...

the point
D,

cide with
in

which these lines intersect.

the AABC coincides with the DEF,
...

therefore equal all respects.

in
to

and
is

it

So that AB DE, and AC DF;

==

and the AABC the DEF in area.

A

Q.E.D.
 CONGRUENT TRIANGLES. 49

EXERCISES.

ON THE IDENTICAL Equality OE 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 in a 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.

ABCD, AB AD, and BC DC shew that

In

quadrilateral,
if

=
a

the diagonal AC bisects each the angles which joins; and that AC
of

it

perpendicular BD.
to

§
is

he of

B he

surveyor wishes river which

a to

11. ascertain the breadth

A

cannot cross. Standing point near the bank, notes an object

at

A
on

immediately opposite He lays down any

of of

the other bank. line AC

a

length right angles AB, facing the middle point AC.

C at

a to

to at

mark
O
a
he

he

From walks along line perpendicular AC until reaches point

a

from which and are seen the same direction. He now measures
O

in
B
D

CD: prove gives

of

that the result him the width the river.

D
H.S. G.
60 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:

all
Two triangles are equal in respects when the following
three parts each are severally equal:
in

7. 4.
Two sides, and the included angle. Theorem
2. 1.

The three sides. Theorem

one triangle

in
Two angles and one side, the side given
3.

Theorem 17.

in
CORRESPONDING that given
to

the other.

all
Two triangles are not, however, necessarily equal

in
to
respects when any three parts one are equal the corre
sponding parts of
of

the other.
For example:
(i) When thethree angles
of

one are
equal the other,
of
to to

the three angles

each each, the adjoining diagram
shews that the triangles need not
be

equal all respects.

in

(ii) When two sides and one angle one are equal
in

to
two
sides and one angle the other, the given angles being opposite
of

equal sides, the diagram below the triangles

to

shews that
all

equal respects.
be

in

need not
É.
C
B

For AB DE, and AC DF, and the ABC the DEF,

it
=
be if

A
4

will the given sides

in
of

seen that the shorter the

lie

triangle DEF may the positions DF

in

or
be of

either DF'.
NOTE. From these data may shewn that the angles opposite
it

Alº

the equal sides AB, DE are either equal (as for instance the ACB,
to

DFE); and that

in Z."

ACB,
or

DF'E) 8wpplementary
is in

(as the the former

case the triangles are equal all respects. This called the
.

9,
p.

ambiguous case the congruence triangles.

in

of

[See Problem 82.]

the given angles are right angles, the ambiguity dis
If

at

and
E
is B

appears. This exception proved the following Theolem.

in
 CONGRUENT TRIANGLES. 51

THEOREM 18.

Two right-angled triangles which have their hypotenuses equal,

all
and one side of one equal to one side of the other, are equal in
Tespects.
A

D
C’
C

É
B

F
two right-angled triangles,
be

Let ABC, DEF

in
which
the "ABC, DEF are right angles,
4

the hypotenuse AC the hypotenuse DF,

==

and AB DE,

all
A"ABC, DEF

in
It

required prove equal

to
is

that the are

Tespects.

Apply the AABC DEF,

so

Proof. that AB falls

to

the
A
on

on

to
the cqual line DE, and DE opposite
of

the side F.
C

the point
be

on

Let
C'

which falls.
its C

Then DEC' represents the AABC new position.

is in

the "DEF, DEC' right angle,

of

Since each
4

EF and EC' are one straight line.

in
...

And the AC'DF, because DF DC' (i.e. AC),

in

5,

Theol.
...

the DFC' the DC'F.

=
Z

the A*DEF, DEC',

in

Hence

the DEF the DEC', being right angles;

= =
Z. 4:

4 4.

because the DFE the DC'E, Proved.

and the side DE common.
is

all all

the A*DEF, DEC' are equal respects;

in in
is, ...

Theor. 17.
that the A*DEF, ABC are equal respects.
Q.E.D.
&
52 GEOMETRY,

19.
*THEOREM [Euclid 24.]

I.

of of
of
If two triangles have two sides the one equal

to
two sides the

by
other, each each, but the angle included the two sides ome

to

by

is of
greater than the angle included the corresponding sides
that which has the greater angle

of
the other; them the base

of
greater than the base the other.

* \,
*
C

G
K
B

F
be

Let two triangles, which

in
ABC, DEF
BA ED,
= =

and AC DF,
but the greater than the EDF.
is A.
BAC
A.

is

required prove BC greater than the

It

base
to

that the
is

base EF.

on
Apply the AABC DEF,

D,
so
to

Proof. the that falls

A

and AB along DE.

Then because AB DE, must coincide with
E.
=

Let DG, GE represent AC, CB their new position.

in

EG passes through greater than EF;

F,

Then EG
is
if

is,

that BC greater than EF.

is

EG does not pass through suppose that DK bisects

F,

But
if

FDG, and meets EG Join

in
K.

the FK.
4

A*

Then FDK, GDK,

in

the
\

FD GD,
is =
|-

to

because DK common both,

FDK the included GDK;
A.

the included
=
/

4.

FK GK. Theor,
=
*.

Now the two sides EK, KF are greater than EF;

is,

that EK, KG are greater than EF.

EG (or BC) greater than EF. Q.E.D.
is
‘.
 CONVERSE OF THEOREM. 19. 53*s

**
“ſhan
conversely, ãf 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

•*.

E
B C

Let ABC, DEF be two triangles in which

BA = ED,
and AC = DF,
but the base BC is greater than the base EF.
It is required to prove that the 4. BAC is greater than the 4. EDF,

Proof. If the A, BAC is not greater than the 4. EDF,

it must be either equal to, or less than the A. EDF.
Now if the Z. BAC were equal to the A EDF,
then the base BC would be equal to the base EF; Theor. 4,
but, by hypothesis, it is not.
Again, if the A. BAC were less than the 4. EDF,
then the base BC would be less than the base EF; Theor. 19.
but, by hypothesis, it is not.
to,
is,

Thät the BAC neither equal nor less than the EDF;
A.
... 4.

is

the BAC greater than the EDF.

A.

A.
is

Q.E.D.
the
be
an
.*

asterisk may postponed

or

at

Theorems marked with omitted

of

discretion the teacher.

* 54 GEOMETRY.

REVISION LESSON ON TRIANGLES,

1. State the properties of a triangle relating to

its interior angles;

to of of
(i)
the sum
(ii) the sum its exterior angles.
What property corresponds polygon With

in
(i)

of
sides?

n
a
what other figures does triangle share the property (ii)

A
2
a

*
TU
a

*
*
*
*
".
.

Classify triangles with regard their angles. any

to
2.

Enumciate
Corollary assumed

in
or

Theorem the classification.

~,

-",
.#
e
3

,
,
*

"
.
.
Enunciate two Theorems which from data relating
in

to
3.

the sides
drawn relating
to
the angles.
is

conclusion
a

#
Q
the triangle ABC,
In

3-6 cm., b=2.8 cm., c=3.6 cm., arrange

if
=
a

the angles their sizes (before measurement); and prove that

in

of

order
the triangle acute-angled.
is

Enunciate two Theorems which from data relating

in

to
4.
u

the
/

drawn relating
to

angles
is

conclusion the sides.

a

the triangle ABC,

(i) In

if
|

A=48° and B-51°, find the third angle, and name the greatest
*

side.
(ii) A=B-62}", find the third angle, and arrange the sides order

in
their lengths.
of

From which the conditions given below may we conclude that

of
5.

the triangles ABC, A'B'C' are identically equal? Point out where
ambiguity arises; and draw the triangle ABC
in

each case.
4:2 cm. A' =36°,
a'
A'

=71°.
A
=
A

B–B'-46°. (ii) b- b'-2'4 (iii) B= B'-

(i)

cm. 121°.
{

{
;

3-7 cm. C' 8.1°. C'—23°.

=
C
a'

=
=

C
a

3.0 cm. B= B’= 53°. C'— 90°.

a'

== =
ac C
=

=
a

(iv) b'=52 cm. (v) =b' 4:3 cm. (vi) cm.

5
=

==

==
{

35
Cb

a' c

= = 4.5 cm. c= 5.0 cm. cm.

=

c’
C
*

º
the last question by stating generally
of
6.

Summarise the results

under what conditions two triangles
are necessarily congruent
(i)

;
be
or

(ii) may may not congruent.

&

If

two triangles have their angles equal, each each, the triangles
to
7.
/

are not necessarily equal all respects, because the three data are not
in

independent. Carefully explain this statement.

 , EXERCISES ON TRIANGLES. 55

(Miscellaneous Eacamples.)

be
drawn

to
is
The perpendicular the shortest line that can
(i)

a
8.
given straight line from given point.

a
(ii) obliques which make equal angles with the perpendicular
*

are
x

equal.

(iii) Of two obliques the less that which makes the Smaller angle with

is
X

the perpendicular.

of
If

two triangles have two sides the one equal

of
to
9.

two sides the

other, each each, and have likewise the angles opposite one pair

of
to

to
equal sides equal, then the angles opposite the other pair equal sides

of
to
or

are either equal 8wpplementary, and the former case the triangles are

in

*;
2.
&
equal all respects.

2
,
*
*
*
*
in

PQ perpendicular length) straight line XY.

in
(4

to
is

10. cm.
a

a
Draw through obliques making with PQ the angles 15°,
of

series
P
a

30°, 45°, 60°, 75°. Measure the lengths these obliques, and tabulate
of

c.
;
the results.
11.

PAB triangle
which AB and AP have constant lengths
is in
is
a

A,
fixed, and AP rotatés about
AB
If

cm. and cm. trace the

4

PB,
of 0°
as

changes angle
in

to

the increases from 180°.

A

Answer this question by drawing series figures, increasing by

A
a

30°. Measure PB each case, and tabulate the results.

in
of

increments

flagstaff AB
D of

12. From the foot horizontal line drawn

B

is
a

a
27

passing two points and which are feet apart. The angles BCA
C

on

and BDA are 65° and 40° respectively. Represent this diagram
a

ft.), and find by measurement the approximate height

10
to

(scale cm.
1

flagstaff.
of

the

PQ, two boats

P,

the top
From lighthouse
of

13. and are

B
A
a

the lighthouse. known that

It
at

of

is

seen anchor line due south

in
a

PQ 126 ft., PAQ =57°, PBQ =33°; hence draw plan which
in
=

a
by

represents 100 ft., and find measurement the distance between

l"

and to the nearest foot.

B

B,

14. From lighthouse two ships and which are 600 yards
A
L
a

apart, are observed directions S.W. and 15° East South respec
in

of

tively. At the same time S.E. direction.

is

observed from
in
B

a
by

Draw plan (scale 200 yds.), and find

1"
to

measurement the distance

a

the lighthouse from each ship.

of
56 GEOMETRY.

PARALLELOGRAMS. zºº - ºr
& "" +
, --, * : DEFINITIONS, º
ca.
2*

'9

>
<2*

*
*
.
.
.
.

&
'
'
'

J
J
quadrilateral plane figure bounded
A
1.

is
a
by

four straight lines. -3

The straight line which joins opposite angular

points encº

2,
quadrilateral diagonal.
in

called
is
a

a
".{...,

,&
".
5

*
.

parallelogram quadrilateral
A

sº
2.

is
a
whose opposite sides are parallel.
,

7
/
[It will proved hereafter that the opposite
be

parallelogram are equal, and that its

of

sides
a

opposite angles are equal.]

f
A g
=

r
a

rectangle parallelogram which

3.

a is
a
its

angles right angle.

of

has one
.
.
.
.

[It will
be

proved hereafter that all the angles CŞ.

of

rectangle are right angles. See page 59.]

a

ty
3.
*
*

square rectangle which has two

A

is
4.

adjacent sides equal.

[It will
be

proved that all the sides square are

of
a

equal and all its angles right angles. See page 59.]
+.

dry
3

”
“,

quadrilateral which
A

rhombus
5.

is
a
its
all

its

has sides equal, but angles are

not right angles.

£, *
-,
,

trapezium quadrilateral
A

which has
6.

is
a

pair parallel sides.

of

one
 PARALLELOGRAMS. 57

20.
THEOREM [Euclid 33.]

I.
."3,
gy

tº
&

of
The straight lines which join the extremities two equal and

the
º.
parallel straight lines towards same parts are themselves equal
and parallel. Yvº,

...
=
c

B
A

D
C

equal and parallel straight lines; and let

be

Let AB and CD
joined towards the same parts by
the straight lines
be

them
AC and BD.

prove that AC and BD are equal and parallel.

It

required
to
is

**.
Join BC.

Proof. Then because AB and CD are parallel, and BC meets

th €II) 2/3/* ce".
3.

...

the ABC the alternate DCB.

=
4.

4.
A*

Now ABC, DCB,

in

the
AB DC,
=
to
{|

because BC common both

is

land the ABC the DCB Proved.

A.
=
4

;
all

the triangles are equal respects;

in
...

DB, .........................
so

(i)

that AC
==

and the ACB DBC.

4

But these are alternate angles;

AC and BD are parallel. ................., (ii)
AC ..'.
is,

That 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.

\!
Let ABCD be a parallelogram, of which BD is a diagonal.
It is required to prove that

*
(i)

AB CD, and AD CB,

Z, A. =

(ii) BAD DCB,

4 A.
= = =

the the
(iii) the ADC the CBA,
(iv) the AABD
in

CDB
A

the area.

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

...

the ABD the alternate 4-CDB.

=
4.

Again, because AD and BC are parallel, and BD meets them,

...

the ADB the alternate CBD.

=
4.

the A"ABD, CDB,

in

Hence
the ABD the CDB,
Z. Z.

= =

A 4

because {the ADB the CBD, Proved.

to

and BD common both

is

17.
all

the triangles are equal respects;

in
...

Theor.
CB; ..................... (i)
so

that AB CD, and AD

=

and the BAD the DCB ........................ (ii)

==
4.

4.

and the AABD the CDB in area. .............. (iv)

A

And because the ADB the CBD, Proved.

Z. Z.

= = =

4. 4.

and the CDB the ABD,

CBA........., (iii)
...

the whole ADC the whole

/
/

Q.E.D.

*****~~
 3.

*
gº*
PARALLELOGRAMS. 59
gº'
one angle parallelogram right

of
If
COROLLARY

1.

is
a

a
all angles are right angles.

its
angle,
In

other words

.
:

.
the
All are right angles.

of
angles rectangle

a
4–2

rt,
a 4.
For the sum two consecutive

of
(Theor. 14.)

;
rt.

be
..., angle, the other must angle.
of

rt.
one these
if

is
a
And the opposite angles the par" are equal;

of
A

all the angles are right angles.

'',

...

CoRollARY All are equal;

2.

the sides

of
square and all
tts *

a
---
aples are right angles.

“,
&Yº

%
The diagonals parallelogram
of
COROLLARY
3.

bisect one

a
another.

C
=
D
+
Let the diagonals AC, BD the par"
of

A-,
O.

ABCD intersect
at

To prove AO OC, and BO =OD.

2"

“…* In the A* AOB, COD,

B
{...}}~
,
,
*
f
\

the Z.OAB the alt. OCD,

=

because the 4-AOB =vert. opp. COD,

{

and AB the opp. side CD

=

OA OC and OB OD. Theor. 17,

...

=
;

EXERCISES.

quadrilateral are equal, the figure

If

the opposite sides

of
1.

is
a
a

parallelogram.

quadrilateral are equal, the figure

of

the opposite angles

If
2.
...

is
a

parallelogram.

the diagonals quadrilateral bisect each other, the figure

of
If
3.
...

is
a
a

parallelogram.

The diagonals right angles.

of

at
4.

rhombus bisect one another

a

its
all

the diagonals parallelogram are equal, angles are

of
If
5.

right angles.
In

parallelogram which not rectangular the diagonals

6.

is

are
a

unequal.
60 GEOMETRY.

EXERCISES ON PARALLELS AND PARALLELOGRAMS,

;
(Symmetry and Superposition.)
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
d iagonal.

2. Prove that the diagonals of a square are aves 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. In a quadrilateral ABCD, AB = AD and CB=CD; but the sides

an
(if
are not all equal. Which of the diagonals

of
is
either) axis
symmetry
:

Prove by the method superposition that

of
(i) 6.

Two parallelograms are identically equal two adjacent sides

of
if

one are equal two adjacent sides of the other, each each, and one
to

to

angle ome equal one angle

of

ºf of
to

the other.
(ii) Two rectangles are equal two adjacent 8ides one are equal
of

to
two adjacent Sides the other, each
of

to

each.

Two quadrilaterals ABCD, EFGH have the sides AB, BC, CD, DA
7.

equal respectively the sides EF, FG, GH, HE, and have also the
to to

be

angle BAD equal the angle FEH. Shew that the figures may
coincide with one another.
to

made

(MIiscellaneous Theoretical Eacamples.)

Any straight line drawn through the middle point diagonal

of
a 8.
at of X

is a
by

parallelogram pair opposite sides,

of

and terminated bisected

a

that point.
In

parallelogram the perpendiculars drawn from one pair

of
9.

a
*

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

X,
sy

ABCD parallelogram, respectively the middle

of If

is

10. and
Y
a

points the sides AD, BC; shew that the figure AYCX paral
is
a

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
the Z-C= 180°– the A-B-H the D.;
(i)

A.
the 4-A
+

(ii) the diagonal AC= the diagonal BD

;
(iii) the quadrilateral symmetrical about the straight line joining

is
the middle points AB and DC.
of

13. AP, BQ are straight rods equal length, turning

of
equal

at
rates (both clockwise) about two fixed pivots respectively.

If
and

B
the rods start parallel pointing opposite senses,

in
but shew that
parallel;
(i)

be

they will always

(ii) the line joining PQ will always pass through certain fixed

a
point.

(Miscellaneous Numerical and Graphical Examples.)

the angles the triangle ABC, having given

of

14. Calculato

:
int. A.A– 3B =4C.
of

ext. 4-A
3

15. yacht sailing due East changes her course successively by 63°,
A

by 78°, by 119°, and by 64°, with sailing round an island.

to

view
a

Easterly
be

an

What further change must

to

made set her once more on

course
?

the interior angles rectilineal figure equal

If

of

16.
of

the sum
is
a

it,

and why?
to

the exterior angles, how many sides has

of

the sum

17. Draw, using your protractor, any five-sided figure ABCDE,

in which
110°, 115°, =93°,
4-

152°.
=

=
C
B

=
D
by A

E
4

Verify construction with ruler and compasses that AE parallel

is
a

BC, and account theoretically for this fact.

to

18. and are two fixed points, and two straight lines AP, BQ,
A

Q,

unlimited towards are piyoted AP, starting from

at

B.

and and
A
P

the direction AB, turns about

at

of

clockwise the uniform rate 74°

A

second and BQ, starting simultaneously from the direction BA, turns
B ;

3;
at

about
of

counter-clockwise the rate second.

a

How many seconds will elapse before AP and BQ are parallel?

(i)

(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.

§º:º
THEOREM * * 22.
§
sº a , sº jº
d §

%
If

are
mºſºftgºgllel straight lines, and intercepts

or
there three

the
by
any transversäl are equal, then

on
made them corresponding

ñº

$
any other transversal are also equal.
on
$ntercepts

§
N
NX

Pl
A

B
+
‘... T-/

's

*)
al

2
!'
N

D
C

M
{^NZ
|

NTF

N
E

I R

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.

XY
It

required prove that YZ.

to

=
is

Through
be

let XM and YN drawn parallel

to
and PR.
X

Proof. Since CD and EF are parallel, and XZ meets them,

XYM the corresponding
...

the YZN.
=
4

#
And since XM, YN are parallel, each being parallel PR,
to

MXY the corresponding

...

the NYZ.
=
4

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 the A*XMY, YNZ,

in

the XYM the YZN,

4. Z.
= = =
A Z

because the MXY the NYZ,

{

and XM YN
;

17.

the triangles are identically equal;

...

Theor.
.*. XY YZ.
=

Q.E.D.
 PARALLELS AND PARALLELOGRAMS. 63
26,

In a triangle ABC, if a lines Pp, Q4, Rr, ...,

set
of
COROLLARY.
drawn parallel base, divide one side AB into equal parts, they

to
the
also divide the other side AC into equal parts.

g
Q

ſ
Yºr
N.
!
R
|
|

3 |
2 ||
1 |

C
B

be
The lengths the parallels Po, Qq, Rr, may thus expressed
of

,
in terms of the base BC.
let pl. q2,
be

Through drawn parl AB.

to
q,
p,

r3

and
r

Then, by Theorem 22, these par” divide BC into four equal parts,

of
which Pp evidently contains one, Qq two, and Rr three.
*S,
In

other words,
\

CTs
→
<

;
*

BG
Po–. BC; Qq=%. BC; Rr=}.
3.

º
2

4
4
4

Similarly the given par” divide AB into equal parts,

if

- BC;
so

BC, Qq=#. BC, and on.

Re-
•

Pp=}
p.

78, should now

be
7,

*...* Problem worked.

º
*

DEFINITION.
-
*\

wº w
G.

º
*
a

from the extre mities straight line AB perpendiculars

If

of
a

straight line PQ indefinite length,

of

AX, BY are drawn

beto
a

the orthogonal projection

to

of

then XY said AB on PQ.

is

-
A

2^T
LT
x

Q
Y
Q
Y
X

P
5

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

Theorem 22. *

In the A ABC, if Z is the middle point of

AB, and ZY is drawn parl to BC, we have to
prove that AY=YC.
Draw YX parl to AB, and then prove the
As ZAY, XYC congruent.]
W

The straight line which joins the

/
2. A
middle points of two sides of a triangle is
parallel to the third side.
[In the AABC, if Z, Y are the middle 2 Y V
points of AB, AC, we have to prove ZY f
parl to BC.
Produce ZY to V, making YV equal to
ZY, and join CV. Prove the A* AYZ, B C
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 identi
cally equal. {

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

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

opposite sides AD, BC: shew that BX and DY trisect the diagonal
§§
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, bisect 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 42 cm. and 5.8 cm., deduce the length
of OX, and verify your result by measurement.
Shew that OX= }(AP+ BQ) or #(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 centi
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.

OX and OY are two straight lines, and along OX five points

12.
Through these points
1,
2,
3,
4,

at

are marked equal distances.

5

parallels are drawn any direction Measure the lengths

in

meet OY.
to

these parallels: take their average, and compare

of of

with the length

it
the third parallel. Prove geometrically that the 3rd parallel the

is
mean of all five.
the corresponding for any odd number

1)
State theorem

of
(2n

+
parallels
so

drawn.

13. From the angular points

of

parallelogram perpendiculars
a

are drawn any straight line which outside the parallelogram

to

is

of :
drawn from one pair
of

shew that the sum the perpendiculars

opposite angles equal
of

those drawn from the other pair.

to
is

the sum
[Draw the diagonals, and from their point suppose
of

intersection
perpendicular drawn tº the given straight line.]
a

the perpendiculars
The sum drawn from any point
in
of

14. the
S

isosceles triangle
an

the perpen
to
of

the equal sides equal

to

base
to is

dicular drawn from either extremity

of

the base the opposite side.

[It follows that any point
of

is of

in

the sum the distances the base

an isosceles triangle from the equal sides
is,
of

constant, that the

same whatever point
in

the base taken.]

is

How would this property the given point were taken

be

modified
if

the base produced .*

in

*,
2

15. The sum drawn from any point within

of

the perpendiculars
an equilateral triangle
to

equal the perpendicular

to

the three sides

is

any
of

drawn from angular points the opposite side, and

to

one the
is

therefore constant.

16. Equal and parallel lines have equal projections

on

any other
straight line.
E
S.

H. G.
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

on
numbered (from 0) 1, 2, 3, We may now read

...
AB

9.
inches and tenths of an inch.
4
3
2
1
A

B
9
8
7
In

read hundredths, ten lines are taken any equal

to

at
order
intervals parallel AB; and perpendiculars are drawn through
to
2,
...
0,
1,

The primary (or inch) division corresponding 0A

on
to

the
tenth parallel now subdivided into ten equal parts; and
is

as

diagonal lines are drawn, the diagram,

in

joining the first point

on

the 10" parallel.

to to to

of

subdivision
2 1 0

y?
,,

the Second 33 3.9

y
,, ,,

the third 92 35 33
;

and so on.

The scale now complete, and its use

in

shewn the
is

is

following example.
Eacample. length
of

To take from the scale 2:47 inches.

a

Place one point AB, and extend them till

at
of

the dividers
in
(i)

the other point reaches We have now

in

the subdivided inch 0A.

4

2'4 inches in the dividers.

(ii) To get the remaining hundredths, move the right-hand point
7

up the perpendicular through till reaches the 7th parallel. Then

it
2

on

extend the dividers till the left point reaches the diagonal also the
4

7th 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 Hº Hº, #,

...
of
the
base, which &
inch.
is
1

Hence these lengths are respectively

4
3
2
1
'01, .02, '03,
...
of

inch.
1

Similarly, by means this scale, the length given

of
of

a
straight line may
be

an
to

of
measured the nearest hundredth
inch.

Again,
on

represent to
one inch-division the scale taken
is
if

on

feet, then 2:47 inches the scale will represent 24-7 feet.
10

the scale represents 100 links, then

on

And one inch-division

if

2.47 inches will represent 247 links. Thus diagonal scale

is
a

preparing plans enclosures, buildings, field

of
in

or
of

service
works, where necessary that every dimension
of

the actual
is
be it

represented by proportional length

on

object must
of

line
a

the plan.

NOTE.
§
be

The subdivision diagonal scale need not

of

decimal.
a

For instance we might construct diagonal scale read centimetres,

to
a

millimetres, and quarters millimetre;

of

which case we should take

in
a

four parallels the line AB.

*
to

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

68 GEOMETRY.

EXERCISES ON LINEAR MEASUREMENTS ,

1. Draw straight lines whose lengths are lºff inches, 272 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 cm. = 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 cm. in inches."

5. Draw lines 2-9 cm. 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
I 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
is,

be
what scale is the map drawn 2 That how many miles will
represented by inch How many kilometres by centimetre
1

; 7

0-3937 inch km. =# mile, nearly.]

[1

cm.
=

35

map
France drawn inch miles, the
to

to
of

11. On the scale

1
a

distance from Paris represented by 4-2 inches. Find the

in to

is

Calais
distance accurately miles, and approximately kilometres,
in

and
express the scale mile, nearly.]
in

[1

metric measure. km. =#

12. The distance from Exeter Plymouth 37% miles, and

to

is
to on be
88 on

appears certain map 24"; and the distance from Lincoln

to

to
a

be

York km., appears map Compare the

to
of is

and another cm.

7

these maps
in

scales miles the inch.

13, Draw diagonal scale, represent yard,

to

centimetres
2

1
a

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

Tesultsof 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.

}f
TROBLEM 1.

To bisect a given angle.

Let BAC be the given angle to be bisected.

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

Proof. Join PO, QC.

In the A*APO, AQO,

AP=AQ, being radii of a circle,

because PO =QO, 25 ,
equal circles,
and AO is common;
all

the triangles are equal respepts;

in
...

7.

Theor.
that the PAO the QAO
so

A.
A.

bisected by AO.
is,

that the BAC

Z.

is
as

PQ has been taken the radius the arcs drawn from the
of

NOTE.
and Q, and the intersection these arcs determines the
of

centres
P

Any radius, however, may PQ, provided

be

point O.
of

used instead
of

that great enough

to

secure the intersection the arcs.

it
is
 PROBLEMS ON LINES AND ANGLES. 71

PROBLEM 2.

To bisect a given straight line.

Sº
,'
Ś\
/ \V.
/ e
\,
*

Aé–Hºb
* *

*
O *

Let AB be the line to be bisected.

Construction. With centre A, and radius AB, draw two
arcs, 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,
| and PQ is common; *
the LBPQ. "Theor.
...

the APQ
A.

7.
=

Again the A*APO, BPO,

in

AP=BP,
because PO common,
is

and the APO the BPO

A.

*
OB;
4.

Theor.
...

AO
=
is,

at

that AB
O.

bisected
is
(i)

as

AB was taken the radius the arcs drawn from the

of

NoTES.

gº*
B,

but any radius may used provided that

be

great
is

centres and
it
A

the arcs which determine the points

of

secure the intersection

and Q.
'P

(ii) From the congruence the A* APO, BPO follows that the
of

it it

AOP=the BOP. As these are adjacent angles, follows that PQ

4.

bisects AB right angles.

at

*
72 GEOMETRY.

PROBLEM 3.

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

#
a given point in
it.

O
><

&A
*
\
\.
/
,’

\\
Af*

*
f&

\*
/ &*

*\
\\
X
A
P

B
Q
the straight line, and the point
be

at
Let AB which

in
it
X
be

perpendicular
to

drawn.
is
a

off

AB
Construction. With centre cut from any two
X

equal parts XP, XQ.

Q,

With centres and and radius PQ, draw two arcs cutting
P

at O.
*

Join XO.
Then XO perp.
to

AB.
is

Proof. Join OP, OG).

In

A*

the OXP, OXQ,

XP=XQ, by construction,
because OX common,
= is

QO, being radii equal circles;

of

and PO
Theor,
...

the Z.OXP the Z.OXQ.

7.
=

And these being adjacent angles, each right angle;

is
a
is,

that XO perp.
to

AB. *
is

the point
or

AB, one
of

'0's.
If

of

near one end other

is
x

be
on

the alternative constructions the next page should 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.
Join DC, and produce it to meet
the circumference of the circle at O.
Join XO. >

Then XO is perp. to AB.

A DSG D2 x B

Proof. Join CX.

Because CO = CX; . . the Z. CXO = the A. COX;
and because CD = CX; . . the Z-CXD = the 4 CDX.
the whole DXO the Z_XOD the A-XDO
== =
‘..

+
/

*
of

180°
}

90°.
to

XO perp. AB.
is
‘.

TROBLEM THIRD METHOD.

3.

Construction. With centre XK

X

and any radius, draw the arc CDE,

cutting AB
C.
at

C,

With centre and with the

arc, cutting
an

'same radius, draw

the first arc at
D.

With centre
D,

and with the

same radius, draw an arc, cut
C
A

ting the first arc

B
at
E.

Bisect the DXE by XO. Prob.

1.
is 4

to

Then XO perp. AB, *

60°;
be

be
to

CXD, DXE may proved

of

Proof. Each the

4
"

half DXE;
is of

and the DXO the

is
... 4

4
#

*.

the CXO 90°.

Z
is,

to

That XO perp. AB.

is
74 GEOMETRY.

}
PROBLEM 4.
, ſº ºvº is . . . .
To draw a straight line perpendicular to a given straight line
from a given eaternal point.
* F ºr ; , , ; } X

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 arcs 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 circle,
because PY = QY, for the same reason,
and XY is common ;
PXY QXY.
7.
...

the the Theor.

A.

A.
=

Again,
A'

PXO, QXO,
in

the
PX QX,
is =

because XO common,
QXO;
|

and the PXO the

= =
Z

Z
...

the A-XOP the A-XOQ.

4,

Theor.
And these being adjacent angles, each right angle,
is
a
is,

to

that XO perp. AB.

is
 PROBLEMS ON LINES AND ANGLES. 75

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. C

...,
With centre C, and radius CX, draw
a circle cutting AB at D and O.
*
Then XO is perp. to AB.
A DNJ’o B

For, as in Problem 3, Second Method, the LXOD is a right

angle.

PROBLEM 4. THIRD METHOD,

Construction. Take any two points

D and E in AB.
te X
.2%
With centre D, and radius DX, draw
an arc of a circle, on the side of AB
opposite to X. F-E-F-5
... /
...”.”

B
With centre E, and radius EX, draw
another arc cutting the former at Y.
`ss
`,
º
Join XY, cutting AB at O.
Then XO is perp. to AB. 2%
XDE, YDE equal
all(i)

Aº

Prove the
respects by Theorem
in

7,

so that the A-XDE the 4-YDE.

=

all

(ii) Hence prove the A'XDO, YDO equal respects

in
by

4,
so

Theorem that the adjacent DOX, DOY are equal.

/
is,

That perp.
to

XO AB.
is
76 GEOMETRY,

PROBLEM 5.

At a given point in a given Straight line to make an angle equal

to a given angle.

A Eë Fo Q G

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 Z. BAC. *

Construction. With centre A, and with any radius, draw

an arc 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 tho
former arc 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 {OQ = AD, for the same reason,
PQ = ED, by construction;
all

the triangles are equal respects;

in
...

so

that the POQ the LEAD. Theor,

7.
A.

=
 PROBLEMS ON LINES AND ANGLES.

PROBLEM 6.

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

straight line.
‘P O

X A |D TY
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 4. AOP equal to the 4 OAY and alternate
to it.
Then OP is parallel to XY.

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

makes the alternate 4 "POA, OAY equal ;
OP par' XY.
to
...

is

*** The constructions

3,

are not usually

of

4,

Problems and
6

followed practical applications. Parallels and perpendiculars

in

Set
by

may more quickly drawn

of
be

the aid Squares. (See LESSONs

IN

ExPERIMENTAL GEOMETRY, pp. 36, 42.)

78 GEOMETRY.

PROBLEM 7.

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 five equal parts.
Construction. From A draw AC, a straight line of unlimited
length, 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, q, r, S.
SS,

Then since the par" Pp, Qq, Rr, TB cut off five equal parts
from AT, they also cut off five equal parts from AB.
(Theorem 22.)

SECOND METHOD,

From any angle with

at

draw AC
A
on

AB, and mark off four equal parts

it

AP, PQ, QR, RS, any length.

of

*
draw BD par
on
to

From AC, and

B

mark off BS', S'R', R'Q', Q'P', each

it

on

q
A

equal the parts marked

to

AC.
f\

Join PP, QQ', RR', SS' meeting AB

Then AB divided into
in
p,
q,
r,

is
s.

w
five equal parts these points.
at
by

20

[Prove Theorems and 22.]

 PROBLEMS. 79

EXERCISES ON LINES AND ANGLEs.

(Graphical Eacercises.)

1. Construct (with ruler and compasses only) an angle of 60°.

By repeated bisection divide this angle into four equal parts.
By means of Exercise 1, trisect a right angle ; that

is,
2. divide

it
into three equal parts.
Bisect each part, and hence shew how

an
to
angle

of
trisect 45°.
[No construction known for exactly trisecting any angle.]
is

Draw line 6-7 cm. long, and divide into five equal parts.
3.

it
of a

the parts inches (to the nearest hundredth), and verify

in

Measure one
your work by calculation. 0-3937 inch.]
[1

cm. =

straight line 3'72" long, cut off one seventh. Measure

4.

From
a

part centinetres and the nearest millimetre, and verify your

in

the
work by calculation. .*

At point straight line AB draw XP perpendicular AB,

to
in
5.

in X
a

making XP 1.8" oblique PQ, 3.0" long,

an
length. From draw
P

meet AB Q. XQ.
in
to

Mcasure

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

a

straight line XY find point which equidistant from two

In
6.

is
a

given points
B.

and
A

When this impossible?

is

straight line XY find point which equidistant from two

In
7.

is
a

intersecting lines AB, AC.

When this impossible?
is

From given point straight line PQ, making with

8.

draw
P
a

given straight line AB

an

angle given magnitude.

of

From two given points straight

on

of
9.

and the same side

Q
P

line AB, draw two lincs which meet AB and make equal angles
in

with it.
From draw PH perp. AB, and produce PH P',
to

to

[Construction.
P

making HFºequal PH. Join PQ cutting AB Join

at

PK.
to

K.

Prove that PK, QK are the required lines.]

10. Through given point draw straight line such that the
P
a

from two points

be
to

perpendiculars drawn and may equal.

it

B
A

this always possible

Is

7
80 GEOMETRY.

THE CONSTRUCTION OF TRIANGLES.

PROBLEM 8.
/2)^2 / CP}2

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

C Ö
©
6
&
B (Z !C X

Let a, b, c 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 part BC equal to a.
a
With centre B, and radius c, draw an arc of a circle.
With centre C, and radius b, draw a second arc cutting the
first at A.
AB, AC. Join
Then ABC is the required triangle, for by construction the
sides BC, CA, AB are equal to a, b, c 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.
*
be
In

order that the construction may possible

(i)

NoTES.
it
is
be

necessary that any two the given sides should together greater
of

than the third side (Theorem ll);

for otherwise the arcs drawn from
the centres and would not cut.
C
B

would,
on
at

(ii) The arcs which cut continued, cut again the

if
A

on

BC. Thus the construction gives two triangles opposite

of

other side
sides of common base.
a
 THE CONSTRUCTION OF TRIANGLES. 81

ON THE CONSTRUCTION OF TRIANGLES.

It
has been seen (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 parts: or, other words,
in
its
To

construct triangle three independent data are required.

a

For example, we may construct a triangle

(b,
(i)

When and the included angle (A) are given.

c)

two sides
The method of construction in this case obvious.
is
(A,

and one side (a) are given.

B)

(ii) When two angles

Here, since
at

and are given, we

once know
C
B B
A

for A+ 180°.
to +

=
C

Hence we have only draw the base equal

26
at
a,

its ends make angles equal

B to

to

and
^c
for we know that the remaining
\
B

and L
C

\
;

Q.
be

angle must necessarily equal

to
A.

of no
B,
A,

(iii) the three angles are given (and side), the

If

C
is,

problem indeterminate, that the number solutions

is

unlimited.
is

For any base we make angles equal

at

to
of

the ends
C, if

the third angle equal

to

and
A.
is
B

This construction indeterminate, because the three data

is

are not independent, the third following necessarily from the

other two.

H. S.G.
F
-
82 GEOMETRY.

PROBLEM 9.

To construct a triangle having given two sides and an angle

opposite to one of them.

B c-c, x

Let b, c be the given sides and B the given angle.

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

the 4 XBY equal to the given A. B.
From BY cut off BA equal to c.
With centre A, and radius b, draw an arc 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 b is less than c but greater than the perp.
from A on BX.

EXERCISE.

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

following cases:
When greater than
(i)

is is is is

c.
b b

(ii) When equal

to to
c.

on

(iii) When equal the perpendicular from BX.

A
b

(iv) When less than this perpendicular.

b
 THE CONSTRUCTION OF TRIANGLES. 83

PROBLEM 10,

To construct a right-angled triangle having given the hypotenuse

and one side.

A O B

Let AB be the hypotenuse and P the given side.

Construction. Bisect AB at O; and with Čentre 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 4. OCA = the Z.OAC.

And because OB = OC;

OCB the OBC.
...

the
A.
=
4.
...

the whole ACB the OAC the Z.OBC

== =

z-

+
/

Theor. 16.
of

180°
}
*

90°.
 84 GEOMETRY.

ON THE CONSTRUCTION OF TRIANGLES.

{Graphical Eacercises.)

1. Draw a triangle whose sides are 7.5 cm., 6.2 cm., and 53 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.]

Draw a triangle ABC, having given a-3:00", b=2'50", c=275".

2.
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
your result with
the value of
É. to two places of decimals. Compare

the value of ;
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 B and C are 47° and 68° respectively. Draw
a plan (scale 1 cm. 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.

Y 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 cm. and, one side a = 5-6 cm. Measure the third side b ; and
find the value of Ncº-a%. Compare the two results.

* 7. Construct a triangle, having given the following parts: B=34°,

b=5-5 cm., c=8'5 cm. Shew that there are two solutions. Measure
the two values of a, and also of C, and shew that the latter are
supplementary.

* 8. In a triangle ABC, the angle A=50°, and b-6-5 cm. Illustrate

by figures the cases which arise in constructing the triangle, when
(i)

=7 cm. (ii) cm. (iii) cm. (iv) cm.

=

=
6

4
a

a
 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 B and 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 cm., and having an

altitude of 6-2 cm. Prove the two sides equal, and measure them to
the nearest millimetre.

X il. 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 6 cm. 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=73 cm., find a and b graphically; and
calculate the value of Na”--b”.

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

18. Construct a triangle ABC from the following data ;

a = 6-5 cm., b + c = 10 cm., and B = 60°.
Measure the lengths of b and c.

19. Construct a triangle ABC from the following data:

a =7 cm., c – b = 1 cm., and B = 55°.
Measure the lengths of b and c.
86 GEOMETRY.

THE CONSTRUCTION OF QUADRILATERALS.

It has been shewn that the shape and size of a triangle are

its
completely determined when the lengths of three sides are
given. quadrilateral, however,
A not completely determined

is
by

its
the lengths

of
four sides. From what follows will

it
to
appear that five independent data are required construct

a
quadrilateral.

PROBLEM 11.
To

quadrilateral, given the lengths

of
construct the four sides,
a

and one angle.

be

the given lengths the sides, and

of

Let
b,
a,

the
c,

A
d

angle between the sides equal

to

and
d.
a

off

Construction. Take any straight line AX, and cut from

AB equal
to
it

a.

to

Make the BAY equal the

Z.

Z.A.
to

From AY cut off AD equal

of d.
an

With
B D,

centre and radius draw arc circle.

c,

With
to
b,

centre and radius draw another arc cut the

C.

former at
Join DC, BC.
is by

Then ABCD the required quadrilateral; for construction

is

d,

to

the sides are equal equal

to

b,
a,

and the DAB the

c,

4.

given angle.
 CONSTRUCTION OF QUADRILATERALS. 87

PROBLEM 12.

To construct a parallelogram having given two adjacent sides and

the included angle.

>JC
.

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 4. BAD equal to the 4-A, 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,
because CB = AD,
and DB is common :
Theor,
7.
...

the CDB the ABD

Z.
=
4

and these are alternate angles,

to

DC par' AB,
is
‘.

Also DC AB;
=

DA and BC are also equal and parallel.

...

Theor. 20.
par".
...

ABCD
is
a
set

Construction (With Draw AB and AD

as

squares.)
2.

before; then with set squares through draw DC par' AB,

to
D

and through draw BC par

to

AD.
B

By construction, ABCD par” having the required parts.

is
a
88 GEOMETRY.

PROBLEM 13.

J
To construct a square on a given side.

: K. C

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 Z. BAD is a right angle, the figure is a rectangle,
its
all

Also, by construction sides are equal.

square.
...

ABCD
is
a

At
set

Construction (With draw AX perp.

to

squares.)
2.

AB, and cut off from AD equal

to

AB.
it

Through draw DC par' AB, and through

to

draw BC
D

par AD meeting DC
in
C.
to

by

Then, construction, ABCD rectangle. page 56.]

3,

[Def.
is
a

Also has the two adjacent sides AB, AD equal.

it

square.
is
it
‘..

a
 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 cm., 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 five 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. In a quadrilateral ABCD,
AB = 5-6 cm., BC = 2.5 cm., CD = 4.0 cm., and DA=3-3 cm.
Shew that the shape of the quadrilateral is not settled by these data.
Draw the quadrilateral when A=30°, (ii) A=60°. Why does the
(i)

construction fail when 100°

=
A

graphically the least value for which the con

of

Determine
A

struction fails.

quadrilateral, having given the lengths

to
7.

Shew how construct

a

one diagonal.
of

the four sides and What conditions must hold

of

be

among the data order that the problem may possible

in

Illustrate your method by constructing quadrilateral ABCD, when"

a

AB =30", BC =1.7", CD=2.5", DA=2'8",

(i)

and the diagonal

BD=2'6". Measure AC.
(ii) AB =3.6 cm., BC=7-7 cm., CD=6-8 cm., DA=5.1 cm., and the
diagonal AC=8-5 cm. Measure the angles
D.
at

and
B
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.

P
Earample 1. Suppose the point P to move so
that its distance from a fixed point O is constant
(say 1 centinetre).
Then the locus of P is evidently the circum
ference of a circle whose centre is O and radius
1 cm.

Eacample 2. Suppose the point P

moves at a constant distance (say 1 cm.)
from a fixed straight line AB.
Then the locus of P is one or other of
B.
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.
 LOCI. 91

PROBLEM 14.

its
To find the locus of a point P which moves so that distances
from two fixed points and are always equal

to
one another.

B
A

which PA PB; B
all

Here the point moves through positions

in

=
P

the moving point O'the middle point

at
one position
of

is
.

of AB.
the moving point
be

Suppose any other position

= to

of
P

:
is,

that let PA PB.

Join OP.
A'

Then POA, POB,

in

the
PO common,
= = is

because OA OB,
and PA PB, by hypothesis;
...

the POA the POB. Theor.

=

7.
A

perpendicular
to

Hence PO AB.
is
is,

every point equidistant from

on

That which and lies

B
P

is

the straight line bisecting right angles.

at

AB

proved that every point the perpen

be

on

Likewise may
it

dicular through equidistant from and

B.
is
O

This line therefore the required locus.

is
92 GEOMETRY.

PROBLEM 15.

its
perpen
To find the locus of a point P which moves so that
straight lines AB, CD are equal

to
dicular distances from two given
one another.

D
A

N
º

M
C.

5
perp. PN.
any point such that thé perp. PM the

=
be

Let
P

AB,
of
O,

CD.
to

Join the intersection

P

PMO, PNO,
A'
in

Then the

the PMO, PNO aré right angles,

4

common,
because {the hypotenuse OP
is

PM o ne side PN
=

and one side

;
all

equal respects; Theor. 18.

in

the triangles are

...

so that the POM the PON.

=
Z.

be

BOD, must on the bisector

it

lies within the

Z.

Hence,
if
; P

that angle
of

of
be

within the Z.AOD, must on the bisector

it

and,
is
if
P

that angle.
pair
of

the required locus the lines which bisect

is

follows that
It

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

all
point must lie. Hence

to
points which are common these
all
is,
two loci, that the points the loci, will

of

of
intersection
satisfy both the given conditions.

ExAMPLE To find point equidistant from three given points

1.

a
C,
(i) B,
A,

which are not the same straight line.

in

The locus points equidistant from

of

the straight line PQ, which

is

and
B
A

right angles.
at

bisects AB
(ii) Similarly, the locus points equi
of

distant from the straight line

is

and
C
B

right angles.
at

RS which bisects BC
Hence the point common to PQ and RS
must satisfy both conditions: that
to
is

the point

B
say, PQ and
of

B,of

intersection
A
§§
X

|
A,

equidistant from
be

will
C.

and
Q

ExAMPLE To construct triangle, having given the base, the

2.

altitude, and the length

of

the median which bisects the base.

be

TNF
Let AB the given base, and and
P

the lengths
of

the altitude and median

Q

E2
C

respectively. D
Then the triangle known its verteac
if
is

known.
is

straight line CD parallel

(i)

to

Draw
a a

AB, and P:
at

equal
to

distance from
it

O
A

B
on
lie

then the required vertea, must CD.

(ii) Again, from the middle point
tº Q P
of
O

Q,
as

centre, with radius equal

to
B

describe circle:
a

*
º
on
to lie

then the required vertea must this circle.

Hence any points which are common CD and the circle, satisfy
both the given conditions: that say, CD intersect the circle
in
to

if
is
E,

the points might

F,

be
of
of

to of

each intersection the vertex the

required triangle.
be

This supposes the length

of

the median
Q

greater than the altitude.

Inso

may happen that the data the problem are

It

to
of

related one
another that the resulting loci do not intersect. this case the
problem impossible.
is
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"..."
1. Find the locus of a point which moves so that its distance ,
(measured radially) from the circumference given circle trº

of

is
constant.

a
\\ ***

º
*.

#
*

/
point moves along straight line RQ find the position in
2.

it A

P
S

;
equidistant from two given points and

B.
which
is

A
are two fixed points within find points

on
3.

and circle the

>

B
A

:
equidistant from How many such points are
B.

circumference and
A

there
2

point moves along straight line RQ find the position

in
4.

it A

which equidistant from two given straight lines AB and CD.

is

by

are two fixed points cm. apart. Find the method,

5.

and
A

A,

loci two points which are cm. distant from

B.
of

and cm. from

4

AB and CD are two given straight lines. Find points

6.

cm.
3

distant from AB, and cm. from CD. How many solutions are there?
4

straight rod given length slides between two straight

of
A
7.

right angles
at

to

rulers placed one another.

its middle point and shew that this locus
of

Plot the locus the

is
;

fourth part
of

of

the circumference circle. [See Problem 10.]

a

right-angled triangles are described.

as

given base hypotenuse

8.

On
a

Find the locus of their vertices.

point, and the point fixed straight

on
A
9.

is

fixed moves
X
S

line BC.
the middle point AX;
P,

Plot the locus and prove the locus

to
of

of

straight line parallel BC.

to

be
a

fixed point, and the point

on
is

10. moves the circumference

X
A
N

given circle.
of
a

the middle point AX;

P,

and prove that this

of

Plot the locus

of

3,
p.

[See Ex.
is

locus circle. 64.]

a
 EXAMPLES ON LOCI. 95

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

from A to any straight line passing through B. BX revolve about B, If
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
constant (=6 cm., say):
(i)
PM PN
+

is is
(ii) PM PN constant (=3 cm., say).
–
And each case give proof the result you arrive
in

of

at
theoretical
a

experimentally.

13. Two straight lines OX, OY intersect right angles O;

at

at
and
from movable point perpendiculars PM, PN are drawn OX, OY.

to
P
a

Plot (without proof) the locus P,

of
when
PM =2 PN:
(i)

(ii) PM =3 PN.
point which given point,
at

14. Find given distance from

is
a

a
and equidistant from two given parallel straight lines.
is

When does this problem admit two solutions, when one only,

of
of

impossible?
is

and when
it

fixed point inches distant from given straight line

is

15.
S

2
a

a
2;

MX. Find two points which are

S,
inches distant from and also
2# inches distant from MX.

points equidistant from given point

of

16. Find series and S

a

given straight line MX. Draw curve freehand passing through all
a

the points
se

found.

given base construct triangle given altitude, having

of

17. On
on a

its vertex given straight line.

a

18. Find point equidistant from the three sides triangle.

of
a

19. Two straight lines OX, OY cut right angles; and

Q
at

and
R

are points OX and OY respectively.

in

Plot the locus

of

the middle
point QR, when
of

OQ --OR constant.
(i)

(ii) OQ OR= constant.

–

are two fixed points. points

of
S'

20. and Find series such

S

P
a

that
SP+S'P=constant (say 3-5 inches).
(i)

(ii) SP S'P= constant (say l'É inch).

–
In

each case draw curve freehand passing through all the points
a

So found.
96 GEOMETRY.

ON THE CONCURRENCE OF STRAIGHT LINES IN A TRIANGLE.

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

middle points are concurrent. -
Let ABC be a A, and X, Y, Z the middle
points of its sides.
From Z and Y draw perps. to AB, AC,
meeting at O. Join OX.

It is required to prove that OX is perp.

to BC.
Join OA, OB, OC. B X C
Proof. , Because YO bisects AC at right angles,
points
tºwn
of

from and

C
g A
‘..

is

the locus
it

;
OA OC.
...

=
Again, because ZO bisects AB right angles,
at

points equidistant from B;

of
...

is

the locus and

it

A
OA OB.
...

Hence OB OC:
=

points equidistant from

on

is,of
...

the locus and

O

C
B
is

;
that OX perp, BC.
to
is

from the mid-points

O.
at
Hence the perpendiculars
of

the sides meet

•

Q.E. D.
II.

the angles triangle are concurrent.

of

of

The bisectors
a

Let ABC be A. Bisect the A." ABC,

a

BCA by straight lines which meet

O.
at

Join AO.
required prove that AO bisects the
It

to
is

BAC.
A.

From OP, OQ, OR perp.

to

draw the
O

sides of the A.
B

C
P

Proof. Because BO bisects the ABC,

A.

points
gºint from BA and BC;
of
...

is

the locus
it

OP= OR
of •.

Similarly CO points equidistant from BC and CA;

is

the locus
OP=OQ.
...

Hence OR OQ.
=

points equidistant from AB and

on

AC:
is of
...

is

the locus
O

that is, OA the A-BAC,

of

the bisector
O.

the angles meet Q.E.D.

at
of

Hence the bisectors

 THE CONCURRENCE OF LINES IN A TRIANGLE. 97

III. The medians of a triangle are concurrent.

Let ABC be a A.
Let BY and CZ be two of its medians, and let
them intersect at O.
Join AO,
and produce it to meet BC in X.
It is required to shew that AX is the remaining
median of the A.
Through C draw CK parallel to BY;
produce AX to meet CK at K.
Join BK.
Proof. In the A AKC,
oecause Y is the middle point of AC, and YO is parallel to CK,
the middle point AK. Theor. 22.

of
...

is
O

Again the AABK,

in

are the middle points AB, AK,

of
since and
... O
Z

ZO parallel BK,
to
is
is,

that OC parallel BK,

to
is

the figure BKCO parin.

...

is
a

But the diagonals

gº. parin bisect one another;
of
a

point
...

is
X

of

BC.
That is; 'AX A.
at of

median the
is
a

the point
O.

Hence the three medians meet Q.E.D.

The point
of

intersection
of

is

DEFINITION. the medians called the

the triangle.
of

centroid
**** --~~
,
*

CoRollary.
in of

triangle cut one another

at

The three medians

a
a

point trisection, the greater segment each being towards the angular
of

potºvt.
\

For the above figure has been proved that

in

it
A

OK;
of

also that OX half

is

OX half of OA
...

is is

of :

that is, OX one third AX.

Similarly OY one third SY,
of
is

and OZ one third of CZ.

is

Q.E.D.

By means this Corollary any triangle

be

may that
in
of

shewn
it

the shorter median bisects the greater side.

be
It

NoTE. will proved hereafter that the perpendiculars drawn

,
,

triangle the opposite sides are concurrent.

of

to

from the vertices

a
S.

H. G. G
98 GEOMETRY.

MISCELLANEOUS PROBLEMS.

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

, 1. A is a 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 by OA and OB, and bisected at P.
terminated
2
2
Z 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?
--"*:A
1*
tº
...

* e
*

*
.

triangle ABC inscribe

In

rhombus having one its angles

of
6.

a
/

coinciding with the angle

A.

Use the properties an equilateral triangle trisect given

to
of
7.

a
straight line.

Triangles.)
of

(Construction

Construct triangle, having given

8.

The middle points

(i)

of

the three sides.

(ii) The lengths
of

of

two sides and the median which bisects the

third side.
(iii) The lengths one side and the medians which bisect
of

the other
two sides.
(iv) The lengths
of

the three medians.

 AREAS. 99

PART II.
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.
NOTE. It is clear that parallelograms or triangles which are between
the same parallels have the same altitude.
G A D H
For let AP and DQ be the alti-
tudes of the A• ABC, DEF, which | |
are between the same parallels BF, . .
! |

Then the fig. APQD is evidently . !

! !
a rectangle ;
AP DQ.
...

Q
B

F
E
P

The area figure

of

of

the amount surface contained

is
3.

within its bounding lines.

square inch
A

of

the area
4.

is

in a

S
on

Square drawn side one inch quare

a

inch
:

length.

Similarly square centimetre

of

the area Sq.

is
5.

on a

square drawn length. CIII.

in

side one centimetre

a

The terms square yard, square foot, square metre are

be
to

understood
.

in the same sense.

on

square
of

Thus the unit

of

area the area side

6.

is

a
a

unit length,
of
100 GEOMETRY.

THEOREM 23.

Area of a rectangle. If
the number of units in the length of a

its
rectangle is multiplied
by the number of units in breadth, the
product gives the number

of

in
square units the area.

C
*
:
- esº *-*
** wº tº-e
rºº *-*
4sº

wº
* - tº wº* - * * *sºtº
i
;
*A i

AB
Let ABCD represent rectangle whose length feet,

is
5
is a

and whose breadth AD feet.

4

Divide AB into equal parts, and BC into equal parts, and

5

4
through the points division each line draw parallels

to
of

of

the other.

The rectangle ABCD now divided into compartments

is

which represents one square foot.

of

cach
Now there are rows, each containing squares,
4

the rectangle contains square feet.

...

×
5

Similarly, the length linear units, and the breadth =b

=
if

linear units
ab

of

the rectangle contains units area.

square linear units,
of

And each side

=
if

a” a

of

the Square contains units area.

thus abridged:
be

These statements may

.
.
.

breadth ........... (i),

of

rectangle length
= =

the area
x
of a

................. (ii).
...

the area square (side)*

a

Q.E.D

COROLLARIES,
(i)

Rectangles which have equal lengths and

equal breadths have equal areas.
(ii) Itectangles which have equal areas and equal lengths have
also equal breadths.
 AREA OF A TRIANGLE. 10]
*_* *
tº
*: * * * *.
NOTATION.

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

its
these adjacent sides fix size and shape.
rectangle whose adjacent sides are AB, AD denoted by
A

is
rect. AB, AD, simply AB AD.
or

sq.
denoted by
on

or
square drawn AB,

on
A

the side AB

is
AB”.

EXERCISES.

Length and Area.)

of

(On Tables

figure shew why

to
1.

Draw
a

sq. yard=3° sq.feet.

(i)
1

(ii) sq. foot 12” sq. inches.

= =
I

(iii) sq., cm. 10° sq. mm.

l

straight line
on
to

Draw figure shew that the square four

2.

is
a

a
on

times the square half the line.

on

Use squared paper shew that the square

to
3.

1"= 10° times the

square on 0.1".

represents miles, what does an area square inches

of
1"
If
4.

represent
2

EXTENSION OF THEOREM 23.

The proof
23

here given supposes that the length and

of

Theorem
by

the given rectangle are expressed but the

of

breadth whole numbers

;

formula holds good when the length and breadth are fractional.
be

This may illustrated thus:

Suppose the length and breadth are 32 cm. and 2.4 cm.; we shall
shew that the area (3.2 2-4) sq. cm.
is

For
32

length 3-2 cm. mm.

=

breadth=2'4 cm. =24 mm.

32 24
×

24) sq. mm. Sq. cm.

...

area (32
=

10?
=(3.2 x2'4) 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.
1. 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 =22", b=1.5". 6. a = l'6", b=2.1".

Calculate the areas of the rectangles in which

7. a = 18 metres, b = 11 metres. 8. a = 7 ft., b=72 in.

9. a = 2.5 km., b = 4 metres. 10. a =# mile, b = 1 inch.

11. The area of a rectangle is 30 sq. cm., and its length is 6 cm.
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 l'5". Draw the rectangle on squared paper; and verify your
work by counting the squares.
rectangle without altering
(i)

When you treble the length

of

13.
a

its breadth, how many times do you multiply the area

7

do
(ii) When you treble both length and breadth, how many times
you multiply the area
7

Draw figure illustrate your answers; and state general rule.

to

a
a

plan rectangular garden the length

In

and breadth
of

14.
a
a

are 3-6" and 2'5", one inch standing for

10

yards. Find the area

of

the
garden.
increased by 300 sq. yds., the breadth remaining the
If

is

the area
same, what will the new length be? And how many inches will repre
your plan
on

sent
it

Find the area rectangular which plan

of

enclosure
of

15.
a
a

20 metres) measures 6-5 cm. by 4-5 cm.

to

(scale cm.
1

rectangle 1440 sq. yds. plan the sides

of

in
If

16. The area

is

a
a

the plan
on

the rectangle are 32 cm. and 4.5 cm., what scale

of

is

drawn
2

ft.

rectangular 52000 sq. plan

of

17. The area field

is of

On
is

a
a

this, drawn 100 ft., the length hat

1"
to

is

3'25".
to

of

the scale
the breadth
?
 EXERCISES ON RECTANGLES. 103

Calculate the areas of the enclosures of which plans are given below.
All the angles are right angles, and the dimensions are marked in feet.

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Calculate the areas rep resented by the shaded parts of the following
plans. The dimensions are mar ked in feet.

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104 GEOMETRY.

THEOREM 24. [Euclid I. 35.]

the
Parallelograms on the same base and between same parallels

in
are equal area.

Dr

F
E
B

C
Let the par” ABCD, EBCF be

on
the same base BC, and
between the same par" BC, AF.

required
It

prove that
to
is

par” ABCD par” EBCF

in
=

the the area.

Proof.
In

A*

the FDC, EAB,

DC the opp. side AB; Theor. 21.
= = = =

because {the ext. FDC the int. opp. EAB; Theor. 14.
A Z.

Uthe int. DFC the ext. Z.AEB;

AFDC AEAB.
...

the the Theor. 17.

Now, from the whole fig. ABCF the AFDC taken, the
if

is

remainder the par" ABCD.

is

And from the whole fig. ABCF the EAB taken, the
A
if

is

remainder the parº EBCF.

is

are equal;
is, ...

these remainders

that the parº ABCD the parº EBCF. Q.E.D.

=

EXERCISE.

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

In

not overlap; (ii) the ends

do

which
(i)

these sides and coincide.

E

Go through the proof with these diagrams, and ascertain applies

it
if

them without change.

to
 AREAS. 105

THE AREA OF A PARALLELOGRAM.

F D E C
Let ABCD be a parallelogram,
and ABEF the rectangle on the
same base AB and of the same
altitude BE. Then by Theorem 24,
A. B
area of par” ABCD = area of rect. ABEF
= AB X BE
=base x altitude.

COROLLARY. Since the area of a parallelogram depends

only on its base and altitude, it follows that

are
Parallelograms on equal bases and of equal altitudes equal
270Qſì'60.

EXERCISES.

(Numerical and Graphical.)

Find the area parallelograms which

in
of
1.

the base=5-5 cm., and the height=4 cm.

(i)

(ii) the base=2'4", and the height=1.5".

Draw parallelogram ABCD having given AB=2}", AD=1#",

2.

AB,
on

and the 4-A 65°. Draw and measure the perpendicular from
=

2 D

and hence calculate the approximate area. Why approacimate

Again calculate the area from the length AD and the perpendicular
of

Obtain the average

of

the two results.

B.

on from
it

30

Two adjacent sides parallelogram are metres and 25 metres,

is of
3.

and the
included angle 50°. Draw plan, cm. representing
1
a

metres; and by measuring each altitude, make two independent

5

Give the average result.

of

aalculations the area.

parallelogram 4.2 sq. in., and the base

of

The area ABCD

4.

is
a

AB Find the height. AD=2", draw the parallelogram.

If
is

2-8".

2", and its area 3-86 sq. in.

of
5.

is

Each side rhombus Calculate

is
a

of ,

an altitude. Hence draw the rhombus, and measure one its acute
angles.
106 GEOMETRY.

THEOREM 25.

the
The Area of a Triangle. The area of a triangle is half area
the same base and having the same altitude.
of

on
the rectangle

E
A
D

** *-*.

E
sºme

wº
*
B

C
B
Fig. Fig.
1.

2.
be

on
Let ABC triangle, and BDEC rectangle the same
a

a
base BC and with the same altitude AF.

required
It

prove that the AABC half the rectangle BDEC.

to

is
is

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

to

of
is

rectangle.
is
a

Because the diagonal AB bisects the rectangle DF,

the AABF half the rectangle DF.
...

is

Similarly, the AAFC half the rectangle FE.

is

adding these results Fig. and taking the difference

in

in
1,
"..

Fig.
2,

*
the AABC half the rectangle BDEC.
is

Q.E.D.

triangle half any parallelogram

on
A

COROLLARY.

º
is

the same
base and between the same parallels.
A
E

For the AABC half the rect. BCED.

is

&

And the rect. BCED=any par" BCHG

on the same base and between the same
par”.
the AABC half the par" BCHG.
...

is
 AREAS. 107

THE AREA OF A TRIANGLE,

If BC and AF respectively contain a units and p units of

length, the rectangle BDEC contains ap units of area.
the AABC =#ap units

of

of
...

the area area.

of be
This result may stated thus:
Triangle

}.
Area altitude.

=
base

x
a

EXERCISES ON THE AREA OF TRIANGLE.

A
(Numerical and Graphical.)
the triangles
of

Calculate the areas which

in
1.

the base =24 ft., the height=15

ft.
(i)

(ii) the base=4'8", the height=3:5".

(iii) the base 160 metres, the height=125 metres.
=

Draw triangles from the following data.

In
each case draw and
2.

base: hence cal

as
measure the altitude with reference given side
to
a

culate the approximate area.

8.4 cm., 6.8 cm., c=4.0 cm.
(i)

= =

= =
c b
b a

(ii) 5.0 cm., 6.8 cm., A=65°.

(iii) =6.5 cm., B=52°, C=76°.
a

triangle right-angled
at

ABC shew that its area =#BC CA.

3.

C
6 is
a

x
;

Given cm., b=5 cm., calculate the area.

=
a

Draw the triangle and measure the hypotenuse draw and measure
c
;

the perpendicular from the hypotenuse; hence calculate

on
C

the
-- approximate ->
<

area,
your approximate
as

Note the érror result, and express pers

in

it

centage
of

the true value.

Repeat process the last question for right-angled
of
4.

the whole
a

triangle ABC, which =2'8" and =4'5"; being the right angle
in

C
b
a

as before.
triangle, given
In
5.

sq. in., base in, calculate the altitude.

80

ft.

Area
(i)

= =

=
1

(ii) Area
Construct triangle ABC, º;
10.4 sq. cm., altitude=16 cm.; calculate the base.

given 3.0", b=2-8",

6.

c=2'6".
=
a
a

Draw and measure the perpendicular from on BC; hence calculate

A
,

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
(ºff'00. D A E 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 AABC = 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 AABC is half the rect. BCED; Theor. 25.
also the A GBC is half the rect. BCED;
the AABC
...

the GBC. A
=

Q.E.D.

Similarly, triangles of
on

equal bases and equal altitudes are

equal
in

area.

*ś
*.
*,
* ,
ºf

wº ***
27.

THEOREM [Euclid 39.]

I.

the
If

on

two triangles are equal area, and stand

in

same base
it,

they are between the same parallels.

of
on

and the same side

ABC, GBC, standing A

on
A*

Let the
.
be

the same base BC, equal

in

area
;

! :

and let AF and GH be their altitudes.

It

required prove AG and

to
is

that
i

H :

BC are par’.
B

C
F

Proof. The AABC half the rectangle contained by BC

is

and AF
:

and the AGBC half the by BC

is

rectangle contained
and GH;
the rect. BC, GH;
the rect. BC, AF
==

GH..*. AF Theor. 23, Cor.

2.

Also AF and GH are par';

hence AG and FH, that BC, are par'. Q.E.D.
is
 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
AXBC YBC;

(i)

AA A
the the

= = =
(ii) the BXY CXY;

AA
the
(iii) the
K, ABY the ACX.
BY CX
If

shew that
at

and cut
(iv) the BKX=the CKY.
A

A
triangle divides
of
that into two parts
2.

Shew median

of
it
a

a
equal area.
How would you divide triangle into three equal parts by straight
a

lines drawn from its vertex?

Prove that parallelogram divided by its diagonals into four

3.

is
a

triangles equal area.

of

triangle whose base BC

4.

X.
ABC

If
at
is

is

bisected

is
any

Y
a

point the median AX, shew that

in

*3.
ABY the AACY in area.
A

the
=

ABCD parallelogram, and BP, DQ are thiºperpendiculars

5.

is
a
on

from and the diagonal AC.

B

A
>*-*.*.*.*
.

Shew that BP= DQ. *

any point AC,
or

Hence AC produced,
in
is
X
if

ADX the AABX;

(i)

prove
AA

= =

the
(ii) the CDX CBX.
A

the
|

Prove by means
of of

26 and 27 that the straight line

6.

Theorems
joining the middle points parallel
of

triangle the third

to

two sides
is
a

side.

The straight line which joins the middle points the oblique
of
7.

parallel the parallel sides.

to
of

trapezium
of

sides
is

each
a

ABCD parallelogram, and X, are the middle points

of
8.

is

the
or Y
a

sides AD, BC; any point XY, XY produced, shew that the
in
is
if
Z

triangle AZB one quarter

of

the parallelogram ABCD.

is

X,

ABCD parallelogram, any points

If

DC and AD
in
9.

and
is

Y
a

respectively: shew that the triangles AXB, BYC are equal

in

area.

10. ABCD parallelogram, any point within

is

is is

and shew
it
P
a

that the sum the triangles PAB, PCD equal half the parallelo
of

to

gram.
 110 GEOMETRY.

EXERCISES ON THE AREA OF A TRIANGLE.

(Numerical and Graphical.)

X 1. The sides of a triangular field are 370 yds., 200 yds., and 190
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. Y.”

X 2. Two sides of a triangular erſäläsäre 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. In a triangle ABC, given that the area=6-6 sq. cm., and the base
BC=5-5 cm., 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 cm.

and 5 cm. 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, increasińg
B by increments of 30°. Find the area in each case and tabulate the
results. Y&

(Theoretical.)

> 6. If two triangles have two sides of one respectively equal to two
sides of the other, and the angles contained by those 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 a quadrilateral 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.
y 10. Two triangles of equal area stand on the same base but on
opposite sides of it: shew that the straight line joining their vertices a cº
is bisected by the base, or by the base produced.
 s
THE AREA OF A TRIANGLE. 111

e. ". ºJ
[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.

ExAMPLE. Find the area of a triangle whose sides measure 21 m.,

17 m., and 10 m.
Let ABC represent the given
triangle.
Draw AD perp. to BC, and
denote AD by p.
We shall first find the length 1O 17
A
of BD.
Let BD = ac metres; then DC
=21 –a metres.
B 32 21 3:

C
D
From the right-angled AADB,

-
we have by Theorem 29
AD2=AB2 – BD2–102 —a2.
And from the right-angled AADC,
AD?=AC” – DC”=17? – (21–2)”;
2–172-(21–2)?
...

102
— –
a a

Of 100 4–289 441 422 2:”;

+
–

–
ac
-6.

whence

Again, AD?= AB2 BD”;

–

OT p?s= 10%–6%=64;
8.
...

=
p

triangle
of

Now Area
(# }.

base altitude
= =

x x
21

sq. m. =84 sq. m.

8)
x

EXERCISES.

Find by the above method the triangles,

of

the area whose sides

are as follows:
20 ft., ft.,
13

13 13 11

yds., yds.,
15

14

13

ft. yds.
1.

4. 2.

11

21 m., 20 m., 30 cm., 25 cm., cm.

5. 3.

m.
37 ft., 30 ft.,
51

37

ft. m., m., 20 m.

6.

the given sides are

ºgº a,

units length, prove

If

+ in
7.

and
b

c2
c2

*-a-(*.*
bº

b2Y
2-tº-º:
a2+
2

2
–
(i)*

(ii)
;

(iii) A=4N(a+b+c)(-a-F5-Fc)(a-b-Fic)(a+b+c).
 112 GEOMETRY.

THE AREA OF QUADRILATERALS.

THEOREM 28,

(i)
To find the area of trapezium.

a
(ii) any quadrilateral.
trapezium, having
be
(i)

Let ABCD
a

C
the sides AB, CD parallel. Join BD,
and from and draw perpendiculars
C

|
CF, DE
to

AB.

!
Let the parallel sides AB, CD measure

l
units length, and let the

B
F
and
of

E
b
a

height CF contain units.

of h

Then the area ABCD ABD DBC

I A

A
=

+
DE
l
+jpc.

CF
=3AB. h
I

I
is,

That
the

=}

trapezium
of

the parallel
of

area height (the Sum sides).

a

any quadrilateral.
be

(ii) Let ABCD

Draw diagonal AC and from
to B
a

and draw perpendiculars BX, DY,

D

*
AC.
*

AC

length, and
of

units
If

contains
#
d

BX, DY and units respectively,

p

the quad'ABCD
of

the area ABC ADC

A

A
=

1
l

Dy

=5Ac. Bx+3AC.
1

=;lp
I

º
q).

d(p
dº
=
+

+
2

2
to

That say,
is

quadrilateral=
of

diagonal
of

the area (sum

;

offsets).
a

x
 EXERCISES ON QUADRILATERALS. 113

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 l'5".

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

offsets from it to B and D are ll
feet and 9 feet. Find the area.

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

measures 8-2 cm., 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
2 y- dimensions are to be in centimetres.
Make any necessary measurements of your
figure, and calculate its area.

A 7.7 B

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

are the parallel sides. AB = 4".; AD = BC =2"; the Z.A= the A. B=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 cm., and AD = BC = 5 cm.
Make any necessary
\, measurement, and calculate the area.

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

that, if the diagonals are at right angles,
area =# (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 RECTILlNEAL FIGURE.

1“METHOD. 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.
Eacample. The measurements re
to find the area of the figure
Äää
BCDE are AC, AD, and the offsets BX,
DY, EZ.

2"|METHOD. The area of a rectilineal figure is also found

by taking a base-line (AD in the diagram below) and offsets
it.

from These divide the figure into right-angled triangles

be
and right-angled trapeziums, whose areas may found after
measuring the offsets and the various of the base-line.
sections
Eacample. ABCDEF the plan
of

Find the area the enclosure from

and measurements tabulated below.
YARDs,
AD 56
= = = = =

VC 12 AV 50
=

||

AZ 40 ZE 18
=
||

YB 20 AY 18
=

AX 10 XF 15
=
|

The measurements are made from

the points
to

alongthe base line

A

from which the offsets spring.

Sq, yas. Sq. yds.

Here
6 16 18 10

AXF =#. AX XF
AAA A

=# 15= 75
x x x x
x

x x x

AYB =#. AYx YB =# 20= 180

18

DZE=#. DZx ZE =} 144

=

DVx VC
}.

DVC= =} 12= 36
x

trap" XFEZ=}. XZx (XF-ZE)=#x30 33= 495

x

trap" YBCV =#. YV (YB+ VC)=#x32X 32= 512

×
...,

by addition, the fig. ABCDEF= 1442 sq. yds.

 EXERCISES ON RECTILINEAL FIGURES. 115

EXERCISES.

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

§ D
*

D
(ii)
(i)

ii
E

| !
C

C
!
:
!
X
A

B
A

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

= =

= =

=
cm.
=

6
5

Lengths offsets figured EY CZ cm.

1
of

diagram. DX=52 cm.

in

Draw full size the figures whose plans and dimensions are given
2.

in

below and calculate the area each case.

;

D
tº
(i) .,

e
E

(ii)
| |

12.5"
| l l |
X

Y.
A

The fig. equilateral AX=1}", XY=2#",

is

:
be

YB=1}".
to

each side 2%".

the figure ABCDEF from the following measure

of

Find the area

3.

ments and draw plan which cm. represents 20 metres.

in

1
a

THE PLAN.
METRES.
C

to
C

180
B

80 to 150
D

40 to 120 50 to
B
F E
to

60 50
F

From
A
116 GEOMETRY.

EXERCISES ON QUADRILATERALS,
*
(Theoretical.)
\
1. ABCD is a rectangle, and PQRS the figure formed by joining in
order the middle points of the sides.
Prove that PQRS rhombus;
(i)

is
of a
(ii) that the area PQRS half that ABCD.

of
is

its
of
Hence shew that the area half the product

of
rhombus

is
a
diagonals.
any quadrilateral whose diagonals cut right angles

at
of
Is

this true

2
Illustrate your answer by diagram.
a
Prove that parallelogram bisected by any straight line which
2.

is
a

passes through the middle point

of
its diagonals.

of
one

be
Hence shew how parallelogram ABCD may bisected by

a
a

straight line drawn

through given point P;
(i)

the side AB;

to
(ii) perpendicular
(iii) parallel given line Q.R.
to
a

ABCD, AB parallel DC; and

In

the trapezium
3.

to
is

is
the

X
\

middle point Through draw PQ parallel

of

BC. AD meet AB
to

to
X

and DC produced and Q. Then prove

at
P

trapezium ABCD=parm APQD.

(i)

(ii) trapezium ABCD=twice the AAXD.

(Graphical.)

The diagonals quadrilateral ABCD cut right angles, and

of
4.

at
a
\

measure 3.0" and 2'2" respectively. Find the area.

Shew by figure that the area the same wherever the diagonals
is
a

they are right angles.

at

cut,
so

as

long

ABCD, AB 8-0 cm., AD=32 cm., and the

In

the parallelogram
5.

=
\

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.
of

parallelogram 2'5", and its diagonals are 3'4"

6.

One side
is
a

and 2'4". Construct the parallclogram and, after making any neces
;

sary measurement, calculate the area.

ABCD parallelogram fixed base AB and

7.

is

on
of

constant
a

area. Find the locus its diagonals.

of

of

the intersection
 EXPERIMIENTAL, EXERCISES. 117

EXERCISES LEADING TO THEOREM 29.

In the adjoining diagram, ABC is a triangle A

right-angled
three sides.
at C ; and squares are drawn on the
Let us compare the area of the square
on the hypotenuse AB with the sum of the squares
T C B
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.

4”,
and ............... the square on BC =

16
or
sq. cm.

the squares on AC, BC 25 sq. cm.

of
‘..

the sum =
Now measure AB; hence calculate the area AB, and

on
the square of
compare the result with the sum already obtained.

Repeat the process the last exercise, making AC= 1.0", and
of
2.

=2'4”.

15, b=8, c=17, shew arithmetically that cº-a2+b^.

If
3.

=
a

triangle ABC, whose sides

17on

squared paper b,
a,
Now draw and
; a

c
8,

are 15, units length and measure the angle ACB.

of

and

Take any triangle ABC, right

4.

on AC,
at

angled and draw squares

on C
;

CB, and the hypotenuse AB.

Through the mid-point the square
of

the dia
of

on CB (i.e. the intersection

gonals) draw lines parallel and perpen-
5

dividing
to

dicular the hypotenuse, thus

the square into four congruent quadri- TIB
C

laterals. These, together with thesquare &

/
< i

on AC, will found exactly

4.
be

fit

>

into
to

the square on AB, the way indicated

in

'S
,'
-o
by

corresponding
*
,

numbers.

These experiñéâts point the conclusion that:

to
In

any right-angled triangle the square

on

the hypotenuse equal

is
of

on

the squares
to

the sum the other two sides.

on

formal proof this theorem given the next page.

A

of

is
118 GEOMETRY.

29.
[Euclid

53I,
THEOREM 47.]
25-)

2
In right-angled triangle the square described

on
the hypotenuse
a

on
of
equal the squares described
to
is

the Sum the other two sides.

H
F

C
-->K

\
A g” Nº.

B
L
E
D

A,

rt.
be

right-angled having the angle ACB

4.
Let ABC
a

a
required
It

on

prove that the Square the hypotenuse AB

to

=
the
of is

AC, CB.
on
the

Sum squares

On AB describe the sq. ADEB; and

on

AC, CB describe the

sqq. ACGF, CBKH.
Through
or
to

draw CL par' AD BE.

C

Join CD, FB.

4-,
Z."

ACB, ACG
rt.

Proof. Because each

of

the
is
a

BC and CG are in the same St. line.

‘.

Now the rt. BAD the rt. FAC

=
A

add to each the CAB

A.

then the whole CAD the whole FAB.

=
/

/
A*

Then CAD, FAB,

in

the
CA FA,
= = =

because AD AB,
and the included CAD the included FAB;
4-
4
...

the CAD the FAB

A

A
=

4.

Theor.
 THEOREM OF PYTHAGORAS. 119

Now the rect. AL is double of the A CAD, 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, ...

be
can shewn that

it
the rect. BL the sq. HB.

=
the whole sq. AE the sqq. GA, HB
is, ...

of
the sum
=

:
on

that the square the hypotenuse

of
AB the sum the

=
on

squares the two sides AC, CB.

Q.E.D.
as

This known Pythagoras. The

the Theorem of
is

Obs.
be

as

result established may stated follows:

BC?--CA2.
=

AB2
is,

That denote the lengths the sides containing

of

and
if
a.

the right angle; and denotes the hypotenuse,

if
c
} c2

a”
=

l2.
+

Hence b”;
b2

a?= and =cº

c”

a”.
–

NoTE The following important results should be noticed.

1.

CL and AB intersect O,
If

in

in

of

has been shewn the course the

it

proof that
the sq. GA= the rect. AL;
by AB, AO. ............
is,

that AC”= the rect. contained

(i)

Also the sq. HB the rect. BL

=

;
is,

that BC*=the rect. contained by BA, BO. ............ (ii)

by
be

on
It

NoTE proved superposition that squares standing

2.

can
equal sides are equal area.
in

Hence we conclude, conversely,

If two squares
on

are equal area they stand equal 3ides.

in
120 , GEOMETRY.

EXPERIMENTAL PROOFS OF PYTHAGORAS'S THEOREM,

I. Here ABC is
the given
rt.-angled A;
and ABED is
square on the hypotenuse
§§B.
By drawing lines par! to the
sides BC, CA, it is easily seen
that the sq. BD is divided
into 4 rt.-angled A*, each
identically equal to ABC, tos
gether with a central square.
Hence
c-4
rt.

sq. on hypotenuse 4°A"

the central square
+

=4. #ab
(a

b)*
+

b%

=2ab-i-a” 2db
+
–

u2+b^.
=

II. Here ABC the given

is

rt.-angled A, and the figs.

CF, HK are the sqq. on CB,
CA placed side by side.
FE
to

made equal DH
is

CA;
or

and the two sqq.

CF, HK are cut along the
lines BE, ED.
Then will be found that
it

ADHE
be

the may placed

fill
so

to
as

up the space ACB

;

BFE
be

may
A

and the made

fill the space AKD.
to

Hence the two sqq. CF,

HK may
be

fitted together
so
as

form the single fig.

to

ABED, which will

be

found
be

perfect square, namely

to

on

the square the hypotenuse

 THEOREM OF PYTHAGORAS. 121

EXERCISES.

(Numerical and Graphical.)

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

cm., cm.;

(i)

= = =

of = = =
3

4
b
a a a
(ii) 2.5 cm., 60 cm.;
-> S

b b
(iii) 1-2", 3'5".
and verify
In

each case calculate the length the hypotenuse

c,

a
your result by measurement.

C,
triangle ABC, right-angled having given

at
2.

Draw
a

:
c-3'4",
(i)

3-0"; [See Problem 10]

=
a

(ii) c-5-3 cm., b=4'5 cm.

side, and verify your result by
In

each case calculate the remaining

measurement. by

(The following examples are

be
to

solved calculation but each

in
;
plan should
be

on

case drawn Some suitable scale, and the calculated

a

*
*
*
*

result verifted by measurement.)

ladder whose foot feet from the front

of
a 3.

is

house reaches
A

window-sill 40 feet above the ground. the length

of
What
to

the
is

ladder
?

miles due South, and then

33

56

ship sails, miles due West.

4.

How far then from its starting point

%
it
is

Two ships are observed from signal station bear respectively

to
5.

N.E. 6-0 km. distant, and N.W. 11 km. distant. How far are they
apart
2

point
to

ladder 65 feet long reaches

in

of
6.

the face house

A

a
63

feet above the ground. How far

is

the foot from the house?

A,

at
of

due East but an unknown distance. due South

B,7.

C
B
is

is
55

73
be
of

and distant AC known Find AB.

to
is

metres. metres.

man travels 27 miles due South then 24 miles due West;

8.

20 A

finally from his starting point?

he

miles due North. How far

is

60

From go West 25 metres, then North metres, then East

9.

metres, finally South

80

12

metres. How far are you then from A2

50

48
so
as

ladder feet long placed window

to
is

10. reach feet

A

turning the ladder over

on

high the street,

of

the other side

to

and
it
;

point
14

feet high. Find the breadth

of

reaches the street.

a
122 GEOMETRY.

THEOREM 30. [Euclid I, 48.]

If the square

the
described on one side of a triangle is equal to
the other two sides, then the angle
of

a on
sum the squares described
by right angle.

is
contained these two sides

D
A
|B

F
E
triangle
be

Let ABC
in

which
a
on

the sq. AB the sum of the sqq. on, BC, CA.

=

required ACB right angle.

It

prove that
to
is

is
a
to
Make EF equal BC.
EF, and make FD equal
to

Draw FD perp
to
CA.
Join ED.

Proof. Because EF BC,

= = = =
on

on

the sq. EF the sq.

...

BC.
And because FD CA,
*
on

on

the sq. the sq.

of ...

FD CA.
EF, FD the sum
on

the sqq. the sqq.

of

Hence the sum

=

on BC, CA.
Z-,

But since EFD rt.

= is
a

EF, FD the sq.

on

on

the sqq. DE:

of
...

the sum Theor. 29.

by

And,
on

on

hypothesis, the sqq. BC, CA the sq. AB,

on =

the sq. on DE the sq.

...

AB.
==

DE AB.
‘.

ACB, DFE,
= = = = A*

Then
in

the
AC DF,
because CB FE,
and AB DE
|

;
...

the ACB the A.DFE.

7.

Theor.
Z

But, by construction, DFE right angle;

is
a

right angle.
...

the ACB
is
4

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 b,
shew that c2– b%- BD2 – DC”.

3. If from any point O within a triangle ABC, perpendiculars OX,

OY, OZ are drawn to BC, CA, AB respectively: shew that
AZ24-BX2 + CY2=AY2+ CX2 + BZ2.

4. ABC is a triangle right-angled at A.; and the sides AB, AC are

by a straight line PQ, and BQ, PC are joined. Prove that
wintersected
-*A*-
BQ2+ PC2=BC2+ PQ2.

5. In a 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 :

50 cm.; *.
20 14

cm., 48 cm.,
(i)

= = =

= = =

=
b b b
a a a

"

cm., c= 41 cm.;
99 10

(ii) 40 cm.,
X

(iii) cm., cm., c=101 cm.

vº

11. ABC triangle right-angled

at

an isosceles deduce from

C
is

Theorem 29 that
AB2 =2AC2.
Illustrate this result graphically by drawing both diagonals
of

the
AB, and one diagonal
on

on

square the square AC.

of

AC BC =2", find AB the nearest hundredth an inch, and

If

to

of
=

by

verify your calculation actual construction and measurement.

}
*
12.

on

square diagonal Calculate,

of

Draw cm. and also

a

6
a

measure, the length Find the area.

of

side.
a
124 GEOMETRY.

PROBLEM 16.

To draw squares whose areas shall be respectively twice, three-times,

four-times, ..., that of a given Square.
*

Hence find graphically approximate values of V2, V3, V4, V5, ...,

Y
P
Take OX, OY at right angles
to one another, and from them
mark off OA, OP, each one
unit of length. Join PA.
O A B C X
Then PA* = OP2 + OA2 = 1 + 1 = 2.
PA

x/2.
to =
...

PB;
off

From OX mark OB equal PA, and join

then PB2 OP2
=

=
OB% 3.
+

+
1

PB V3.
=
...

From OX mark off OC equal PB, and join PC;

to

then PC” OP24 OCº-1 +3=

–

4.
...

PC
=

W4.
PC

be

by

The lengths PA, PB, may now

of

found measurement;
by

we

continuing the process may find V5, V6, V7,

...

and
.
30
on

ExERCISES THEOREMS 29, (Continued).

13. Prove the following formula: sº

Diagonal square=side
of

N2.
x

on
to

the diagonal square

of

Hence find the nearest centimetre

a

side of 50 metres.
plan (scale
10
to

as

Draw cm. metres) and obtain the result

1
a

you can by measurement.

as

nearly

units,
an

14. ABC equilateral triangle

to of

which each side=2m

is

and the perpendicular from any vertex the opposite side=p.

Prove that
p

mx/3.
=

Test this result graphically, when each side cm.

=
8
 THEOREM OF PYTHAGORAS. 125

15. If in a triangle a=m”— n”, b = 2mm, c=m^+n”; prove algebraic

w
ally that cº–a4+b”.
Hence by giving various numerical values to m and m, 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.
25 cm., p=12 cm., BD =9 cm.; find
If If
(i)

and

c.
=

b
a

(ii) =41", c=50", BD=30"; find

a.
and

p
b

And prove that Nbº-pº-i-Vcº-º-a.

17.

the triangle ABC, drawn perpendicular

In

AD BC.

to
is
Prove that
c2

BD2–b?

—
CD2.
–

cm.; find BD.

37
51

cm., b=20 cm.,

If

of =
=
a

AD, and the area the triangle ABC.

p,

Thence find the length

of
by
18.

the last example the areas the triangles

of

Find

of
the method
whose sides are as follows:

25 ft., ft.,

12
17
ft.
(i)

17”, 10", c=9". (ii)

=
=
= =
=
=

b
a a

c
b
a

yd., b=37 yd., c=13 yd.

40
15

(iii) =41 cm., b=28 cm., c= cm. (iv)

a

straight rod PQ slides between two straight rulers OX, OY

§º.
at A

19.
*

right angles one position

In

of
the rod
to

laced one another.

another position OP=4.0 cm.,
If

cm., and OQ=3-3 cm.

in

by
find OQ graphically; and test the accuracy your drawing
of

calculation.
C,

triangle right-angled
at

20. ABC the length

of
p

the
is

and
is
a

perpendicular from AB. By expressing the area the triangle

on

of

in
C

two ways, shew that

pc=ab.
||
1

Hence deduce
+

=a.
2

52'
126 GEOMETRY.

PROBLEMS ON AREAS.

PROBLEM 17.

To describe a parallelogram equal to a given triangle, and having

of
its

angles equal given angle.

to
one

F/

G
A
be

Let ABC the given triangle, and the given angle.

D

'
required parallelogram ABC,
It

equal and
to

to
is

describe
to a

having one angles equal

of
its

D.

Construction. Bisect BC at
E.

At CE, make the CEF equal

in

to
D
E

through BC;
to

draw AFG par

A

and through draw CG par" EF.

to
is C

Then FECG the required par”.

Proof. Join AE.

on
A*

ABE, AEC are equal bases BE, EC, and

of

Now the
the same altitude;
... the AABE the AAEC.
is =

the AABC double of the AAEC.

...

But FECG par" by construction;

is is
a

the AAEC,
of

and double
it

the same base EC, and between the same par" EC

on

being
and AG.
the par" FECG the AABC;
...

its angles, namely CEF, the given

of

and one
-

A.D.
-
 PROBLEMS ON AREAS. I27

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°.
by calculation, (ii) by measurement the length
(i)

Find

of
an oblique
parallelogram.
of

side the

Draw any parallelogram ABCD

2.

which AB=2#" and AD=2";

in
and on the base AB draw

of
rhombus equal area.
a
In

DEFINITION. parallelogram
a

ABCD, through any point

if

in
the
K

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
be
to

their
complements.
In

the diagram
of
3.

the preceding definition shew by Theorem

21
that
the complements EG, HF are equal
in

area.
Hence, given parallelogram, EG, and straight line HK, deduce
a

a
for drawing
on

HK
as

construction one side parallelogram equal and

a
to

equiangular the parallelogram E.G.

4.

Construct rectangle equal given rectangle CDEF,

in

to

area
a

and having one side equal

to

given line AB.

a

AB=6 cm., CD=8 cm., CF

If

cm., find by measurement the

=
3

remaining side the constructed rectangle.

of

Given parallelogram ABCD, which AB=2'4", AD=1-8", and

5.

in
a

an

the 4-A=55°. Construct equiangular parallelogram

of

equal area,
the greater side measuring 27". Measure the shorter side.
Repeat the process giving any other value; and compare your
to
A

results. What conclusion do you draw:

6.

on

Draw rectangle
of

equal
to

side
in

cm.
a

area an
of a

equilateral triangle on side cm.

a

the remaining
of

Measure side the rectangle, and calculate its

approximate area,
128 GEOMETRY.

PROBLEM 18.

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

\\
D

A E; X

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 AXDB = the A CDB in area.
To each of these equals add the AADB ;
then the A DAX = the fig. ABCD.

CoroLLARY. 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. D
For example, in the adjoining dia E
gram the five-sided fig. EDCBA is equal C
in area to the four-sided fig. EDXA.
The fig. EDXA may now be reduced
to an equal A DXY. Y A B X
 PROBLEMS ON AREAS. 129

PROBLEM 19.

To draw a parallelogram equal in area to a given rectilineal

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

K D H

A G B F
\
Let ABCD be the given rectil. fig, and E the given angle.
It is required to draw a par” equal to ABCD and having an
angle 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 parº AGHK equal to the AADF, and having the
4. KAG equal to the A. E. Prob. 17.
Then the parº KG = the AADF
= the fig. ABCD ;
and it has the A. KAG equal to the A. E.

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

EXERCISES.

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

S. Draw a quadrilateral ABCD from the following data :

1.
AB = BC = 5.5 cm.; CD = DA=4-5 cm.; the Z_A=75°.
Reduce the quadrilateral to a triangle of equal area.

\
Measure the
base and altitude of the triangle ; and hence calculate the approximate
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.
r &
3. On a 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 its
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.,
and 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 its

base 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 is a given triangle, and X a given point. Draw a triangle

X,
its

its

equal in area to ABC, having

at

in

vertex and base the same

straight line
as

BC.

Construct triangle equal the quadrilateral ABCD,

X in

to
8.

area
a
its

given point DC, and its base

at

having
in

in

vertex the same

a

straight line
as

AB.

triangle may equal parts by straight

be

divided into
9.

Shew how
n
a

lines drawn through one its angular points.

of
 PROBLEMS ON AREAS. } 31

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

of
its
in one sides.
A,

A
be
[Let ABC the given and the

P
given point the side AB.
in
Bisect AB Z; and join CZ, CP.
at
Through draw ZQ parallel CP.

to
Z

Join PQ.
Then PQ bisects the A.]

Q
B

C
by
triangle straight lines drawn from given point

in
1]... Trisect
a

a
,
its
of

one sides.
A,
be

[Let ABC the given and the

X
given point the side BC.
in

the points
P,
Q Q.

Trisect BC Prob.
at

7.
Join AX, and through and draw PH and
P

QK parallel AX.
to

Join XH, XK.

These straight lines trisect the
as

may
A
;

shewn by joining AP, AQ.]

be

C
X
given triangle fourth, fifth, sixth, P any part

of or
12. Cut off from
a

required by straight line drawn from given point its sides.

in

one
a

13. Bisect quadrilateral by straight line drawn through an angular

a

point.

[Reduce the quadrilateral triangle equal area, and join the

of
to
a

the middle point

to

of

vertex the base.]

given quadrilateral third, fourth, fifth,

or

14. Cut off from

a

any part required, by straight line drawn through given angular

a

point.
132 GEOMETRY.

AXES OF REFERENCE, COORDINATES.

EXERCISES FOR SQUARED PAPER.

If we take two fixed straight lines XOX, YOY" cutting one

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.

i
F{ S

The lines XOX', YOY" are usually drawn horizontally an

º
vertically.
In practice the distances of P from the axes are estinated
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 a.
PM ordinate
3/.

95 35 35 99
The abscissa and ordinate taken together are called the
its (a,
y).
by

the point
P,
of

coordinates and are denoted

We may thus find the position point
of

coordinates
if
a

are known.
ExAMPLE. Plot the point whose coordinates are (5,4).
Along OX mark off OM, units length.
4 in
5

At draw MP perp. OX, making MP units length.

in
M

to

Then the point whose coordinates are (5,4).

is
P

The axes reference divide the plane into four regions XOY,
of

YOx', X'OY", Y'OX, known respectively

as

the first, second,

third, and fourth quadrants.
 AXES. COORDINATES. 133

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 : &
Abscissae measured along the a-axis to the right of the
origin are positive, those measured to the left of the origin

is,
lie
are negative.

lie in
Ordinates which above the a-axis (that
the first and second quadrants) are positive; those which
is,

the third and fourth quadrants)

in
below the 2-axis (that
are negative.

R,
the points Q,
of

Thus the coordinates are

4) S
(–5, 4), (–5, 4), and (5,
– respectively.
–

NoTE. The coordinates the origin are (0, 0).

of
In

practice use squared paper. Two

to

convenient
it
is

axes, and slightl

be

as

intersecting lines should chosen

aid the eye, then one the length
or
to

of

thickened more
be

is as

divisions may the linear unit. The paper used

in
taken
the following examples
an
to

of

ruled tenths inch.

the points are (7,

of

ExAMPLE
8)
1.

The coordinates and and

B
A

(–5, 3): plot the points and find the distance between them.
After plotting the points
as

the diagram, we may find

in

º
AB approximately by direct
measurement.

Or we may proceed thus:

Draw through line parl
B
a

to XX' to meet the ordinate

Then ACB
= is

of at C.
A

which BC 12,
in
A

and AC
5.
=

Now AB2 BC? --AC2

= = = = =

5°

122
+

144+25
169.
..".

AB 13.
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 these
are the vertices.

Having plotted the points as in

the diagram, we may measure AB,
and draw and measure the perp.
from C on AB. Hence the approxi
mate area may be calculated.

Or we may proceed thus:

Through A and B draw. AP, BQ
parl to YY".
C draw PQ parl to
xºrough

Then the AABC = the trap" APQB – the two rt.-angled A* APC, BQC
= #PQ(AP+BQ)–3. AP. PC – #. BQ. QC
= } x 13 × 19 –3 x 12x2 – 4 × 7 × 11
=73 units of area.

EXERCISES.

1. Plot the following sets of points:

(6, 4), (–6, 4), (–6, –4), (6, –4);
(i)

0), (0, -8);

8,

(ii) (8, 0), (0, 8),

–
(

5,

(iii) (12, 5), (5, 12), (— 12, 5), 12).

-
(

Plot the following points, and shew experimentally that each set
in 2.

lie one straight line.

7), (0, 0), (–9, -7);
(9,

(0,

(ii) (–9, 7), 0), (9, -7).

(i)

Explain these results theoretically.

Plot the following pairs points; join the points each case,
in
of
3.

the mid-point the joining line.

of

of

and measure the coordinates

(4, 3), (12, 7); (ii) (5,4), (15, 16).
(i)

Shew why the mid-point are respec

of
in

each case the coordinates

tively half the sum
of
of

of

the abscissae and half the sum the ordinates

the given points.

Plot the following pairs points; and find the coordinates

of

of
4.

the mid-point their joining lines.

of

(0, 0), (8, 10); (ii) (8, 0), (0, 10);

- -
(i)

(iii) (0, 0), (–8, 10); (iv) (–8, 0), (0, 10).
 EXERCISES FOR SQUARED PAPER. H35

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:

(5, 0), (5, 2), (5, 5), (5, 1), (5, –4);
(i)

–
4,

1,
(ii) 8), 8), (0, 8), (3, 8), (6, 8).

–
(

(
Shew that they lie on two lines respectively parallel

to

of
y,
the axis
Find the coordinates the point which they

in
of

of
2.

and the axis

intersect.

Plot the following points, and calculate their distances from the
7.

origin.
(15, 8); -8);
(i)

(ii) 15, (iii) (2-4”, “7”); (iv) ’7”, 2'4").

–
–
(

(
Check your results by measurement.

Plot the following pairs points, and

of

in
8.

each case calculate the

distance between them.
(4,0), (0, 3); (ii) (9, 8), (5, 5);
(i)

(iii) (15, 0), (0, 8);

5,
(iv) (10, 4), 12);

– –
(
(v) (20, 12), 15, 0); (vi) (20, 9), 15, -3).
–

(
(

Verify your calculation by measurement.

Shew that the points (–3, 2), (3, 10), (7, are the angular
2)
9.

points
an

isosceles triangle. Calculate and measure the lengths

of

of
the equal sides.
10. Plot the eight points (0, 5), (3, 4), (5,0), (4, -3), (–5, 0),
-5), (–4, 3), (–4, -3), and shew that they all lie
on

(0, circle whose

a

the origin.
is

centre

11. Explain by diagram why the distances between the following

a

pairs points are all equal.

(i) of

(b,

0), (0, a);

b).

(a, 0), (0, b); (ii) (iii) (0, 0), (a,

12. Draw the straight lines joining
and (0, a); (ii) (0, and (a, a);
at 0)

(a,
(i)

0)

and prove that these lines bisect each other right angles.
13. Shew that (0, 4), (12, 9), (12,
4)
—

of

are the vertices an isosceles

ac.
by

triangle whose base

of

the axis
is

bisected
14. Three vertices rectangle are (14, 0), (14, 10), and (0, 10);
of
a

find the coordinates the fourth vertex, and the intersection

of

of

of

the
diagonals.

15. Prove that the four points (0, 0), (13, 0), (18, 12), (5, 12) are the
angular points Find the length each side, and the
of

of

rhombus.
a

the diagonals.
of

of

coordinates the intersection

point which moves
so

16. Plot the locus that its distances from

of
a

and (4,-4) are always equal

(0,

the points Where

0)

to

one another.
,

does the locus cut the axes

2
I 36 GEOMETRY.

17. Shew that the following groups of points are the vertices of
rectangles. Draw the figures, and calculate their areas.
(4, 3), (17, 3), (l7, 12), (4, 12);

(i)
(ii) (3, 2), (3, 15), (– (–6, 2);

6,
15),
-8), (5, 8).

8,

8,
(iii) (5, 1), 1),

–
(

1",
(0,
Join order the points (1", 0), 1"), 0), (0, -1"). Of
in
18.

–
(
the quadrilateral

so
what kind Find its area.
is

formed

?
second figure formed by joining the middle points the first,
If

of
is
a

find its area.

19. Plot the triangles given by the following sets points; and find

of
their areas.
(10, 10), (4, 0), (18, 0); (ii) (10, -10), (4, 0), (18, 0);
(i)

(iii) (— 10, 10), (–4, 0), (— 18, (iv) (— 10, tºº10), (–4, 0), (— 18, 0).
0)
;

20. Draw the triangles given by the points

(0, 0), (5, 3), (6, 0); (ii) (0, 0), (3, 0), (0, 6).
(i)

Find their areas; and measure the angles the first triangle.
of

Plot the triangles given by the following sets points.

of
21. Shew
parallel ence find the
to

of

that
in

each case one side one the axes.

is

8,I'68,.
(0, 0), (12, 10), (12, -6); (ii) (0, 0), (5, 8), 15, 8);
(i)

–
-8); (-6, -8), (
(iii) (0, 0), 12, 12,), 12, (iv) (0, 0), (20, -8).
–
–
(

of
the following triangles shew that two sides
In

22. each are

parallel Find their areas.
to

the axes.
(5, 5), (15, 5), (15, 15); (ii) (8, 3), (8, 18), (0, 18);
(i)

(iii) (4, 8), 16, 4), (4, -4); (iv) (1, 15), 11, 15), (1, -7).
–

–
-

(
(

Shew that (–5, 5), (7, 10), (10, 6), (–2, are the angular
1)

23.
points parallelogram. Find its sides and area.
of
a

the following sets points gives trapezium.

of

of

24. Shew that each

a

Find the area of each.

tº-º-º: ſº +-tº-
3), (0, sº3);
2,
5,

(3, 0), (3, 3), (9, 0), (9, (ii) (0, 3), 3),
(i)

6)
;

(-1, (-4,
1),

(0,

5),

(3,-1); (–8,
0).

(iii) (8,4), (4,4), (11, (iv) 0), 5),

-

25. Find the area the triangles given by the following points:
of

(5, 5), (20, 10), (12, 14); (ii) (7, 6), (— 10, 4), (–4, –3);
(i)

(iii) (0, 6), (0, –3), (14, 5); 6), (–2, +---
7,

(iv) (6, 4), (— 15).

–
–

(-5,
(7,

Shew that 0), (7, 5), (19, 0), angular

5)

26. are the

-

points Find its sides and its area.

of

rhombus.
a
 EXERCISES FOR SQUARED PAPER. 137

27. Jointhe points (0, -5), (12, 0), (4, 6), ( – 8, -3), in the order
given. Calculate the lengths of the first three sides and measure the
fourth. Find the areas of the portions of the figure lying in the first
and fourth quadrants.

28. The coordinates of four points A, B, C, D are respectively

(-4, –4), (— 10, 4), (— 10, 13), (5, 5).
Calculate 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

(scale 1"= 100 yds.). On the plan the coordinates of A, B, §
8.Te
1",

(– -3"), (3", 4"), (–5", -2") respectively. Find the area the field,

of
the length the side represented by BC, and the distance from this
of

the opposite corner

of

of

side the field.

31. Shew that the points (6, 0), (20, 6), (14, 20), (0, 14) are the
square. side and hence find the approximate
of

vertices Measure
a

by drawing circumscribing
(i)

area. Calculate the area exactly

a

square through its vertices; (ii) by subdividing the given square

as
in
on

the first figure page 120.

 138 GEOMETRY.

MISCELLANEOUS EXERCISES.

1. AB and AC are unequal sides of a triangle ABC; AX is the

median through A, AP bisects the angle BAC, and AD is the perpen
dicular from Å to BC. Prove that AF is intermediate in position and
magnitude to AX and AD. ſov v.

2. In a triangle if a perpendicular is drawn from one extrºity of

will make with

an (i)
the base to the bisector of the vertical angle,

it
either the sides containing the vertical angle angle equal half

to
of

the angles the base; (ii) will make with the base an
at
of

the sum

it
angle equal half the difference the angles
to

at
of
the base.

any triangle the angle contained by the bisector

In

of
3.

the
vertical angle and the perpendicular from the vertex

to

is
the base

at
equal half the difference the angles
to

of

the base.
*.*
*
& right-angled triangle having given the hypotenuse
Construct
4.

and the difference of the other sides.

Construct triangle, having given the base, the difference

of
5.

the
N

angles the base, and the difference, (ii) the sum the remaining
(i)
at

of
sides.

Construct an isosceles triangle, having given the base and the

6.

the equal sides and the perpendicular from the vertex

to
of

of

sum one
the base.

given straight line

so

divide that the square on

to
7.

Shew how
a

one part may

be

the square on the other.

of

double

ABCD parallelogram, any point without the angle

8.

is

and
O
is
a

BAD its opposite vertical angle; shew that the triangle OAC equal
or

is

the triangles OAD, OAB.

to

of

the sum
to or

within the angle BAD its opposite vertical angle, shew

If

is
O

that the triangle OAC equal the triangles

of
is

the difference
OAD, OAB.

quadrilateral equal triangle

of

The area
to

of
9.

the area
is
a

having two
to

its sides equal the diagonals the given figure, and

of

of
to

the included angle equal the angles between the diagonals.

of

either
10. Find the locus triangles
of

of

of

the intersection the medians

on

given base and given area.

of

described
a

given triangle construct second triangle

of

11. On the base

a

in a

the first, and having its vertex given straight

to

equal
in

area
a

line.

ABCD parallelogram made rods connected by hinges.

If
of
is

12.
a

AB fixed, find the locus the middle point CD.

of

of
is
 PART III.
THE CIRCLE.

DEFINITIONS AND FIRST PRINCIPLES. -T

**
*rº.
\\
1.A circle is a plane figure contained by a line traced out
by point which moves so that its distance from a certain
a
fixed point is always the same.
The fixed point is called the centre, and the bounding line
is called the circumference. \N

NOTE. 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. Aradius of a circle is a straight line drawn from the

centre to the circumference. It follows that all radii of a
circle 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.
a7” ( ...ºf rz22

5. Circles that have the same centre are said to be

concentric. &* Abe a 3.6%
140 GEOMETRY.

From these definitions

*J
we draw
~ the following inferences:
(i) closed curve;

so
circle that

is
A
the circumference

if
a
by
crossed straight line, this line produced will cross
is

if
a
second point.

at
the circumference

a
(ii) The distance point from the centre

of

of
circle

a
a
greater less than the radius according the point

as
or
is

is
without or within the circumference.

point

or
(iii) circle according its
is

outside
A

inside

as
a
greater

or
distance from the centre less than the radius.
is
(iv) Circles equal radii are identically equal. For by
at of of

on
superposition one centre the other the circumferences
must coincide every point.

N
º

(v) Concentric circles unequal radii cannot intersect, for

of

every point

on
the distance from the centre
of

the smaller
circle less than the radius the larger.
of
is

(vi) two circles have

If

the circumferences
of

common

a
point they cannot have the same centre, unless they coincide
altogether.

An any part
of

arc circle
of

the circumference.
6.

is
a

straight line joining any two

A

of

chord circle
7.

is
a

§
on

points the circumference.

be

NoTE. From these definitions may

it

seen
that chord circle, which does not pass through
of
a

the centre, divides the circumference into two un-,

equal arcs; these, the greater called the major
of

is

arc, and the less the minor arc. Thus the major
greater, and the minor arc less than the semi
is

arc
circumference.
The major and minor arcs, into which cir-
*

YS
a

cumference divided by chord, are said `-----~~

to

be
is

a
to

conjugate one another.

 SYMMETRY OF A CIRCLE. 141

$.” *, *
-
Ža
* , SyMMETRY. . .

& Some elementary properties of circles are easily proved by

Čonsiderations of symmetry. For convenience the definition
given on page 21 is here repeated. ‘a -

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.

- Q

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 4. AMP = the 4-AMQ, and
MP = MQ.

The points P and Q are said to be symmetrically opposite

s 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. 'S
tº the

It is required to prove that symmetrical about AB.

is
circle

two radii making any equal

be

Proof. Let OP and OQ

on
4."

AOP, AOQ opposite sides

of

OA.

be
Then the figure folded about AB, OP may

to
made
is
if

-
fall along OQ, since the AOP= the AOQ.
4.

4
Q,

And thus will coincide with since OP OQ. =

P

Thus every point the arc APB must coincide with some
of ; in

is,

point the arc AQB that the two parts the circum
in

of

ference on each side AB can be made to coincide.

symmetrical
...

the circle about the diameter AB.

is

Q, M,

drawn cutting AB on
on at

COROLLARY. PQ then
If

is

folding the figure about AB, since falls MP will

P

-
coincide with MQ, -
MP MQ;
=
•.

and the OMP will coincide with the UMQ,

4.

4
.

';

these angles, being adjacent,

rt.
...

are
s

points are symmetrically opposite

...

the and with-

Q
P

‘Sregard
to

AB.
Hence, conversely, given point
P,
if

circle passes through

a

also passes through the symmetrically opposite point with regard

it

any diameter.
to

DEFINITION. The straight line passing through the. centres

two circles called the line of centres.
of

is
 SYMMETRICAL PROPERTIES. 143

II.

by
Two circles are divided symmetrically

of
their line centres.
O,

be

st.
Let two circles, and let the
at of
O, O'

the centres line

of A,
through cut the O*

B'.
A,
O'

and Then AB and

B
symmetry their

of
A'B' are diameters and therefore axes
is,

respective circles.

of
That the line centres divides each
circle symmetrically.

III. one point,

If

at

at
two circles cut they must also cut

by a
point; and right angles
at
is

second the common chord bisected

of

the line centres.

O,

the point
P.

Let the circles whose centres are

O'

at

cut
Q,
to

so

Draw PR perp. OO', and produce that

to

it

RQ RP.
=

Then are symmetrically

and opposite points with
Q
P

regard centres OO';

to

of

the line
on the O* both circles, follows that
of

since
it
is

is
Q
‘.

[I.
on

the O* Cor.]
of

also both.
And, by construction, the common chord PQ
at

bisected
is
by

right angles OO'.

144 GEOMETRY. .

ON CHORDS.

31.
III.

3..]
If THEOREM [Euclid
straight line drawn from the centre

it of
circle bisects

at a
a

a
chord which does not pass through the centre, cuts the chord
Tight angles.

it.
right angles, bisects

it
at
Conversely, if cuts the chord
it
be

Let ABC circle whose centre and let OD bisect

is
O
a

;
chord AB which does not pass through the centre.
a

required
It

prove that OD perp. AB.

to

to
is

is

Join OA, OB.

Proof.
by A*

Then ADO, BDO,

in

the

AD BD, hypothesis,
is =

because OD common,
and OA=OB, being radii circle;
of

the
the 4ADO BDo; Theor.
7.
...

the
A.
=

and these are adjacent angles,

perp. Q.E.D.
...

to

OD AB.
is
be

Let OD perp.
to

Conversely. the chord AB.

It

required prove that OD bisects AB.

to
is

Proof.
In

A*

the ODA, ODB,

.

. ODA, ODB are right angles,

the
4:

because {the hypotenuse OA the hypotenuse OB,

=

and OD common;
is

-
Theor. 18,
...

DA DB
=

;
is,

that
at

Q.E.D.
D.

OD bisects AB
 CHORD PROPERTIES. -
145

CoROLLARY 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.

O
is st.
For suppose a line meets

*
the points

at
circle whose centre
O
and
B.
A

Draw OC perp.
A-C-E---B
AB. to
Then AC CB
=

AC
AB
third point

D,
Now

in
to
the circle were cut
if

a
be

equal CD, which impossible.

to

would also is

CoroLiary lies wholly within

it.
A

of
3.

chord circle
a

EXERCISES.

(Numerical and Graphical.)

In

the figure Theorem 31, AB cm., and OD=3 cm., find

of
1.

if

=
8

OB. Draw the figure, and Yºrity your result by measurement.

{K}
N&as
3
a

at
of

5"
Calculate the length chord which stands distance
a

a
2.

circle whose radius

of

from the centre 13".

is
a
In

length. .
of
1"

circle radius draw two chords 1.6" and 12"

in
3.

a
*
-

Calculate and measure the distance of each from the centre.

Draw circle whose diameter 8'0 cm. and place

in
4.

is

chord
it
a

length. Calculate the nearest millimetre the distance

in

to

of

6.0 cm.
by

the chord from the centre; and verify your result measurement.

length
10
ft.

in.

Find the distance from the centre

in
of
a 5.

chord
5
a

circle whose diameter yds. in. Verify the result graphically

byin

is
1 2

drawing figure which cm. represents

in

10".
a

AB chord 2'4" long circle whose centre

in
6.

is

and whose
O
is
a

13"; find the area the triangle OAB square inches.

of

in
is

radius

Two points apart. Draw circle with radius 1.7"

3"
'?
7.

and are
Q
P

its
byQ.

pass through
of

Calculate the distance

to

and centre from the

P

chord PQ, and verify measurement.

H.S. G.
K
146 GEOMETRY.

THEOREM 32.

One circle, and only one, can pass through any three points not
in the same straight line.

TB

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. -
st.

Then since AB and BC are not in the same line, DF and

EG are not par'.
O.

Let DF and EG meet

in

right angles,
at

Proof. Because DF bisects AB

every point equidistant from and
on

DF
B.
...

is

Prob. 14.
Similarly every point
on

equidistant
C.

EG from and
is

B
A, O,

the only point common DF and EG, equidistant

to
...

is
B,

from and
no C
;

other point equidistant from

B,
A,

C.

and there and

is

its

and radius OA will pass

at

circle having centre

...

O
a

C.;

through and and this the only circle which will pass
is
B

through the three given points. Q.E.D.

 CHORD PROPERTIES. 147

S COROLLARY 1. The size and position of a circle are fully

Šdetermined 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 entirelift för 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 to pass through the vertices of

any triangle. -

DEFINITION. The circle which passes through the vertices

its

of a triangle is called

is be
circum-circle,

to
and said

of is
circumscribed about the triangle. The centre the circle
the triangle, and the radius
of

called the circum-centre called

is
the circum-radius.

EXERCISES ON THEOREMS 31 AND 32.

.”
..
..

(Theoretical.)
,

The parts straight line intercepted between the circum

of
1.

ferences of two
Congºngjo girdles are equal,
.
..
B, ..
.
.

D;
C,
at

at

Two circles, whose centres are

is 2.

and intersect and

A

the middle point Shew that AM and BM

of

the common chord.

M

the same straight line.

in

are
of

Hence prove that the line right

at

centres bisects the common chord

angles.

AB, AC are two equal chords circle; shew that the straight
of
3.

line which bisects the angle BAC passes through the centre.
all

circles which pass through two

of

. of

Find the locus

4.

the centres
gºver points. .xºn
Y.
.

&
(.

.
.

.
(
.

.
.
.

.
..

..
.

circle that shall pass through two given points and have
5.

Describe
a

-
.
its

centre given straight line.

is in
a

When this impossible?

of

given radius pass through two given points.

to
6.

Describe circle
a

.
.

When this impossible?

is
 - GEOMETRY.
148

*THEOREM 33. [Euclid III. 9..] .

If from a point within a circle more than two equal straight

limes can be drawn to the circumference, .* that point is the centre
of the circle.

Let ABC be a circle, and O a point within it from which

the O*, namely
st.

more than two equal lines are drawn

to
- - -
OA, OB, OC.

of
required the circle ABC.
It

prove that
to

the centre
is
O
is

Join AB, BC.

be

the middle points AB and BC respectively.

of

Let and
E
D

-
Join OD, OE.
A*

Proof.
In

the ODA, ODB,

DA DB,
is =

because DO common,
by

and OA= OB, hypothesis;

7.

Theor.
the ODA the ODB
=
4.
...

;
4

rt.

these angles, being adjacent, are

...

".
A
AB
at

Hence Do bisects the chord right angles, and therefore

passes through the centre. Theor. 31, Cor.
1.
be

Similarly may shewn that EO passes through the

it

-
...

centre.
"

... the only point common DO and EO, must

O,

to

which
is

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.

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 1 cm. Calculate and measure the radius.
it its

7. Shew on squared paper that if a circle has any point

at

centre
(6,

the 2-axis and passes through the point

on

5), also passes through

the point (6, 5). [See page 132.]
-

Theoretical
..)
(

The line joining the middle points two parallel chords

of

of
8.

circle passes through the centre.

the middle points parallel chords
of
of

Find the locus

9.

circle.
in
a

10. Two intersecting

of

chords circle cannot bisect each other

a
is

unless each diameter.

a

circle, the point

be

parallelogram inter
If

in

of

11. can inscribed

a

of a
be
at

its diagonals must

of

section the centre the circle.

that rectangles are the only parallelograms

be

12. Shew that can -

.

inscribed in circle.
a
150 * GEOM ETRY.

THEOREM 34. [Euclid III. 14.]

Equal chords of a circle are equidistant from the centre.

Conversely, chords which are equidistant from the centre are

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.

IProof. Because OF is perp. to the chord AB,

OF bisects AB; Theor. 31.
...

AF half
of

AB.
is

..'.

Similarly CG
of

half CD.
is
by

But, hypothesis, AB CD,

= =

.*. AF CG.
A*

Now OFA, OGC,

in

the

the "OFA, OGC are right angles,

4

because {the hypotenuse OA the hypotenuse OC,

= =

and AF CG
;
all

the triangles are equal respects;

= in
...

Theor. 18.
OG;
so

that OF
is,

that AB and CD are equidistant from

O.

Q.E.D.

*ść 3.
&
 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,

ſº
4 "OFA, OGC are right angles,
because {the hypotenuse OA = the hypotenuse OC,
l and OF = OG;
AF Theor. 18.
...
CG

;
these are equal;
...

the doubles of
is,

that AB = CD.
Q.E.D.

EXERCISES.

(Theoretical.)

º
the
of

middle points equal chords

of

Find the locus of

1.

circle.
a
circle cut one another, and make equal
of

two chords
If
2.

angles with the straight line which joins their point

of

to
intersection
the centre, they are equal.
two equal chords
If
3.

circle intersect, shew that the segments

of
a

a
the one are equal respectively the segments
of

to

of

the other.

In given circle draw chord which shall

be

equal one given

to
4.

straight line (not greater than the diameter) and parallel

to

another.

PQ circle, and AB any diameter

of in
5.

is

fixed chord shew

is
a

A :

that the sum the perpendiculars let fall from and

or

difference
B
is,
on

PQ constant, that the same for all positions AB.

of
is

Jr.
9,
p.

i... [See Ex. 65.]

º.
,

,
.

(Graphical.)

radius 4:1 cm. any number

In

of

of

circle chords are drawn

6.

each 19 cm. length. Shew that the middle points

in

of

these chords
lie
on

length its radius, and

of

all circle. Calculate and measure the

a

draw the circle.

apart, their common chord

4"

The centrestwo circles are

of
7.

is

length,
and the radius the larger circle Give con
in

of

is

2'4" 3.7".
a

struction for finding the points the two circles, and

of
of

intersection
of

find the radius the smaller circle.

15 2 # GEOMETRY.

TIIEOREM 35. [Euclid III. 15.]

Of any two chords of a circle, that which is nearer to the centre

is greater than one more remote.º.º. 23, F
Conversely, the greater of two chords is nearer to the centre than
the less.

A C

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

CD;
(i)

OF less than OG, then AB greater

if if

is

than
is

(ii) AB greater than CD, then OF less than OG.

is

is

Join OA, OC.

Proof. Because OF perp. the chord AB,

to
is
...

OF bisects AB;
of AB.
AF

half
of
...

is

Similarly CG half CD.

is

Now OA OC,
= =
on

on

the sq. the sq. OC,

...

OA
But since the ZoPA
rt.

angle,
is
a
on

on

the sq. the sqq. OF, FA.

...

OA
=
on sq.

Similarly the
on

on on

OC the sqq. OG, GC,

= =

OF,
FA

the sqq. the sqq. OG, GC,

...
 CHORD PROPERTIES. * 153

given less than OG;

(i)
Hence OF

is is is is is
if
... ... on on

on
the sq. OF less than the sq. OG.

on
... the sq. FA greater than the sq. GC

;
FA greater than GC:
AB greater than CD.

(ii) But if if AB given greater than CD,

is is is is is
is,

that FA greater than GC

;
... on on

on
then the sq. FA greater than the sq. GC,
OG;

on
OF
...

the sq. less than the sq.

OF less than OG. Q.E.D.

CoroLLARY. The greatest chord

in

is
a circle diameter.

a
à

EXERCISES.

(Miscellaneous.)
Y,
,
,

Through given point within circle draw the least possible

1.

a
*

chord.

Draw triangle ABC which =3:5", b=1:2", c=3.7". Through

in
2.
$

the ends of the side draw circle with its centre on the side
c.
a

Calculate and measure the radius.

Draw the circum circle triangle whose sides are 2-6", 2-8",
of
3.
Y.

and 3'0". Measure its radius.

circle, and XY any other chord having

of

AB
4.
S.

fixed chord
is
a

a
.

its middle point AB;

on

what the greatest, and what the least

, is
Z

length that XY may have

2

". -
t

Shew that XY increases, the middle point

as

approaches AB.
of
Z
on

squared paper that

at

circle whose centre

5.

Shew
is

the
a

origin, and whose radius 3'0", passes through the points (2-4", l'8"),
is

(1:8", 24").
Find length the chord joining these points, (ii) the co
(i)

of

the
point, (iii) its perpendicular distance from the
its

middle
of

ordinates
origin.
154 GEOMETRY.

III.

7.]
*THEOREM 36. [Euclid
If from any internal point, not the centre, straight limes are

of
drawn circle, then the greatest
to

is
the circumference that

a
which passes through the centre, and the least the remaining part

is
of

that diameter.
any other two such lines the greater
of

And that which Sub

is
at
tends the greater angle the centre.
be

Let ACDB circle, and from any internal point, which

P
a

the O*,
be

not the centre, let PA, PB, PC, PD

to
drawn
so is

O,
of PA

that passes through the centre and PB the remain is

ing part that diameter. Also let the POC the centre
at
4.

by
subtended by PC
be

greater than the POD subtended PD.

4.

required
st.
It

prove that
of

lines
to
is

these
(i)

PA the greatest,
is

(ii) PB the least,

vs is

(iii) PC greater than PD.

Join OC, OD,
(i)

Proof. POC, the sides PO, OC are together

In

the
A

greater than PC. Theor. 11.

But OC OA, being radii;
=

PC;
..".

PO, OA are together greater than

is,

that PA greater than PC.

is

Similarly PA may
to be

be
to

shewn greater than any other

the C*;
st

line drawn from

P

all

the greatest
of

PA such lines.
is
.
 DISTANCE OF A POINT TO THE CIRCUMFERENCE. 155

(ii) In the A OPD, the sides OP, PD are together greater

than OD.
But OD = OB, being radii;

are
PD
OP, together greater than OB.
...
Take away the common part OP;
then PD greater than PB.

is
Similarly any other

st.
O*

to
line drawn from the may
*

P
greater than PB;
be

be
to

shewn
PB the least of all such lines.
is
‘.

A'
In

(iii) the POC, POD,

PO common,
= is

because OC OD, being radii,

but the POC greater than the POD;
4.

4.
is

PC greater than PD. Theor. 19.

...

is

Q.E.D.

EXERCISES.

(Miscellaneous.)

Allcircles which pass through fixed point, and have their centres
a 1.

a
on

given Straight line, pass also through second fiaced point.

a

two circles which intersect are cut by straight line parallel

If
2.

the common chord, shew that the parts intercepted

to

of

between
it

the circumferences are equal.

two circles cut one another, any two parallel straight lines
If
3.

drawn through the points

to

cut the circles are equal.

of

intersection
two circles cut one another, any two straight lines drawn
If
4.

through point section, making equal angles with the common

of
a

chord, and terminated by the circumferences, are equal.

\

Two circles diameters 74 and 40 inches respectively

of
5.

have
a

length find the distance between their centres.

in

common chord feet

2

represent 10") and verify your result by

to

Draw the figure

(1

cm.
measurement. *

Draw two circles radii l'0" and l'7", and with their centres
of
6.

2:1" apart. Find by calculation, and by measurement, the length

of

the common chord, and its distance from the two centres.
,
156 GEOMETRY.

III.

8.]
*THEOREM 37. [Euclid
If any easternal point Straight lines are drawn
from

to
the

of
circle, the greatest that which passes through

is
circumference

a
the centre, and the least when produced passes through

is
that which
the centre.

any other two Such lines, the greater

of

And which sub

is
that
the

the
angle

at
tends greater centre.

circle, and from any external point

be

Let ACDB let the

P
a

O, be

to

so

lines PBA, PC, PD drawn the O*', that PBA passes

by
so

through the centre and that the POC subtended PC

4.

greater than the by

at

the centre POD subtended PD.

is

St.

required
of
It

prove that lines

to
is

these
*
(i)

PA the greatest,
is is is

(ii) PB the least,

(iii) PC greater than PD.
Join OC, OD.

the sides Po, OC are together

(i)
In

Proof the POC,

A

greater than PC.

But OC OA, being radii;
=

PO, OA are together greater than PC;

...

is,

that PA greater than PC,

is

Similarly
PA

be

be

st.
to

may shewn greater than any other

line drawn from the O*;
to
P

all
is,

that the greatest

of

PA such lines.
is
 DISTANCE OF A POINT TO THE CIRCUMFERENCE. 157

(ii) In the A POD, the sides PD, DO are together greater

than PO.
But OD = OB, Toeing radii;
greater than the remainder
...

the remainder PD PB.

is
Similarly any other

be
st.
line drawn from O* may

to
the

P
PB;
be

greater than
to

shewn is,
PB all such lines.

of
that the least

POC, POD, is
(iii)
In

the
A

PO common,
= is

because OC OD, being radii;

|but the Focis greater than the POD;
.

A
PC greater than PD. Theor. 19.
is
.

Q.E.D.

EXERCISES.

(Miscellaneous.)

Find the greatest and least straight lines which have one ex
1.

tremity
on

two given circles which do not intersect.

of

each

from any point circle straight lines

on

of
If
2.

the circumference
is a

the circumference, the greatest that which passes

to

are drawn
through the centre; and any two such lines the greater that
of

is
at

which subtends the greater angle the centre.

Of

all straight lines drawn through point

of

of
3.

intersection two
a

circles, and terminated by the circumferences, the greatest that which

is

parallel the line

to

of
is

centres.
Draw on squared paper any two circles which have their centres
4.

the point (8, -11).

at

on the ac-axis, and cut Find the coordinates

of

their other point

of

intersection.
on

Draw squared paper two circles with centres the points

at
5.

(–6, respectively, and cutting the point (0, 8).

at

(15, Find
0)

0)

and
the lengths their radii, and the coordinates their other point
of

of

of

*
intersection.

isosceles triangle OAB with

an

an

at

Draw angle its vertex

of
6.

80°
With circle, and on its circum
O.

centre and radius OA draw

O

a
Q,
R,

points
P,

on

any
of

at of
...

ference take number the same side

as

AB the centre. Mgasure the angles subtended by the chord AB

R,

points Q,
P,

the 2.47 Repeat the same exercise with any other given
do

angle you draw

O.

What inference
at

2
sº
º,

2 X,
ºn

{
*
158 GEOMETRY.

`s.
*
*.
*.
*~s.
*.
r
*.
\s
ON ANGLES IN SEGMENTS, AND ANGLES AT THE
CENTRES AND CIRCUMEERENCES 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 arc.

Fig. I. Fig. 2.

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 arc BC.

It is required to prove that the 4. BOC is twice the A. BAC.

Join AO, and produce it to D.

Proof. In the A OAB, because OB = OA,

of ...

the Z.OAB the OBA.

A.
=

"OAB, OBA
...

the sum the twice the OAB.

=
4

But the ext. "OAB, OBA.;

of

BOD the sum the

= =
A

4
...

the BOD twice the OAB.

A

Similarly the DOC twice the OAC.

=
4

..., adding these results Fig. and taking the difference

in

1,

Fig. follows case that

in
2,
it

each
in

the BOC twice the BAC.

A.

A.

Q.E.D.
=
 ANGLE PROPERTIES. 159

-
A A

B /2^N.º
I C ×1N.
D
E E
Fig. 3. Fig. 4.

Obs. If the arc BEC, on which the angles stand, is a semi

circumference, as in Fig. 3, the A. BOC at the centre is a
straight angle ; and if the arc BEC is greater than a semi
circumference, as in Fig. 4, the A. BOC at the centre is reflea.
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 4 BOC = twice the A. BAC, on the same arc BEC.

DEFINITIONS.

A segment of a circle is the figure bounded

by a chord and one of the two arcs into which
the chord divides the circumference.

NoTE. The chord of a segment is sometimes called * ...”

its base.

An angle in a segment is one formed by two

straight lines di ºwn from any point in the arc
its

of the segment to the extremities of chord.

in 32

We have seen
be
in

Theorem that circle may drawn

a

through any three points not straight line. But

it
is
a

only under certain conditions that

be

circle can drawn

a

through more than three points.

.
.
.
.
.

more points are

or

DEFINITION.
so

four placed that

If

be a
be

circle may drawn through them, they

to

are said
concyclic.
160 GEOMETRY,

THEOREM 39. [Euclid III, 21.]

Angles in the same segment of a circle are equal.

t
Fig.1. Fig.2.

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 4. BDC.

-
Join BO, O.C.

Proof. Because the 4 BOC is at the centre,- and the 4. BAC

at the O*, standing on the same arc BC,
the A-BOC twice the BAC, Theol. 38.
=

4.
•'.

Similarly the BOC twice the BDC.

A.

= =

4.
...

the BAC the BDC.

4.

4.

Q.E.D.
be

as

NoTE. The given segment may greater than

in

semicircle
a
2;

Fig. Fig.
1,

as

the latter case the angle

in
or

less than semicircle

in
a

But by virtue
be

of

of

BOC will reflex. the extension Theorem 38.

#. the preceding page, the above proof applies equally

on

to

both
gures.
‘.

**...Y.
Śrī ..
A

ºn
‘Y

tº
 ANGLE PROPERTIES. 16]

CONVERSE OF THEOREM 39.

Equal angles standing on the same base, and on the 8ame side of

it,
on

of
an arc circle,

of
have their vertices which the given base the

is
a
chord.

Let BAC, BDC

be
two equal angles standing on EAD

A
the same base BC, and on the same side

it.
of
on
required prove that

lie
It

to

and an arc
is

D
of circle having BC
its
as

chord.
a

be

Let ABC the circle which passes through

B,
at A,

the three points and suppose cuts BD

C
it
C

B
;

the point
or

BD produced -
E.

Join EC. -

Then the BAC= the BEC the same segment.

in
Z.

Proof.
Z

But, by hypothesis, BAC= the BDC;

Z.

the

!, 4.
BEC= the BDC
Z.
...

the
Z

which impossible coincides with

D
is

unless
E

;
B,
A,

the circle through must pass through

D.
‘..

thé vertices
of on
of

of

CoRoDLARY. The locus triangles drawn the same

of

8ide given base, and with equal vertical angles, an arc circle.
is
a

EXERCISES ON THEOREM 39.

In

Fig.
1.
&

1,

the angle BDC 74°, find the number degrees

of

in
if

is

the angles BAC, BOC, OBC.

of

each
let

Fig.
In

the angle DXC=40°,

2,

at
X.

BD and CA intersect
If
2.
*

and the angle XCD=25°, find the number

§§§
degrees the angle BAC
in
of

angle
in

and the reflex BOC.

Fig.
In

the angles CBD, BCD

1,
3.

43° and 82°,

of if
*

find the number degrees the angles BAC, OBD, OCD.

in

Shew that Fig. the angle OBC always

in

less than the angle

4.

is
2
by

BAC right angle.

a

on

39

[For further Exercises Theorem see page 170.]

H. S.G.
L
 162 GEOMETRY.

* THEOREM

40.
[Euclid III. 22.]
any quadrilateral

of
The opposite angles

in
inscribed circle

a
are together equal two right angles.

to
be

Let ABCD quadrilateral

in

G)
inscribed the ABC.
to a

-
required prove that
It
is

(i)

"ADC, ABC together

ri. ri.
angles.
= =
the two
4

- (ii) the "BAD, BCD together two angles.

4

Suppose
of

the centre the circle.

O
is

Join OA, OC.

Proof. Since the ADC at the O* half the AOC at the

=
4

Z
on

centre, standing the same arc ABC

;

the O*
at

and the half the reflex 4-AOC the centre,

at

ABC
=
4
on

standing
º

the same arc ADC

= ;

the "ADC, ABC together half the sum

of

the 4-AOC and

...

the reflex AOC. -

/

up

But these angles make angles.

rt.

four
rt. rt,

"ADC, ABC together angles.

...

the two
= =
4

Similarly the "BAD, BCD together two- angles.

4

Q.E.D.
39

carefully
40

be
of

Note. The results Theorems and should

compared. -

º ...
39

From Theorem that angles same segment

in

we learn the are

equal. A&so
,
40

From Theorem angles cºnjugate segments

in

we learn that are

supplementary.

DEFINITION. quadrilateral called cyclic when

A

circle
is

--
drawn through its four vertices.
-
be

can
 ANGLE PROPERTIES. 163

ſº
CoNVERSE of THEOREM 40.

If
a pair of opposite angles of a quadrilateral are supplementary,

its
vertices are concyclic.
quadrilateral
be
Let ABCD which the

in

A
B a
to at
opposite angles and are supplementary.

E
D

D
C,
B,
A,
required prove that the points
It
is

D
are concyclic.
be

Let ABC the circle which passes through

B,
in A,

the three points and suppose cuts AD

it
C

C
B
;

the point
or

AD produced
E.

Join EC.
Proof. Then since ABCE cyclic quadrilateral,
is
a

AEC the supplement

of
A.
‘..

the the 4–ABC.

is

But, by hypothesis, the A-ADC the supplement ABC;

4-
of
the
is

AEC the A-ADC

Z.
‘..

the
=

impossible unless

D.
which coincides with
is

A, E
B,

the circle which passes through must pass through

...

E. D
:
is,

C,
B,
A,

that are concyclic. Q. D.

D

EXERCISES ON THEOREM 40.

quadrilateral ABCD, making

In

circle
of
1.

1-6" radius inscribe

a

a
to

the angle ABC equal 126". Measure the remaining angles, and
hence verify this case that oppºsite angles are supplementary.
in

{}}
§
iá

Theorems 39 and 16, after first

of

Prove Theorem 40 by the

2.

joining the opposite vertices the quadrilateral.

of

the parallelo
be

parallelogram,
If

S3. circle can described about

a

a
be

gram must rectangular.

triangle, and XY drawn parallel

an

ABC
to
4.

is

isosceles the
is
Y:

C,
B,

base BC cutting the sides shew that the four points

in

and
X
X,

lie on circle.
Y

cyclic quadrilateral produced, the exterior angle

of
If
5.
is Y.

one side
is
a

equal the opposite interior angle the quadrilateral.

of
to
164 GEOMETRY.

THEOREM 41. [Euclid III. 31.]

The angle in a semi-circle is a right angle.

wº

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 to prove that the 4. ACB is a

ri.
is required angle.

The the O* half the straight angle

at

1st Proof. ACB

is
4

the centre, standing

at

AOB on the same arc ADB

angles: ;
rt.

and straight angle two

is =
a

rt.

angle. Q.E.D.
...

the ACB
4.

2nd Proof. Join OC.

Then because OA OC,
= =
...

the OCA the Z.OAC.

Z.

Theor.
5.

And because OB OC,

= = =

... the OCB the OBC.

4. A.
4
...

the whole ACB the OAC the OBC.

A.
+
/

rt.

But the three angles the AACB together angles;

of

two
=

rt.

angles
...

of

the ACB one-half two

= =
4

rt.

one angle. Q.E.D.

 ANGLE PROPERTIES. 1.65

CoROLLARY. 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.

O .
D

The 4. ACB at the O* is half the 4 *AOB at the centre, on the

same are ADB.
(i)

the segment ACB greater than semi-circle, then

If

is

a
ADB minor are
;
is
a

angles;

rt. rt.
two
... ...

the AOB less than

4. 4.

is is

the ACB less than one angle.

(ii) the segment ACB less than semi-circle, then ADB

If

is

major arc a *
is
a

greater rt. rº, angles;

... ...

the 4-AOB than two

is is

the ACB greater than one angle.

4

EXERCISES ON THEOREM 41.

on

circle described the hypotenuse

A
I.

as
right-angled triangle
of
a

diameter, passes through the opposite angular point.

Two circles intersect and B; and through
at

two diameters
2.

A
*

AP, AQ are drawn, one that the points B,

P,
each circle:
in

Q
shew
are collinear.
on

an

the equal sides

of

circle
of
3.

is

described one isosceles

A

triangle passes through the middle point

as

of

diameter. Shew that

it

the base.
s

triangle
as

Circles described on any two sides

of
4.

diameters
a

on the third side, the third side produced.

or

intersect
straight rod given length slides between two straight rulers.
of to of
at A
5.
.

a its

placed right angles one another; find the locus middle point.
of
of

the middle points

of

Find the locus circle drawn

6.

chords
through fixed point. Distinguish between the cases when the given point
a

or

within, on,
...

without the circumference.

is

...•** * ~
,’

figure
of

DEFINITION.
A

sector circle
is
º

\
&

a
by

bounded two radii and the arc intercepted

*

*!

-
*

between them.
I66 GEOMETRY.

THEOREM 42. [Euclid III. 26.]

In equal circles,
arcs which Subtend equal angles, either at the
centres or at the circumferences, are equal.

Let ABC, DEF be equal circles, and let the 4. BGC = the 4 EHF
at the centres; and consequently , s \sº

38.
the ABAc=the LEDF at the O*. Theor.

required BKC ELF.

It

prove that the arc the arc

to

=
is

Proof. Apply the DEF,

so

that the centre

to
G)

G)

ABC the

G
H,
on

falls the centre and GB falls along HE.

Then because the BGC the EHF,
=
A

GC will fall along HF.

...

And because the circles have equal radii,

on
E,

will fall
B
F,

the circles will coincide

of

and on and the circumferences

C

entirely.
ELF;
...

the arc BKC must coincide with the arc

... the arc BKC the arc ELF.
=

Q.E.D.
In

COROLLARY. equal circles Sectors which have equal angles

t

are equal.
-

clear that any theorem relating

to
It

arcs, angles,
in is

Obs.
be

true
in

and chords equal circles must also the same circle.

 ARCS AND ANGLES. 167

THEOREM 43. [Euclid III. 27.]

In equal circles angles,
either at the centres or at the circum
ferences, which stand on equal arcs are equal.

Let ABC, DEF be equal circles;

and let the arc BKC = the arc ELF.

It is required to prove that

the

the

the

LBGC EHF centres;

at at
= =

A. 4

also the BAC the EDF the O*.

4

Proof. Apply the DEF, that the centre

so
G)

to

G)

ABC the
G
H,
on

falls the centre and GB falls along HE.

Then because the circles have equal radii,

and the two O* coincide entirely.
on
E,

falls
B
‘.

And, by hypothesis, the arc BKC the arc ELF.

=
on

on

and consequently GC
F,

falls HF
...
C

EHF.
...

the BGC the

=
A

And since the O* half the the centre;

at

at

BAC the
A.

BGC
A.
=

and likewise the EDF half the EHF

A.

;
the
.

2BAC-the 2EDF, Q.E.D.

.

SN
~...
3.
f-
*,
*.
cos•.

-
168 GEOMETRY.

THEOREM 44. [Euclid III. 28.]

by
In

off
epual circles, arcs which are cut equal chords are equal,
the major arc equal the major arc, and the minor the minor.

to

to
D
be

Let ABC, DEF equal circles whose centres are and

H
;
and let the chord BC the chord EF.
=

required
It

prove that
to
is

the major arc BAC the major arc EDF,

= =

and the minor arc BKC the minor arc ELF.

Join BG, GC, EH, HF.

Proof. the A* BGC, EHF,

In

EH, being radii equal circles,

of

BG
= = =

because GC HF, for the same reason,

and BC EF, by hypothesis;
|

7.

EHF Theor.
...

the BGC the

=
A

the are ELF; Theor. 42.

...

the arc BKC

=

and these are the minor arcs.

But the whole O* ABKC the whole O* DELF;
=

the remaining arc BAC the remaining

...

arc EDF:
=

and these are the major arcs. Q.E.D.

 ARCS AND CHORDS. 169

45.
THEOREM [Euclid III. 29.]
!

off
In

equal circles chords which cut equal arcs are equal.

be

Let ABC, DEF equal circles whose centres are and

H
;
and let the arc BKC the arc ELF.
=

EF.
It

required prove that the chord BC

to

=
is

the chord

Join BG, EH.

*

Proof. Apply the G)ABC DEF,

on
to

that falls
G)

so

the
G

H
and GB along HE.

Then because the circles have equal radii,

and the O* coincide entirely.
on
E,

falls
...
B

And because the arc BKC the arc ELF,

=

falls on
...

F.
C

the chord BC coincides with the chord EF;

...

the chord EF.

...

the chord BC
=

Q.E.D.
170 GEOMETRY.

EXERCISES ON ANGLES IN A CIRCLE.

1. P is any point on the arc of a segment of which AB is the chord.

Shew that the sum of the angles PAB, PBA is constant.

X 2. PQ and RS are two chords of a circle intersecting at,X: prove

that the triangles PXS, RXQ are equiangular
A
to one another.
****)
> 3. Two circles intersect at A and B; and through A any straight
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 arcs PX, QY subtend equal angles at B.

7 5. P is any point on the arc 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 arcs cut off by two chords of a circle at right
angles to one another is equal to the semi-circumference.

* 9. If
AB is a faced chord of a circle and P any point on one of the
it,

arcs cut off by the angle APB cuts the conjugate arc
of of

then the bisector

the same point for all positions
in

P.

AB, AC are any two chords circle; and

P,
Q
of

10. are the middle

a

points the minor arcs cut off by them joined, cutting AB

of

PQ
if

is
;
Y,

and AC shew that AX=AY.

in

in
X

triangle ABC circle, and the bisectors

X,in

of

11. inscribed the

A

is

Y,a
at

angles meet the circumference Show that the angles

Z.

of

the triangle XYZ are respectively

2. B
5’ A

C
e

90 90 90
T2'
~
T

and B; and through these points

at

12. Two circles intersect

P A
*

lines are drawn from any point

on

of

of

the circumference one the

circles: shew that when produced they intercept on the other circum
ference an arc which constant for all positions
P.
of
is
 EXERCISES ON ANGLES IN A CIRCLE. 171

13. The straight lines which join the extremities of parallel chords
in a circle towards the same parts, (ii) towards opposite parts, are

(i)
equal.

A,
14. Through point two equal circles, two

of

of
intersection

a
straight lines PAQ, XAY are drawn; shew that the chord PX equal

is
to

the chord QY.

15. Through the points two circles two parallel

of
of
intersection
straight lines are drawn terminated by the circumferences: shew that
the straight lines which join their extremities towards the same parts
equal.
re

and B; and through any

at
16. Two equal circles intersect

A
straight line PAQ drawn terminated by the circumferences:
is

shew
that BP= BQ.

17. ABC an isosceles triangle inscribed circle, and the

is

in
S

at a
the base angles meet the circumference

Y.
Shew
of

bisectors and

X
that the figure BXAYC must have four its sides equal.
of

What relation must subsist among the angles the triangle ABC,

in
equilateral? of
be

order that the figure BXAYC may

18. ABCD cyclic quadrilateral, and the opposite sides AB, DC

to is
a

and CB, DA
P,
at

are produced
to

at

meet meet the circles

Q

R, if
at :
circumscribed about the triangles PBC, QAB intersect shew that
R,

points
P,

the are collinear.

Q

Q,
of P,

are the middle points triangle, and

of

of

19. the sides

X
: is
R

on a
let

the foot the perpendicular fall from one vertear the opposite side
R,

Q,
P,

-shew that the four points are concyclic.

X
10,
-. 2.;

p.

[See page 64, Ex. also Prob. 83.]

e * *
preceding points
of
to

20. Use the eacercise shew that the middle the

let

triangle perpendiculars fall from

of

of

sides and the feet the the

on a

vertices the opposite sides, are concyclic.

triangles are drawn standing on fixed base, and

of
If

21. series
a

having given vertical angle, shew that the bisectors the vertical
of
a

angles all pass through fixed point.

a

22. ABC triangle inscribed circle, and the middle point

in

E
is
a

a
on

the are subtended by BC through

of a of

the side remote from

E
if
A
:

diameter ED drawn, shew that the angle DEA half the difference
is
B is

C.
at

the angles and

172 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 :

P
the points
at

Let secant cut the circle

(i)

and Q, and suppose

to

recede from the

it
P

Ö
centre, moving always parallel its original
P to

position then the two points and will

Q
;

clearly approach one another and finally coin-

Q
cide.

Q
In

the ultimate position

Q

when and
P

become one point, the straight line becomes

a
at

tangent that point.

to

the circle

the points
at

(ii) Let secant cut the circle

a

and Q, and suppose

be

turned about
to
P

it

the point
so

that while remains fixed,

Q
P

moves on the circumference nearer and nearer

Then the line PQ its ultimate
P.

in
to

P,

position, with
is

when coincides
Q

tangent the point

P.
at

two points only,

at

Since secant can cut circle clear

is
it
a

that tangent can have only one point common with the
in
a

namely the point

at

circumference, contact, which two

of

as

points Hence we may define tangent

of

section coincide.
a

follows:
tangent straight line which meets
at to

circle
is
A
3.

the circumference one point only and though produced

;

indefinitely does not cut the circumference.

 TANGENCY, 173

TNC T I

\\

Fig.1. Fig. 2. 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 two circles cannot intersect in more than two points,

two circles, 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.

NoTE. When each of the circles which meet is outside the other, as
in Fig. 2, they are said to touch ome 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, T2P 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 Q T

Ilet PT be a tangent at the point P to a circle whose centre

is O.

It is required 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.
greater than the radius OP.
...

OQ
is is

true for every point PT;

in

And this
Q

OP PT.
to

the shortest distance from

...

O
is

Hence OP perp.
1,
to

PT. Theor. 12, Cor.

is

Q.E.D.

only one perpendicular

be

COROLLARY Since there can

1.

the point follows that one and only one tangent

to

P,

OP
at

it

given point
on
be

at

can drawn
to

circle the circumference.

a
a

only one perpendicular

be

COROLLARY Since there can

2.

the point follows that the perpendicular

to

at

P,

PT
to
it

a
its

point contact passes through the centre,

of

tangent
at

only one perpendicular

be

COROLLARY Since there can

3.

the line PT, follows that the radius drawn perpen

to

from
it
O

the tangent passes through the point

of

dicular
to

contact.
 TANGENCY, 175

THEOREM 46. [By the Method of Limits.]

The tangent at any point of a circle is perpendicular to the

Tadius drawn to the point of contact.

(Q)
Fig.1. Fig.2.

point
be

on

O.
Let circle whose centre

is
P

required perpendicular
at
It

prove that the tangent

to

to
is

is
P
the radius OP.

Let RQPT (Fig. secant cutting the circle

be

at
and

P.
Q
1)

Join OQ, OP.

Proof. Because OP OQ,

=

OQP
...

the the OPQ

=
Z

the supplements these angles are equal;

of
...

is,

that the OQR the OPT,

=
4

and this
to

true however near

is

is

P.
Q

turned about the point

let

be

P;so

Now the secant QP that

P

continually approaches and finally coincides with then

in Q

the ultimate position,

RT
(i)

at

the secant becomes the tangent Fig.

As
º
2
8.

(ii) OG) coincides with OP;

and therefore the equal "OQR, OPT become adjacent,
is /

to

OP perp.
...

RT. Q.E.D.

proof employed here

as

NotE. The method known

of

the Method
is

of 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 G) PQR in two points, since T is
without, and O is within, the G) PQR. Let P and Q be these
points.
Join TP, TG); OP, OQ.

Proof. Now each of the 4 "TPO, TGO, being in a semi

angle;
rt.

circle, is a
TP, TG, are perp. the radii OP, OQ respectively.
...

to

TP, TG, are tangents

..".

Q.
at

and Theor. 46.

P

Q.E.D.

The two tangentsº

an

COROLLARY. circle from eſºternal

a

point are equal, and subtend equal aºles

at

the centre.

For the A*TPO, TGO,

in

the "TPO,
TCAO are right angles,
4

because {the hypotenuse TO common,

is

and OP=OQ, being radii;

TP TQ,
Z, ...

= =

and the TOP the TOQ. Theor. 18.

4.
 TANGENCY, 177

EXERCISES ON THE TANGENT.

(Numerical and Graphical.)

1. Draw two concentric circles with radii 5-0 cm. and 3:Q 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 l'6" in
length. Shew that they all touch a concentric circle, and find its
radius.

3. The diameters 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

on
lie

the angle between them.

of

lines must the bisector

º
to

AB and AC are two tangents circle whose centre

7.

is
shew
O
a

;
that AO bisects the chord contact BC right angles.
at
of

PQ joined the figure Theorem 47, show that the angle

If

in

of
8.

is

PTQ double the angle OPQ.

is

Two parallel
on
9.

to

circle intercept any third tangent

a

segment which subtends right angle

at

the centre.
a

circle bisects all chords which are parallel

to
of

10. The diameter

a

either extremity.
at

the tangent
given
of

of

11. Find the locus the centres all circles which touch
a

8traight line given point.

at
a

12. Find the locus of theºcontres all circles which touch each of
of

two parallel straight lines.

13.

all
of

of

of

Find the locus the centres circles which touch each two
,

intersecting straight limes length.

of

unlimited
any quadrilateral circumscribed about circle, the sum
of
In

14. one
a

pair opposite sides equal the other pair.

of

of
to

the sum
is

State and prove the converse theorem.

quadrilateral circle, the angles sub
If

15. described about

is
a

the centre by any two opposite sides are supplementary.

at

tended
II. 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.
*
T

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.

2.

Theor.
...

O,

st.
is,

the points
P,

in

That and are one line. Q.E.D.

Q
is (i)

two circles touch eaternally the distance

If

COROLLARIES.
of

between their centres equal their radii.

to

the sum

(ii) two circles touch internally the distance their

If

between
of

equal their radii.

to

the difference
is

centres
 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 l'7" and 0:9"
respectively. Why and where do these circles touch one another ?
If circles of the above radii are drawn from centres 0-8" 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 cm., b=7 cm., and c=6 cm.
From A, B, and C as centres draw circles of radii 2.5 cm., 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 cm. a circle is drawn. What must be
the radius of a circle drawn from centre B to touch the first circle 2

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

(Theoretical.)

straight line drawn through the point

of

P of

contact two circles

A

is
6

B,

cutting the circumferences re

at

whose centres are and and

Q
A

spectively; shew that the radii AP and BQ are parallel.

#

Two circles touch externally, and through the point

of
7.

contact
a

straight line drawn terminated by the circumferences; shew that the

is

tangents its extremities are parallel.

at

of

X8. Find the locus all circles

of

the centres
given point
at

which touch given circle

(i)

(ii) which are given radius and touch given circle.

of

From given point

as

circle given
to
9.

centre describe touch

a

circle. How many solutions will there be?

10.

of

circle radius
to

Describe touch given circle

of

radius
a

given point. How many solutions will there be?

at
a
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 Segments of the circle.

E -5 F

Let EF touch the G) ABC at B, and let BD be a chord drawn

from B, the point of contact.

It is required to prove that

BAD;
(i)

FBD the angle the alternate segment

in in
4-

= =

the
(ii) the EBD the angle the alternate segment BCD.
be 4

the diameter through any point

B,

Let BA

in
and the
C

the segment which does not contain

of

arc
A.

Join AD, DC, CB.

rt.

Proof. Because the LADB angle,

in

semi-circle
is
a

the "DBA, BAD together angle.

rt.
...

=
4

But since EBF tangent, and BA diameter,

is

a
= A. a

rt.

angle.
...

the FBA
is
a

FBA the "DBA, BAD together.

the
...

Take away the common DBA,

A.

FBD the BAD, which the alternaté segment.

in

then the
A.
4-

is
=

Again because ABCD cyclic quadrilateral,

is
a

the supplement
of of
...

the BCD the BAD

A.

4- A.
= = =

the supplement the FBD

the EBD
A.

BCD, which the alternate segment.

in
...

the EBD the

is
A.
=
4

D.

Q.E
 ALTERNATE SUCGMENT. 181

EXERCISES ON THEOREM 49.

1. In the figure of Theorem 49, if the 4- FBD =72°, write down the
values of the 4 "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, (ii) 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 arc 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
which AB is the chord; and let PAT’ be any
secant through A. Join PB.
Then the A BCA= the Z-BPA;

º
Theor. 39.
this is true however mear P approaches
to A.
If P moves up to coincidence with A,
then the secant PAT’ becomes the tangent
AT, and the A. BPA becomes the A- BAT.
..., ultimately, the A. BAT = the A. BCA, in
the alt. segment.]
2. From Theorem 31, prove by the
Method of Limits that
its

The Straight line drawn perpendicular to the diameter of a circle at

extremity tangent.
is
a

from the property that the line

48

of
3.

Deduce Theorem centres

right angles.
at

bisects common chord

a

49

5,

from Ex. page 163.

5. 4.

Deduce Theorem
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
Tesults 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 assuming 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. -

its
Given a circle, or an arc of a circle, to find centre.

is be

an

of
Let ABC arc circle

a
whose centre to be found.

Construction. Take two chords

AB, BC, and bisect them right

at
by

angles the lines DE, FG, meeting

at O. Prob.

2.
Then the required centre.
is
O

Proof. Every point DE equi.

in

is

distant from and Prob. 14.

B.

>~
A

And every point FG equidistant

in

C.
from and
is

C. B
equidistant B,
A,
... ...

from and
is is
O O

of

the centre the circle ABC. Theor. 33.

PROBLEM 21.

To bisect given arc.

a

the given arc

be

be
to

Let ADB bisected.

g
it ©

e
&

Construction. Join AB, and bisect

at
by

right angles
at

CD meeting the arc

D.

Prol. 2.,
Then the arc
A

B
D.

bisected at
is

Proof. Join
X

DA, DB.
B;

Then every point

on

CD equidistant from and

is

Prob. 14.
-

...

DA DB
= =

6.

Theorem
...

the DBA the DAB

4.
4

the arcs, which subtend these angles the O*, are equal
at
...

;
is,

-
that the arc DA the arc DB.
=
184 GEOMETRY.

PROBLEM 22.

To draw a tangent to a circle from a given external point.

of

let
to its

at
Let PQR be the given circle, with centre and

T
the point from which
be

be
tangent drawn.
is
a

Construction. Join TO, and on describe semi-circle TPO

it

a
to cut the circle at
P.

Join TP.
-

Then TP the required tangent.

is

Proof. Join OP.

Then since the TPO, being semi-circle, rt. angle,
in

is
Z

a
a

TP right angles
to

the radius OP.

... at
...

is

TP Theor. 46.
at

tangent
P.
is
a

TO,
be

on

Since the semi-circle may

in of

described either side

T,
be

as

second tangent TQ can drawn from shewn the

a

figure.

NoTE. Suppose the point approach the given circle, then the
to
T

angle PTG gradually increases. When reaches the circumference,

T

the angle PTQ becomes straight angle, and the two tangents coincide.
a

be
no

p.

When enters the circle, tangent can drawn. [See Obs. 94.]
T
 COMMON TANGENTS. 185

PROBLEM 23.

To draw a common tangent to two circles.

its its
Let A be the centre of the greater circle, and a radius;
be

and let the smaller circle, and

of

the centre radius.

B

b
at

E.
Analysis. Suppose DE and

D
to

touch the circles

Then the radii AD, BE are both perp. DE, and therefore
to

º
to

par' one another.

Now BC were drawn par' DE, then the fig, DB would --
to
if
be

so

rectangle, that CD BE =b.

=
a

AD, BE are AB,

of
us b, on

And
if

the same side

rt.

then AC and the ACB angle.

is
=
–

4
a

These hints enable

to

draw BG first, and thus lead

to

the
following construction.
A,

With centre
to

Construction. and radius equal the

the radii the given circles, describe circle,
of

of

difference
a

and draw BC to touch it.

-

Join
in at

AC, and produce

to

D.

meet the circle (A)

it

Through BE par'
to

draw the radius AD and the same

B

Sense. Join DE.

Then DE common tangent the given circles.
to
is
a

be
as

Since two tangents, such BC, can general

in

Obs.
drawn from construction, this method will
to

of

the circle
B

to

furnish two common tangents the given circles. These are

,

called the direct common tangents.

186 GEOMETRY.

PROBLEM 23. (Continued.)

<2

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 rt.
AC = AD + DC = a + b ; and, as before, the 2 ACB is a angle.

Hence the following construction.

A,

With centre and radius equal

to

Construction. the sum

the radii the given circles, describe circle, and draw
of

of

BC to touch it.
Then proceed the first case, but draw BE
in

in
as

the sense
to

opposite AD.
be

As before, two tangents may

to

Obs. drawn from the

B

be

hence two common tangents may

of

circle construction
;

the given circles.

to

thus drawn These are called the

transverse common tangents.
an
as

[We leave the arrangement the proof

to

of

exercise the student

synthetic form.]
in

*
 COMMON TANGENTS. 187

EXERCISES ON COMMON TANGENTS.

(Numerical and Graphical.)

1. How many common tangents can be drawn in each of the

following cases?
when the given circles intersect;
(i)

(ii) when they have external contact;

(iii) when they have internal contact.
Illustrate your answer by drawing two circles

0"
radii l’4” and

of

1
respectively,
the centres;
(i)

with l'0" between

(ii) with 2'4" between the centres; ºf
(iii) with 0'4" between the centres;
(iv) with 3'0" between the centres.

Draw the common tangents each case, and note where the general
in

construction fails,
or

*
is

modified.

Draw two circles with radii 2.0" and 0.8", placing their centres
2.

2:0° apart. Draw the common tangents, and find their lengths between
the points contact, both by calculation and by measurement.
of

Draw all the common tangents to two circles whose centres are
3.

l'8" apart and whose radii are 0.6" and 12" respectively. Calculate and
measure the length the direct common tangents.
of

radii l'7" and 10" have their centres 2-1" apart.

of

Two circles
4.

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 bisects the common tangents.
it

Draw two circles with radii l'6" and 0.8" and with their centres
5.

3'0" apart, Draw all their common tangents.

Draw the direct common tangents two equal circles.
to
6.

Theoretical.
(

the two direct,

or

the two transverse, common tangents are

If
7.

of to

drawn two circles, the parts the tangents intercepted between the
of

points contact are equal.

four common tangents are drawn two circles external
If

to

to
8.

one another, shew that the two direct, and also the two transverse,
tangents intersect on the line
of

centres.
A,

Two given circles have external contact direct common

at
9.

and
a

shew that PQ subtends

at

tangent drawn
Q
to

touch them and

P
is

a
,

right angle the point

at

A.
188 GEOMETRY.

ON THE CONSTRUCTION OF CIRCLES.

(i)
In order to draw a circle we must know the position

of
the centre, (ii) the length

of
the radius.
(i)

To find the position the centre, two conditions are

of
needed, each giving

in on

so
locus which the centre must lie

;
that the one more points
or which the two loci intersect

on
as
are possible positions the required centre, explained

of
page 93.

(ii) The position the centre being thus fixed, the radius
of

we know (or can find) any point

on
determined the
if
is

circumference.

circle three independent

in

to

Hence order draw data are

a

required.

For example, we may draw circle we are given

if
a

the circumference;
on

three points
(i)

(ii) three tangent lines;

or or

(iii) one point and its point

on

the circumference, one tangent,

of contact.

be
It

will however often happen that more than one circle can drawn
satisfying three given conditions.
*

Before attempting the constructions

of

the next Exercise the

student should make himself familiar with the following loci.

through two
of

of
(i)

The locus the centres circles which pass

given points. *

given straight
of

of

(ii) The locus the centres circles which touch

a

given point.
at

line
a

of

of

(iii) The locus the centres circles which touch given circle
a

given point.
at
a

given straight
of

of

(iv) The locus the centres circles which touch

a
#

line, and have given radius.

a

of

of

(v) The locus the centres circles which touch given circle,
a

and have given radius.

a

the
of

of

(vi) The locus centres circles which touch two given

straight lines.
 THE CONSTRUCTION OF CIRCLES. 189

EXERCISES.

1. Draw a circle to pass through three given points.

2. If
a circle touches a given line PQ at a point A, on what line
must its centre lie Ż
If a circle passes through two given points A and B, on what line
must its centre lie Ż
Hence draw a circle to touch a straight line PQ at the point A, and
to pass through another given point B.

3. If
a circle touches a given circle whose centre is C at the point A,
on what line must its centre lie Ż
Draw a circle to touch the given circle (C) at the point A, and to pass
througha given point B.

4. A point P is 4.5 cm. distant from a straight line AB. Draw two
circles of radius 3:2 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 6-0 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 2

6. If a circle touches two straight lines OA, OB, on what line must
its centre lie 2
Draw OA, OB making an angle of 76°, and describe a circle of radius
l'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. [See 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 2

[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 line, and the given angle.

st.

required circle containing

of
on
It

AB segment
to
is

describe a
a
an

C.

angle equal
to

Construction. At BA, make the BAD equal

in

to the Z.O.
4.
A

From draw AG perp.

to

AD.
A

by

angles FG, meeting AG

rt.

G.
in
at

2.
Bisect AB Prob.

Proof. Join GB.

B;

Now every point equidistant from

in

FG and
is

Prob. 14.
GA GB.
=

.*.

With centre circle, which must

G,

and radius GA, draw

a

Theor. 46.
B,

pass through
at

and touch AD
A.

an
to

Then the segment AHB, alternate the BAD, contains

4.

angle equal Theor. 49.

C.
to

the particular case when the given angle

rt.
In

NOTE. angle, the

is
a
be

on

as

segment required will the semi-circle AB diameter. [Theorem.41.]

 PROBLEMS. 191

to off
COROLLARY. To cut from given circle segment containing

a
a
given angle, enough draw tangent the circle, and from

to
is
it

a
a

an
the point of draw chord making with the tangent

to
contact

a
angle equal to
the given angle.

was proved on
page 161 that
It

triangles which stand

of

of

on
The locus the vertices the same base
the segment standing

of
and have given vertical angle, the arc

is
a
on

an
this base, and containing angle equal the given angle.

to
The following Problems are derived from this result by the
Method Loci [page 93].
of

of

Intersection

EXERCISES.
on

triangle given base having given vertical angle

1.

Describe
a

a
on
its

and having verteac given Straight line.

a

triangle having given the base, the vertical angle, and

2.

Construct
a
(i)

one other side.

(ii) the altitude.
(iii) the length
of

the median which bisects the base.

the perpendicular from the verteac the base.
of

(iv) the foot

to

triangle having given the base, the vertical angle, and

3.

Construct
a

by

the point the vertical angle.

it, of
at

which the base cut the bisector

is

the given point

be

[Let AB the base, the given angle.

in

and
X

circle containing K;
an

On AB describe segment angle equal

is at to
of
a

a
by

gomplete the Oce drawing the are APB. Disect the are APB P:
join PX, and produce meet the O* Then ABC
C.
to

at

the
it

required triangle.]
triangle having given the base, the vertical angle, and
4.

Construct
a

the remaining sides.

of

the sum
be

[Let AB the given base, the given angle, and line equal
to
H
K

segment containing
an

angle
K,of

the sum the sides. On AB describe

a

also another segment containing

an

equal angle equal half the

to

to
H,
at A,

With centre circle cutting the are

K.

of
A.

and radius describe

a

the latter segment and Y. Join AX (or AY) cutting the arc
of

the
X

first segment Then ABC the required triangle.]

C.
at

is

5. Construct triangle having given the base, the vertical angle, and
a

the remaining sides.

of

the difference
. 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,

sia, sides Hexagon,
25 92
;} Seven sides 25 Heptagon,
33 eight sides 9) Octagon,
53 ten sides 75 Decagon,
22 twelve sides 33 Dodecagon,
25 fifteen sides » Quindecagon.
all

A Polygon is Regular when

its

2. sides are equal, and

all

its angles are equal.

in
be
to

rectilineal figure said

is
3.

its
of all

circle, when angular points

in

scribed
a

are on the circumference the circle and

a
;

circle said to be circumscribed about recti

is

lineal figure, when the circumference

of

the
all

circle passes through the angular points

of

the figure.

circle said to be inscribed in

A
4.

is

of a

rectilineal figure, when the circumference

by

the figure;
to of

the circle touched each side

is

be

and rectilineal figure said circum

is
a

circle, when each side

of

scribed about the

a

sº-º
figure tangent
to

the circle.
is
a
 PROBLEMS ON TRIANGLES AND CIRCLES. 193

PROBLEM 25.

To circumscribe a circle about a given triangle.

Let ABC be the triangle, about which a circle is to be

drawn.
rt.
Construction. Bisect AB and AC at angles by DS and
Prob.

2.
ES, meeting
at
S S.

Then the required

of

the centre circle.

is

Proof. Now every point equidistant

in

DS from
is

A
and B; Prob. 14.

and every point equidistant

in

ES from and
is

C
C. A

;
B,

equidistant from
A,
...

and
is
S, S

With centre and radius SA describe circle this will

a

;
is,

pass through
C,

and and therefore, the required circum

B

circle.

the given triangle

be

will found that acute

It

if

is

Obs.
angled, the centre the circum-circle falls within it;
of

it
if
on

right-angled triangle, the centre falls the hypotenuse:

if is
it a

an

obtuse-angled triangle, the centre falls without the

is

triangle.
94

to

NoTE. From page seen that joined the middle

is

is
it

if
S

point BC, then the joining line perpendicular

of

BC.
to
is

Hence the perpendiculars triangle from their

of
to

drawn the sides

a

middle points are concurrent, the point intersection being the centre
of

of the circle circumscribed about the triangle.

S.

H. 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.

the 4 " ABC, ACB by the

st.
Construction, Bisect lines Bl,
Prob.

1.
at

Cl, which intersect

1.

Then the required circle.

of

the centre
is
I

Proof. From draw ID, IE, perp. BC, CA, AB.

to
IF
1

Then every point equidistant from BC, BA; Prob.

BI

15.
is
in

.*. ID = |F.
And every point equidistant from CB, CA;
Cl
in

is

ID |E.
=
"
.

ID, IE, are all equal.

IF
‘.

With
ID

centre and radius draw circle

E a

F. ;
I

this will pass through the points and

Also the circle will touch the sides BC, CA, AB,
are right angles.
E,
D,

because the angles

at

the AABC.
in

DEF
G)

the inscribed
is
...

From II., joined, then

Al
96

Al
p.

seen that bisects

is
if
it

NOTE.
is

the angle BAC hence follows that

it
:

of

of

the point
of

The bisectors the angles triangle are concurrent,

a

intersection being the centre of the inscribed circle.

DEFINITION.
triangle and the other
of

circle which touches one side

A

a
an

two sides produced the triangle.

of

called escribed circle

is
 PROBLEMS ON TRIANGLES AND CIRCLES. 195

PROBLEM 27.

To draw an escribed circle of a given triangle.

i
!
A B F 'TD
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 AECBD, BCE by the

st.
lines Bl,
11.

Cli which intersect

at

Then the required

of

the centre circle.

is
li

IG, perp. AD, BC, AE.

F,
11

Proof. From draw

to
I,
H
is I,
Bl,

Then every point equidistant from BD, BC;

in

Prob. 15,
liF lig.
=
...

Similarly 1,6=l,
all H.

equal.
G,

11F, 11B are

I,
.

With and radius 11F describe circle;

1,

centre
a

this will pass through the points

H.

and
G

Also the circle will touch AD, BC, and AE,

G,

because the angles angles.

rt.
F,
at

are
H

an escribed circle of the AABC.

G)

... the FGH

is

NoTE clear that every triangle has three escribed circles.

It
1.

is

as

Their centres are known the Ex-centres.

II., page 96, that Ali joined,
be

as

NoTE may shewn,

in
It
2.

is
if

then Ali bisects the angle BAC; hence follows that

it

two eacterior angles triangle and the bisector

of

of

ofof

.The bisectors the

a

third angle are concurrent, the point 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

Let ABC be the given circle, and DEF the given triangle.

Analysis. A A ABC, 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 A-B = the 4-E, and
the 4 C = the A. F.; for then the Z A = the A. D. Theor. 16.

Now the A- B, 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 A. HAC = the A. E.;
and similarly, the 4. GAB = the A. F.

Reversing these steps, we have the following construction.

Construction. At any point A on the O* of the G) ABC

draw the tangent GAH. Prob. 22.
At A make the A. GAB equal to the 4- F,
and make the A. HAC equal to the A. 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 circumscribe a triangle equiangular to

a given triangle.

C
A #

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 A. M = the 4-E, the A N = the 4. F, and consequently,
the A. L = the A. D.

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

is,
drawn if we knew the relative positions of KA, KB, KC, that
we knew the BKA, BKC.
z*
if

rt.

Now from the quad' BKAM, since the and are

×
B

',
A
4

the BKA
=

180° 180°
M
– –

– –
E
A

similarly the BKC

F.
=

180° 180°
N
4

Hence we have the following construction.

Construction. EF both ways

H.
to

Produce and
G

Find ABC,
G)
of

the centre the

K

and draw any radius KB.

At make the BKA equal
to to

the DEG
Z. Z.
K

Z. A

and make the BKC equal the DFH.

Through draw LM, MN, NL perp.
B,

KA, KB, KC.

A,

to
C

Then LMN the required triangle.

is

[The student should now arrange the proof synthetically.

]
198 GEOMETRY.

EXERCISES.

ON CIRCLES AND TRIANGLEs.

(Inscriptions and Circumscriptions.)

1. In a circle of radius 5 cm. 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 cm., and find by

calculation and measurement (to the nearest millimetre) the radii of
the inscribed, 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 :

=2'5", B=66°, C=50°;
(i)
a

(ii) =2'5", B=72°, C=44°;

a

(iii) =2'5", B=41°, C=23°.

a

Circumscribe circle about each triangle, and measure the radii

a

to the nearest hundredth of an inch. Account for the three results

being the same, by comparing the vertical angles.

cm. inscribe an equilateral triangle.

In

circle radius
of
4.

4
a

the nearest millimetre; and verify

to

Calculate the length its side

of

by measurement.
Find the area the inscribed equilateral triangle, and shew that

it
of

one quarter equilateral triangle.

of

the circumscribed
is

the triangle ABC, the length

of In

the centre, and of the

is
5.

if

r
I

radius the in-circle, shew that

15C =#ar; ICA=$br; IAB =#cr.
A
A
A

Hence prove that ABC =# +b+c)

A

(a

r.

Verify this formula by measurements for triangle whose sides are

a

cm., cm., and cm.

7
9

A,

opposite prove that

of

to
If

the radius the ex-circle

6.

ri
is

AABC=# (b+ a)ri.

–
c

cm., b=4 cm., c=3 cm., verify this result by measurement.

If
=
5
a

by

the triangle ABC

of

in

Find measurement the circum-radius

7.

which 6.3 cm., 3.0 cm., and c=5.1 cm.

=
=

b
a

B,
A,

Draw and measure the perpendiculars from the opposite

to
C

p, p, ps, verify the following

by

their lengths are represented

If

sides.
statement

*
:

ab

*-.
bc

ca

circum-radius =::--
=

2p, 2p2 2ps

 PROBLEMS ON CIRCLES AND SQUARES. 199

EXERCISES. e

ON CIRCLES AND SQUAREs.

(Inscriptions and Circumscriptions.)

1. Draw a circle of radius l'5", and find a construction for inscribing

it.

a square in
Calculate the length an inch,
of
the nearest hundredth

to

of
the side
and verify by measurement.
the inscribed square.
of

Find the area

square about circle radius l'5", shewing all

of
2.

Circumscribe
a

a
lines of construction.
Prove that the area the square
of

is
circumscribed about circle

a
double that square.
of

the inscribed
on

square 7.5 cm., and state for

of
3.

Draw side construction

a

a
inscribing
it.

circle
in
a

Justify your construction by considerations symmetry.

of

Circumscribe square whose side

4.

circle about is cm.

6
a

millimetre, and test your

to

Measure the diameter the nearest

drawing by calculation.
In

rectangle
of

of

circle radius l'8" inscribe which one side

5.

3 a

Find the approximate length

of

measures 0". the other side.

i.
Of

all rectangles inscribed the circle shew that the square has the
in

greatest area.

square and an equilateral triangle are inscribed

in
a 6.

circle.
A

a
If

and the lengths their sides, shew that

of
b

3a*=2b°.

circle, and any point

on

ABCD square inscribed the

in
7.

is

is
P
a

are AD; shew that the side AD subtends

an
at

as

angle three times

P

by any one
at
as

great
of

that subtended the other sides.

P

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

a

given circle.
8.

Circumscribe rhombus about

a

given square ABCD,

so

Inscribe square that one its

a in

of
9.

angular points shall given point

be
at

AB.
in
X
In

given square inscribe the square

of

10. minimum area.

a

ll. circle, (ii) square about given rectangle.

(i)

Describe
a

Inscribe circle, (ii) square given quadrant.

in
(i)

12.
a

a
 *
*

200 GEOMETRY.

ON CIRCLES AND REGULAR POLY GONS.

PROBLEM 30.

(i)
To draw a regular polygon (ii) about

in
given circle.

a
Let AB, BC, CD,

...
be consecutive
regular polygon inscribed

in
of

sides
a

circle whose centre O.

is
a

*
Then AOB, BOC, COD,

...
are con
gruent isosceles triangles. And

if
the polygon has sides, each

of
the
n

"AOB, BOC, COD, sº

...
=

º
A

Thus inscribe polygon given circle,"

(i)

in
to

of
sides

n
a

a
360°
an

This gives
at

angle AOB the centre equal

to
draw

side AB; and chords equal

be
the length AB may now

to
of
a

set off round the circumference. The resulting figure will

be

clearly equilateral and equiangular.

(ii) To circumscribe polygon sides about the circle,
of
n
... a

be
to C,
A,

the points before, and

D,
B,

must determined as
tangents drawn these points. The resulting
at

the circle
figure may readily
be

proved equilateral and equiangular.

NOTE. This method gives strict geometrical construction only when

a
sº
be

the angle can drawn with ruler and compasses.

EXERCISES.

Give strict constructions for inscribing

in
2. a 1.

circle (radius cm.)

4
a

regular hexagon; (ii) regular octagon; (iii) regular dodecagon.

(i)

About circle of radius 1:5" circumscribe

(i) a

regular hexagon; (ii) regular octagon,

a

Test the constructions by measurement, and justify them by proof.

An equilateral triangle and regular hexagon are inscribed
in
3.

given circle, and denote the lengths their sides: prove that
of

and
b
of a

triangle=# (area hexagon); (ii) a”=3b*.

(i)

of

area
By means your protractor inscribe regular heptagon
in
2".of
4.

its angles; and

of

circle radius Calculate and measure one

of
of .

measure the length side.

a
 *
PROBLEMS, ON CIRCLES AND POLYGONS. 201

PROBLEM 31.

(i)
To draw a circle (ii) about

in
regular polygon.

a
Let AB, BC, CD, DE,

E
be
...
con

of
secutive sides regular polygon

a
of sides.
n

Bisect the "ABC, BCD by BO,

at 4

D
ſ
CO meeting
O.

Then
is

the centre both

of
O

the
inscribed and circumscribed circle.

C
Join OD; and from the
of

Outline Proof. congruent

:

A*OCB, OCD, shew that OD bisects the

CDE.

Z.
Hence we
conclude that
All
of

of

the bisectors the angles

O.
the polygon

at
meet
...;
(i)

Prove that OB OC OD from Theorem

=

is =

6.
Hence the circum-centre.
O

(ii) Draw OP, OQ, OR,

...

perp. AB, BC, CD,

to

...
Prove that OP= OQ OR ...; from the congruent
.
=

OBP,
is =

OBQ, ... A.
Hence
.

the in-centre.
O

EXERCISES.
regular hexagon
Draw
on
1.

of

side 2:0". Draw the inscribed

a

and circumscribed circles. Calculate and measure their diameters to

the nearest hundredth of an inch.

Shew that the area

2.

of

regular hexagon inscribed

in

circle
a

is
a

three-fourths
of

that
of

the circumscribed hexagon.

Find the area
of of

hexagon inscribed
10
in

to

circle radius
of
a a

cm.
a

the nearest tenth sq. cm.

ABC
an

isosceles triangle inscribed

If
3.

circle, having each

is

in
a
of

the angles and the angle

in of

double
B

shew that BC
A

of
is

side
;

regular pentagon inscribed the circle.

a

of
4.

On side cm. construct (without protractor)

a

4
(i)

regular hexagon; (ii) regular octagon.

a

In each case find the approximate

of

area 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

l
=
diameter

by all
and it
can be proved that this is the same for circles.

be
this ratio theory
A

more correct value

of

to
found

is
3.1416; while correct places to

of
decimals 3.1415926.

it
is
7
3}

Thus the value (or 3:1428) too great, and correct

to
is

2
places only.

its
The ratio which the circumference any circle bears

of

to
denoted by the Greek letter-T

so
diameter that
is

;
T.
= of =

circumference diameter
Or, denotes the radius the circle, ×
if
r

2irr;
2r

circumference
of x
t

Twe are
to

give one

in or
the values 3}, 3.1416,
to

where
3.1415926, according
to

the degree accuracy required

of

the final result.

NoTE. The theoretical methods by which tris evaluated any

to
be

required degree accuracy cannot

at

explained this stage, but its

of
be

value may easily verified by experiment two decimal places.

to

For example: round cylinder wrap strip

of

so

paper that the

a

ends overlap. At any point the overlapping area prick pin

in

through both folds. Unwrap and straighten the strip, then measure
the distance between the pin holes: this gives length the circum
of

the
ference. Measure the diameter, and divide the first result by the second.

Ex. From these data CIRCUMFERENCE, DIAMETER.

tr.
1.

VALUE OF
find and record the value
m.

of 16'0 cm. 5°l cm.

Find the mean of the 8.8% 2-8”
three results. 13.5" 4'3"

Ex. wound evenly round cylinder, and

2.

fine thread
is

it
is
a
20

found that the length required for complete turns The

is

75'4".
cylinder 1:2": find roughly the value
of

tr.

diameter the
is

of

Ex. bicycle wheel, diameter, makes 400 revolutions

in
3.

28"
travelling over 977 yards.
in

From this result estimate the value

tr.
of
 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 r. Then
we have
Area of polygon
= m,. /\ AOB
= n. #AB x OD
= }. mAB x r

= }(perimeter of polygon) x

r);
and this true however many sides the polygon may have.
if is

Now the number increased without limit, the

of

sides
is

be
perimeter and area
of

the polygon may

to
made differ from

by
the circumference and area quantities smaller
; of

the circle
of be

than any that can named hence ultimately

W
Area
}.
= = =

circle circumference x
r
27trix
,

r
,

r^2.

ALTERNATIVE METHOD,

Suppose the circle divided into any even number sectors having
m.of

equal central angles: denote the number

by
as of

sectors
by
of be

Let the sectors placed side represented the diagram

in

side
;

then the area the circle= the area the fig. ABCD
of

and this true however great may be.

as is

Now
of

the number sectors increased, each arc decreased;

is
is

so that AB, CD tend

(i)

the outlines become straight, and

to
,

rt.

(ii) the angles angles.

at

to

and
D

tend become
B
204 GEOMETRY.

Thus when n is increased without limit, the fig, ABCD ultimately

becomes a 2.ectangle, whose length is the semi-circumference of the circle,
and whose breadth is its radius.

of
Area circle=#. circumference radius

...

x
=#.2irrx ratrº.

.
THE AREA OF SECTOR.

A
C

1°,
an

two radii angle they cut off

of

circle make of
If

a
an
(i)

arc whose length sāg the circumference;

D of of
= =

and (ii) sector whose area sān the circle;

a

-:
the angle AOB contains degrees, then
‘..
if

-;
(i)

the circumference;
of

the arc AB
the

the
of

(ii)
of

the sector AOB area circle

*
(;

==
of

circumference radius)
x
}.

arc AB radius.
=

THE AREA OF SEGMENT.

A

by

minor segment subtracting from

of

The area found

is
a

the corresponding sector the area the triangle formed by

of

the chord and the radii. Thus

Area Segment ABC triangle AOB.
of

Sector OACB
=

by

The area major segment most simply found

of

is
a

subtracting the area the corresponding minor segment from

of

the area of the circle.

 CIRCUMIFERENCE AND AREA OF A CIRCLE. 205

EXERCISES.

[In each case choose the value of ºr 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 (ii) 100 cm.
(i)
4-5 cm.

Find square inch the areas

to

of

of
2.

the nearest hundredth the

a
circles whose radii are (ii) 10'6".
(i)
2.3".

two places
of
Find
to

of
3.

decimals the circumference and area

a
square whose side
in

circle inscribed 3:6 cm.

is
of a
In

square described: find

to
4.

is
circle radius 7-0 cm. the
a

a
nearest square centimetre the difference between the areas

of
the circle
and the square.

square inch the area

of

Find
to

of
5.

the nearest hundredth the

a

circular ring formed by two concentric circles whose radii are 5’7” and

Shew that the area ring lying between the circumferences

of

is of
6.

equal
of

two concentric circles

to

the area circle whose radius

is

the inner circle from any point on the outer.

to

the length tangent

of
a

rectangle 80

in
7.

is

whose sides are cm. and 6.0 cm. inscribed

A

a
Calculate square centimetre the total
to

of

circle. the nearest tenth

a

the four segments outside the rectangle.

of

area

Find an inch the side square whose

to

of

of
8.

the nearest tenth

a
5".

equal that
to

of

of
is

area circle radius

a

circular ring formed by the circumference

of

two concentric
9.

is
A

The area the ring 22 square inches, and its width 10";
of

circles.
is

is

taking
as

*.*, find approximately

of

the radii the two circles.

r

10. Find square inch the difference

to

of

the nearest hundredth

a

and inscribed circles of an

of

between the areas the circumscribed

4".

equilateral triangle each

of

is

whose sides
on

squared paper two circles whose centres are

'7"at

11. Draw the

(0,

points (l'5", '8"), and whose radii are respectively

0)

and and
1-0". Prove that the circles touch one another, and find approximately
their circumferences and areas.

radius l'0" having the point (1.6", l.2")

as
of

12. Draw circle

a

Also draw two circles with the origin

as

radii
of

centre. centre and

1'0" and 3.0" respectively.
of

Shew that each 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 I, S are the centres of the inscribed and

circumscribed circles; if A, ), S are collinear, shew that AB = AC.
4. The sum of the diameters of the inscribed and circumscribed
circlesof 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, 90-5.
6. If
I is the centre of the circle inscribed in the triangle ABC,
and I, the centre of the escribed circle which touches BC; shew that
11,

l, B, are concyclic.
In C

any triangle the difference two sides equal the

at to
of
7.

is is

the segments into which the third side divided

of

difference the
point
of

of

contact the inscribed circle.

the triangle ABC,
In

of

the inscribed and

8.

and are the centres

S
I

to
circles: shew that an angle cqual
IS

at
A.

circumscribed subtends
at

the angles the triangle.

of

half the difference

of

the base
drawn perpendicular BC, then
Al

Hence shew that AD

to

is

the
if

is

the angle DAS.

of

bisector
The diagonals quadrilateral ABCD intersect O; shew
at
of
9.

that the centres about the four triangles

of

the circles circumscribed

AOB, BOC, COD, DOA are the angular points parallelogram.
at

of
a

any triangle ABC,

In

the inscribed circle,

of

10. the centre

if
is
I

O; shew that
at

produced
Al

to

and meet the circumscribed circle

if

is

the circle circumscribed about the triangle BIC.

of
is

the centre
O

11. Given the base, altitude, and the radius

of

the circumscribed
circle; construct the triangle.
A,
B,

12. Three circles whose centres are touch one another

C
E,

externally two by two

D,

of

shew that the inscribed circle

at

F
:

the triangle ABC the triangle DEF.

of

the circumscribed circle

is
 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 are concurrent.

In the AABC, let AD, BE be the

perpº drawn from A and B to the opposite
sides; and let them intersect at O. A
F
Join CO; and produce it to meet AB at
F. E
O
It is required to shew that CF is perp.
to AB.
&
Join DE.
B D C
Z_*

Then, because the OEC, ODC are rt.

angles,
O,

C,
E,

the points are concyclic

...

in D

:
DEC= the 4-DOC, the same segment;
A.
‘..

the
the vert. opp. FOA.
=

/
Z."

Again, because the AEB, ADB are rt. angles,

E,
D,
A,

the points are concyclic

...

in B

DEB=the Z-DAB, the same segment.

A.
...

the
Z_*

Z_*

FOA, FAO DEC, DEB

of

of
...

the sum the the sum the

=

=a rt. angle
:
rt.

the remaining Z-AFO angle Theor. 16.

‘..

=
to a

:
is,

that CF perp. AB.

is
g

Hence the three perpº AD, BE, CF the point

O.
at

meet
E.

Q. D.

DEFINITIONS.
(i)

The intersection the perpendiculars

of

drawn from the

its

triangle the opposite sides

to
of

vertices called
is
a

orthocentre.

(ii) The triangle formed by joining the feet the perpen

of

diculars called the pedal orthocentric triangle.

or
is
208 GEOMETRY.

II. In 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.
A
In the acute-angled AABC, let AD, BE,
CF be the perpº drawn from the vertices to
the opposite sides, meeting at the Ortho

be
centre and let DEF the pedal triangle.
O

E
N
;

It is required to prove that

E
AD, BE, CF bisect respectively

(\
DEF, EFD.

%
the Z = F DE,

It may be shewn, as in the last theorem,

e
B

C
that the points O, D, C, E are concyclic ;
the Z.OCE, the same segment.

in
the 4-ODE
‘..

=
O,
D,
B,

Similarly the points are concyclic;

F

OBF, the same segment.

in
the 4-ODF
A.
...

the
=

OBF, each being the compt BAC.

A.
But the 4-OCE

of
A.

the the
=

ODE ODF.
A.
A.
‘..

the the
=

DEF, EFD
Z."

Similarly are bisected by

be

may shewn that the

it

BE and CF.

E.
Q. D.
of
(i)

CoRoDLARY. Every two sides the pedal triangle are equally

the original triangle which they meet.
of
to

in

inclined that side

ODE
of of

For the 4-EDC= the compt

4.

the
the compt the Z.OCE
= =

BAC.
A.

the

Similarly BAC,
be

may shewn that the Z-FDB

A.

the
=
it

EDC the FDB the Z.A.

A.
4-
...

the
=
=
be
In

like manner may proved that

it

DEC= the FEA the A-B,

A- A.
A. A.

the
= =

and the DFB the EFA the A.C.

=

CoRoll ARY. (ii) The triangles DEC, AEF, DBF are equiangular
to

the triangle ABC.

to

one another and

the angle BAC obtuse, then the perpendiculars BE, CF

If

NoTE.
is

bisect eacternally the corresponding angles the pedal triangle.

of
 THE ORTHOCENTRE OF A TRIANGLE. 209

EXERCISES.

l. If O is the orthocentre of the triangle ABC and if the perpendicular

AD is produced to meet the circum-circle in G, prove that OD=DG.

2. In an acute-angled triangle the three sides are the eacternal 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
ing angles of the pedal triangle.

3. If O is the orthocentre of the triangle ABC, shew that the angles

BOC, BAC are 8wpplementary.

4. If O is the orthocentre of the triangle ABC, them any one of the

four points O, A, B, C is the orthocentre of the triangle whose vertices are
the other three.

5. The three circles, which pass through two vertices of a triangle of and
its

orthocentre are each equal the triangle.

to

the circum-circle
D,

of
a 6.

are taken on the circumference semi-circle described

E

given straight line AB the chords AD, BE and AE, BD intersect

on

F :

perpendicular
at

(produced necessary) shew that FG

is
and
G
if

to AB.

ABC triangle, its orthocentre, AK the

of
7.

: is

and diameter
O
is
a

circum-circle shew that BOCK parallelogram.

is
a

triangle joined the middle point

of

The orthocentre
to

of
8.

is

the
is a

base, and the joining line produced meet the circum-circle prove
to

will meet the same point

as

that the diameter which passes

at
it

it

through the vertex.

The perpendicular from the vertex triangle

on

the base, and

of
9.

to a

the straight line joining the Orthocentre the middle_point

of

the
base, are produced meet the circum-circle shew that
to

at

and
Q
P

PQ parallel
to

the base.
is

of

of

10. The distance each vertea triangle from the orthocentre

to is
a

of

the perpendicular
of

double drawn from the centre the circum-circle

the opposite side.

ll.Three circles are described each passing through the Orthocentre

triangle and two its vertices: shew that the triangle formed by
of

of
a

joining their centres the original triangle.

to

equal all respects

in
is

12. Construct triangle, having given vertex, the orthocentre,

a

and the centre of the circum-circle.

H. S.G.I. -III,
O
210 GEOMETRY

LOCI.

III. Given the base and vertical angle of a triangle, find the locus of
its 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 Z. A
equal to the Z-X. F.
Draw the perp BE, CF, intersecting at E
the orthocentre O.
O
It is required to find the locus of O.
A_*

Proof. Since the OFA, OEA are rt.

C
angles,
O,

the points
A. F,
A,

are concyclic
‘..

;
the FOE the supplement

of
...

the 4-A
is

:
the vert. opp. BOC the supplement
4-

of
...

is
the 4-A.
But the Z.A constant, being always equal the A-X;

to
is

its supplement
...

is
constant

;
BOC has fixed base, and constant vertical angle;
is,

that the
A

segment

of
its vertex which BC
of

of

is
hence the locus the arc
O
is

the chord.

IV. Given the base and vertical angle

of
triangle,
of

find the locus

a

the in-centre.

any triangle on the given

be

Let BAC

/\
base BC, having its vertical angle equal
to

the given 4-X; and let Al, Bl, A

X
be
is Cl

the
its angles. Then
of

bisectors the in-

l

Centre.
of

required find the locus

It

to
is

l.

Denote the angles the AABC

of

Proof.
B,
A,

by and let the BIC be denoted

4.
C
;

by
B
I.

Then from the BIC,

A

rt,

angles; Theor. 16.

(i)

#B $C two
+

=
|

and from the AABC,

angles;
rt.

A+B+C two
=
so

(ii) that AA++B+ #C= one rt. angle.

..., taking the differences and (ii),
(i)

the equals
in
of

#A one rt. angle

=
= –
| |

A. :
rt.

Or, one angle

+

But constant, being always equal the A-X;

: to
A
is

of is

constant
i

on

segment the fixed chord BC

of
‘..

is

the locus the arc

a
l
 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

parallel 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

a 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 arc of constant length.

: 6.

and QB.

7.
A and B are two fixed points on the circumference of a circle,

§§ is any diameter: find the locus of the intersection of PA

BAC is any triangle described on the fixed base BC and having

a 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
anglesto 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 is
any other straight line similarly drawn : find the locus of the inter
Section of HP and QK.
212 GEOMETRY,

iº
SIMSON'S LINE.

V. of the perpendiculars
The feet drawn to the three sides of g
triangle from any point on

its
circum-circle are collinear.

any point
; be

on

of
Let the circum-circle
P
and let PD, PE, PF

F
be
the ABC the perps.
A

P
drawn from P to the sides.

D,
E,
required prove that the points

E
It

to
is

F
are collinear.
Join FE and ED

:
FE ED will be shewn to be

in
then and the
same straight line.
Join PA, PC.
*N 978
4–5

Proof. Because the PEA, PFA are rt. angles,

A,
E,
the points
P,

are concyclic
...

:
PEF=the PAF, inthe same segment
A.

4.
...

the
the suppº PAB
of

4.
the
==

PCD,
4A A, A.

the
C,

since the points

P,

are concyclic.
B

Again because the PEC, PDC are rt.

angles,
D,
E,

the points
= = P,

are concyclic.
... ...

PED the suppº

of of

the PCD
A.

4. 4.

the
the suppº the PEF.
FE and ED are in one st. line.
‘.

Obs. The line FED

or
as

of
known the Pedal Simson's Line the
is

triangle ABC for the point

P.

From any
º EXERCISES.

the triangle ABC,

on

the circum-circle
of
1.

perpendiculars PD, PF are drawn BC and AB; FD,

to

FD
or
to if
of E,

shew that PE perpendicular

at

produced, cuts AC AC.

is

point which moves

so

Find the locus that perpendiculars

2.

if
a

given triangle, their feet are

of

are drawn from

to

the sides
it

collinear.
ABC and AB'C' are two triangles with common angle, and
3.

meet again P; shew that the feet perpen

of

their circum-circles
at

diculars drawn from the lines AB, AC, BC, B'C' are collinear.
to
P

triangle circle, and any point

on

inscribed the circum

in
A
4.

P
to is

joined the triangle: shew that this

of
is

ference the orthocentre

by

joining line the pedal the point

of

P.
is

bisected
 THE TRIANGLE AND ITS CIRCLES, 213

THE TRIANGLE AND ITS CIRCLES.

VI. D, E, F are the points of contact of the inscribed circle of the

triangle ABC, and D1, E1, F1 the points of contact of the escribed circle,
which touches BC and the other sºdes produced: a, b, c denote the length
of the sides BC, CA, AB; s the semi-perimeter of the triangle, and r, r,
the radii of the inscribed and escribed circles.

Prove the following equalities:

AE AF –a,
(i)

= = =

= = =
8 8 8

BF
c. b,

BD
– –

CD CE
(ii) AE, =AF, F S.

(iii) CD1 CE1 =s—b,

–

(iv) CD BD, and BD=CDr.

=

(v) EE =FF1=a.
ABC=rs
of

(vi) The area

A

the
=ri (8-a).
(vii). Draw the above figure right angle, and
c.; in

is

the case when

C

prove that
r= r1=8–b.
–
8
214 GEOMETRY.

VII. In the triangle ABC, l is the centre of the inscribed circle, and
1, la,
mºmeº. the escribed circles touching respectively the sides

of
the centres
is

Ağ
36,85,

laws, *
\\
>
*
.
^|
\

Prove the following properties:

12;

C,
B,
11

80
(i)

C, le.
A,
1,

The points and

I,

are collinear are

11, l,
:

[1;
la,
B,
A,
lz,

80

lo

(ii) The points are collinear are and

is

;
C,

(iii) The triangle3 Bl, Cl,A, AlsB are equiangular

to

one another.
by

(iv) The triangle lila's equiangular the triangle

to

formed
of is

joining the points

of

contact the inscribed circle.

11,

12,

is,

of

(v) Of the four points

1,

each the orthocentre the

is

triangle whose vertices are the other three.

(vi) The four circles, each passes through

of

of

which three the

is,
ll,
la,

points are all equal.

l,
 THE TRIANGLE AND ITS CIRCLES, 215

EXERCISES.

1. With the figure given on page 214 shew that if the circles whose
11,
12,

D,
centres are 1, touch BC D1, D2, D3, then

at
is
(i)

DD2= DiD3–b. (ii) DD3- DiD2–c.

(iii) D2Ds=b+c. (iv) DD1=b

c.
-
of
Shew that the orthocentre and vertices triangle are the centres
2.

a
of

the pedal triangle.

of
the inscribed and escribed circles

of
Given the base and vertical angle

of
triangle, find the locus
3.

the

a
centre of theescribed circle which touches the base.

Given the base and vertical angle

of
triangle,
4.

shew that the centre

a
of

the circum-circle faced.

is

Given the base BC, and the vertical angle the triangle, find

of
5.

A
the locus of the centre of the escribed circle which touches AC.

Given the base, the vertical angle, and the point with
6.

of
contact
the in-circle; construct the triangle.
of

the base

Given the base, the vertical angle, and the point

of
contact with
7.

the base, base produced,

of

an escribed circle construct the triangle,

or

11,
ſ2,
of

triangle, and
8.

the centre the circle ºnscribed

in

the
ofis

byIs
a

Ill, lla,
Ils

centres the escribed circles shew that are bisected the

;
of

circumference the circum-circle.

le,

ABC triangle, and

of
9.

is

the centres the escribed circles

la
a

which touch AC, and ÅB respectively: shew that the points 12,
C,
B,

lie upon circle whose centre the circum is

is

of

on the circumference
a

circle the triangle ABC.

of

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 the three escribed circles; construct the
of

triangle.
12. Given the centre circle, two
of

the inscribed
of

and the centres

escribed circles; construct the triangle.

13. Given the vertical angle, perimeter, and radius the inscribed
of

circle; construct the triangle.

14. Given the vertical angle, the radius the inscribed circle, and
of

the length the perpendicular from the vertex the base; construct
of

to

the triangle.
triangle ABC,
In

the inscribed circle; shew

of

15.
is

the centre
a

that the centres about the triangles BIC,

of

the circles circumscribed

Cl/A, AlB lie on the circumference
of

the circle circumscribed about the

given triangle.
216 GEOMETRY

THE NINE-POINTS CIRCLE.

VIII. In any triangle the middle points of the sides, the feet of the
perpendiculars from the vertices to the opposite sides, and the middle
points of the lines joining the orthocentre to the vertices are concyclic.

In the AABC, let X, Y, Z be the

middle points of the sides BC, CA,
AB; let D, E, F be the feet of the
erp" to these sides from A, B, C ; let
§ be the orthocentre, and a, 8, 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
Wre concyclic.
Join XY, XZ, Xa, Ya, Za.
Now from the A ABO,
B X C
since AZ= ZB, and Aa = a C, D

º
parl
2,
p.

BO. Ex.
to
...

Za 64.
is

And from the AABC, since BZ=ZA, and BX=XC,

parl
to

ZX AC.
...

is

rt,rt.

but BO produced makes angle with AC

a a

the A-XZa angle.

...

is

Similarly, the A-XYa

rb, angle.
is
a
X,
Z,

the points
a,

are concyclic
‘..

Z;
X,
Y,
the O* the circle which passes through
is,

on

of

that lies
is a

and Xa diameter of this circle.

a

Similarly the O*
be

on

may
of

shewn that and lie this circle.

it

y
8

- Again, since DX angle,

rt.
is
a

diameter passes through

D.
on

as

the circle Xa
...

Similarly the O* this circle;

be

on

may
of

shëwn that and lie

it

E,E

F
Y,

D,
X,

Z,

the points
F,

8,
a,

are concyclic.
...

Q.E.D.
Y

Obs. From this property the circle which passes through the middle
points triangle called the Nine-Points Circle; many
of

the sides
of

is
bea

its properties may its being the circum

of

of

derived from the fact

the pedal triangle. .
of

circle
 THE NINE-POINTS CIRCLE, 217

To prove that
. . .(i) the centre of the nine-points circle is the middle point of the
straight line which joins the orthocentre to the circum-centre.

†- (ii) the radius of the nine-points circle is half the radius of the
circum-circle. -
(iii) the centroid is collinear with the circum-centre, the nine-points
centre, and the orthocentre.

In the AABC, let X, Y, Z be the

middle 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) To 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.
B

Similarly the perp. to EY at its

middle point bisects SO :
is,

the middle point SO:

of at

that these perp" intersect

of
And since XD and EY are chords the nine-points circle,

rt.
the intersection the lines which bisect XD and EY
at
angles
...

of

1, is
its centre Theor. 31, Cor.
:

the middle point SO. Q. D.

E.
...

the centre
is

of
N

w"
of

(ii) To prove that the radius the nine-points circle half the
is
of

rodeus the circum-circle.

By the last Proposition, Xa the nine-points circle.
of
is

diameter
a

the middle point Xa its centre

of

is
‘.
.

but the middle point SO the nine-points circle.

of

is

of

also the centre

(Proved.)
Hence Xa and SO bisect one another at N.
Then from the A• SNX, ONa,
SN=ON,
because and NX Na,
=

and the Z-SNX= the A-ONa

:

SX
...

Oa.
=

=Aa.
And SX also parl Aa,
to
is

..". =Xa.
But SA
of

radius the circum-circle

is
a

and Xa the nine-points circle;

of

diameter
is
a

nine-points
of

the radius
j,

circle
of

the half the radius the circum

2 is

267, Examples Q. D.,

p.

circle. [See also

E.

and 3.]
218 GEOMETRY.

(iii) To prove that the centroid is collinear with points S, N, O.

Join AX and draw ag parl to SO.

Let AX meet SO at G.

Then from the AAGO, since Aa = a C,

and ag is parl to OG,
Ag-gG. Ex.

1,
p.
... 64.

And from the AXag, since NX,

N
=
a
and NG parl ag,

to
is
..".

GX
g\i \ix.
*

=
=

AG=# AX;
of
...

the triangle ABC.

of

the centroid
...

is
G

Theor. III., Cor.,

X
p.

C
97.

D
B
is,

That collinear with

is

the centroid
S,

the points N,
O.

Q.E.D.

EXERCISES.

Given the base and vertical angte triangle, find the locus
of

of
1.

the
a

the nine-points circle.

of

centre

The nine-points circle

is 2.

any triangle ABC, whose orthocentre

is
of
O,

also the nine-points circle the triangles AOB, BOC,

of
of

each .
COA.

:
11,
12,

are the centres

If

of
a 3.

the inscribed and escribed circles

la
I,

triangle ABC, then the circle circumscribed about ABC

of

is
the
nine-points circle triangles formed by joining three
of

the four
of

each
11,
12,

points
1,

is:
of

the

All triangles which have the same orthocentre and the same
4.

circumscribed circle, have also the same nine-points circle.

Given the base and vertical angle triangle, shew that one
of
5.

angle and one side the pedal triangle are constant. -

of

Given the base and vertical angle triangle, find the locus
of
of
6.

the circle which passes through the three escribed

of

the centre
centres.

For some other important properties the Nine-points

of

NOTE.
Circle see Ex. 54, page 310.
APPENDIX.
 APPENDIX.
ON THE FORM OF SOME SOLID FIGURES.

(Rectangular Blocks.)

A B

The solid whose shape you are probably most familiar with
is that represented by a brick or slab of hewn stone. This
solid is called a rectangular block or cuboid. Let us examine
its form more closely.
How many faces has it? How many edges? How many
corners, or vertices?
• The faces are quadrilaterals: of what shape?
Compare two opposite faces. Are they equal? Are they
parallel?
We may now sum up our observations thus:
A cuboid
six

has faces; opposite faces being equal rectangles

parallel planes.
in

has twelve edges, which fall into three

It

groups, corresponding
to

the length, the breadth, and the height

of

the block. The four edges each group are equal and
in

parallel, and perpendicular the two faces which they cut.

to

The length, breadth, and height

of

rectangular block are

a

called its three dimensions.

If

Ex. two dimensions rectangular block are equal, say,

of
1.

the breadth AC and the height AD, two faces take particular
a

shape. Which faces? What shape?

the length, breadth, and height
If
2.

Ex. rectangular block

of
a
do

are all equal, what shapes the faces take

3
 SOLID FIGURES, CUBES, 3

E G

C F

A B

A rectangular block whose length, breadth, and height are

all equal is called a cube. Its surface consists of six equal
Squares.
We will now see how models of these solids may be
constructed, beginning with the cube, as being the simpler
figure..
Suppose the surface of the cube to be cut along the upright
edges, and also along the edge HG; and suppose the faces to
be unfolded and flattened out on the plane of the base. The
surface would then be represented by a figure consisting of six
-
squares arranged as below.

G E C F G

This figure is called the net of the cube: it is here drawn

on half the scale of the cube shewn in outline above.
To make a model of a cube, draw its net on cardboard.
Cut out the net along the outside lines, and cut partly through
along the dotted lines. Fold the faces over till the edges come
together; then fix the edges in position by strips of gummed
paper.

Ex. 3. Make a model of a cube each of whose edges is 60 cm.

4 * APPENDIX ON SOLID FIGURES.

Ex. 4. . Make a model of a rectangular block, whose •length is

4", breadth 3", height 2". -

First draw the net which will consist of six rectangles arranged
as below, and having the dimensions marked in the diagram.
*

Now cut the net out, fold the faces along the dotted lines,
and secure the edges with gummed paper, as already explained.

(Prisms.)

Let us now consider a solid whose side-faces (as in a rect

angular block) are rectangles, but whose ends (i.e. base and
top), though equal and parallel, are not necessarily rectangles
Such a solid is called a prism.

The ends of a prism may be any congruent figures:

these may be triangles, quadrilaterals, or polygons of any
number of sides. The diagram represents two prisms, one on
a triangular base, the other on a pentagonal base.

Ex. 5. Draw the net of a triangular prism, whose ends are

equilateral triangles on sides of 5 cm., and whose side-edges measure
7 cm.
 SOLID FIGURES. PYRAMIDS, 5

(Pyramids.)

A B

The solid represented in this diagram is called a pyramid.

The base of a pyramid (as of a prism) may have any number
of sides, but the side-faces must be triangles whose vertices are
at the same point.
The particular pyramid shewn in the Figure stands on a

all
square base ABCD, and its side-edges SA, SB, SC, SD are
In

equal. this case the side faces are equal isosceles triangles;
and the pyramid
be

right, for placed on

to

said the base

is

is
an if

level table, then the vertex lies upright line through

in
a

the mid-point
of

the base.

right pyramid standing

on

Ex. square
of

Make model
6.

measure 3", and each side-edge

to

Each edge
to of

base. the base

is
4".

the pyramid
be
of

is

To make the necessary net, draw

a

square
on

3". This will form

of

side
of a

the pyramid. Then

on

the base the

square
of

sides this draw isosceles

triangles making the equal sides
in

each
triangle long.
4"

Explain why the process folding

of

w
about the dotted lines brings the four
vertices together.
6 APPENDIX ON SOLID FIGURES.

Another important form of pyramid has as base an equi.

all
lateral triangle, the side edges are equal the edges

to
and
of the base.

2.
FIG.

How many faces will such pyramid have? How many

a
edges? What sort triangles will the side-faces be? Fig.
of

3
shews the net on reduced scale.
a

pyramid this kind regular tetrahedron

of
A

caked
is

a
(from Greek words meaning four-faced).
regular tetrahedron, each edge
of
7.

Ex. Construct model

a
a

long.
of

3"

which
is

What the smallest number of plane faces that will

a 8.

Ex.
is

What
of

space the smallest number curved surfaces

is

enclose
7

that will enclose space

#
a

(Cylinders.)
B
C

9
D

FIG.
2.

FIG.
I.

The solid figure here represented called cylinder.

is

a
 SOLID FIGURES. CYLINDERS. 7

On examining the model of which the last diagram is a

drawing, you will notice that the two ends are plane, circular,
equal, and parallel.
The side-surface is curved, but not curved in every
direction; for it is evidently possible in one direction to rule
straight lines on the surface: in what direction!

2),
Let us take a rectangle ABCD (see Fig. and suppose

it
to

as
rotate about one side AB fixed axis.

a
What will BC and AD trace out, they revolve about AB!

as as
parallel

be
so
Observe that CD will move always

to

to
the axis AB, and pass round the curve traced out by As
to

D.
CD moves, will generate (that say, trace out)
to
surface.
it

is

a
What sort of surface?
why parallel

to
We now one direction, namely
in

see the

on
to

axis AB, possible rule Straight lines the

it
is

curved
cylinder.
of

surface
a

to

to

easy plane surface represent

It

find the curved

is

cylinder.
of

surface
a

Q
R

P
S

Cut rectangular strip

paper, making the width PQ
of
to a

equal the cylinder. Wrap the paper round

the height
of

the cylinder, and carefully mark off the length

PS

that will
go

all

make the paper exactly once round. Cut off that

overlaps; and then unwrap the covering strip. You have now
rectangle representing the curved surface the cylinder,
of
a

and having the same area.

IH.S.G. I.-III.
P
8 APPENDIX on solid FIGUREs.

(Comes.)

FIG. T.

We have now to examine the model of a cone, of which a

drawing is given above.
Its surface consists of two parts; first a plane circular base,
then a curved surface which tapers from the circumference of
the base to a point above it called the vertex. Thus the form
of a cone suggests a pyramid standing on a circular instead of
a rectilineal base.
2),
Let us take a triangle ABC right-angled at B (Fig. and
suppose What
to

as

rotate about one side AB fixed axis.

it

will BC trace out as the triangle revolves? Notice that AC

always pass through the fived point
A,

will and move round

the curve traced out by As AC moves, will generate
C.

it

-
of

surface. What sort surface?

-

We now see that the kind cone represented

in
of

the
by

diagram solid generated the revolution right

of
is
a

angled triangle about one side containing the right angle.

Why must the AABC, rotating about AB, be right

at 9.

Ex.
$
B,

angled generate cone

to

order
in

7
a
by

an

generated
be

What would obtuse-angled

7 of

the revolution
triangle about one side forming the obtuse angle
by
10.

an

generated parallelogram
be

Ex. What would oblique

revolving about one side
3
 SOLID FIGURES. CONES. SPHERES. 9

The curved surface of a cone may be represented by a plane

figure thus:
A

Taking the slant-height AC of the cone as radius, draw a

circle. Cut it out from your paper; call its centre A; and cut
it along any radius AC. If you now place the centre of
the circular paper at the vertex of the cone, you will find that
you can wrap the paper round the cone without fold or crease.
Mark off from the circumference of your paper the length CD
that will go exactly once round the base of the come ; then cut
through the radius AD. We have now a plane figure ACD
(called a sector of a circle) which represents the curved surface
of the cone, and has the same area.

(Spheres.)

The last solid we have to consider is the sphere, whose

shape is that of a globe or billiard ball.
B
*

C O

A
FIG. 2.

We shall realise its form more definitely, if we imagine

a semi-circle ACB (Fig. 2) to rotate about its diameter as a
fixed axis. Then, as the semi-circumference revolves, it
*
generates the surface of a sphere.
H. S.G. I.-III. P2
 -
10 APPENDIX ON SOLID FIGURES.

all
points

on
Now since

in
the semi-circumference are

O,
all positions constant distance from its centre

at
we

a
on
see that all points sphere are

at
of
the surface constant

a
of it,
distance from fixed point within namely the centre.

a
This constant distance the radius the sphere. Thus

is
all

straight lines through the centre terminated both ways by

the surface are equal: such lines are diameters.
11.

cylinder

on
We have seen that

of
Ex. the curved surfaces

a
possible (in certain ways only) rule straight lines.

to
and come
is
it

there any direction straight

on
Is

which we can rule line the

in
.

a
sphere
of

surface
a

Ex. 12. Again we have cut out plane figure that could

be
of a
wrapped round the curved surface cylinder without folding,

a
creasing, stretching.
as a or

The same has been done for the curved

be
flat piece paper wrapped about
of

of
surface cone. Can

a
a

.
sphere all over the surface without creasing?
fit
to
so
13.

Suppose you were sphere straight through the

to

Ex. cut
a

centre into two parts, way that the new surfaces (made by
in

such
a

cutting) are plane, these parts would every way alike. The
be
in

parts into which sphere divided by plane central section are

is
a

a
Of

called hemispheres. what shape the line which the plane

in
If is

surface meets the curved surface the section were plane but
7

not central, can you tell what the meeting line

of

the two surfaces

would be
--
7

by

cylinder were cut plane parallel-

to

Ex. 14. the base,

If
a

what shape would the new rim be?

of

cone were cut by plane parallel

If

Ex. 15. the base,

to
a

a
.

what would be the form of the section

?
 ANSWERS TO NUMERICAL EXERCISES.
Since the wtmostcare cannot ensure absolute accuracy in graphical work, results so
obtainedare likely to be only approximate. The answers here given are thosefound by
calculation, and being true sofar as they go, furnish a standard by which the student
may testthe correctnessof his drawing and measurement. Results within one per cent.
of thosegiven in theAnswers maywavally beconsideredsatisfactory.

Exercises. Page 15.

30°; 126°; 261°; 85°. 11 min. ; 37 min.
:

1123°; 155°; 5 hrs. 45 min. 3. 50°; 8 hrs. 40 min.

(i)

4. 145°, 35°, 145°. (ii) 55°, 55°. 86°, 94°.

Exercises. Page 27.

22", 50°, 73° nearly.
7. 2,

68°, 37°, 75° nearly.

5, 1.

8. 4.
6-0 em.
v.

37 ft. 27 ft. 424 yds., nearly; N.W.

6.

101 metres.
281 yds., 155 yds., 153 yds. 10. 214 yds.
9.

Exercises. Page 41.

15

125°, 55°, 125°. 12. secs., 30 secs.

1.

Exercises. Page 43.

-
3.

21°. 92°, 46°. 67°, 62°.

4.

27°.
5.

6.

Exercises. Page 45.

(i)

30°, 60°, 90°. 36°, 72°, 72°; (ii) 20°, 80°, 80°.
6. 3. 1.

2.

34°; (ii) 107°.

(i)
7. 4,

40°. 51°, 111°, 18°.

8. 5.

68°. 120°. 36°, 72°, 108°, 144°.

.

5,

11. 15.
9.

165°.
-
Exercises. Page 47.
(i)
(i)

45°; (ii) 36°. 12; (ii) 15.

3.
2.

Exercises. Page 54.

(i)

(ii) 55°.
4.

81°
c.
;

Degrees 15° 30° 45° 60° 75°

|

10.
|

Cm. 4-1 4-6 5-7 8-0

|lsº
||

90'
60

120 150 180°

30

Degrees
a

||
0
||

Cm. 1-0 2:0 3-6 5-0 6-1 6-8 7-0

||

||

||

||

||

||

12. 37 ft. 13. 112 ft. 14. 346 yds. 693 yds.
H. S.G.
H GEOMETRY.

Exercises. Page 61.

14. 54°, 72°, 54°. 15. 36°. 16, 4.

16; (ii) 45°; (iii) 113° per sec.

(i)
18.

Exercises. Page 68.

3,
9. 2,

8,
10-6 cm,

5.
4.
6-80 cm. 2.24”. O'39. 2'54.
3-35". 10, 20 miles; 12.6 km.
11. 147 miles; 235 km. cm. represents 22 km.

1
15

20
represents represents
1"

1"
12. mi. mi.

;
4. Exercises. Page 79.

5.
cm. 2'4”.
3.

0-53 in.
1
3

Exercises. Page 84.

7. 5. 2.
4-3 cm., 5-2 cm., 6-1 cm. 10.

3,
200 yards,
9. 8. 6. 4, 1.

S,

E,
65°, 77 m., 61 m., 56 m. 6.04 knots. 15° nearly,
Results equal. cm. . 4.3 cm.; 9.8 cm., 60°; 120°.
9

(iii) one, right-angled

(i)

One solution (ii) two (iv) impossible.

;

;
380 yds. 10. 6-5 cm. 11. 6-9 cm.
Two solutions; 10.4 cm. 2.8 cm., 4-5 cm., 53 cm.
or

12. 4-5 cm. 16.

18,

19. cm., cm.

5.8 cm.,
7

8
4:2 cm.

Exercises. Page 89.

2' 12".
3.

4.
6. 2.

60°, 120°. 3-54". 4'4 cm.

l,

90°. 4'25"; (ii)

(i)

16:4 cm., 3.4%.

7.

90°.
D
5.

=
B

Exercises. Page 102.

sq.

sq.

sq.
m. in.
in.

in.
6. 2,

8. 4,

sq. in. 280

7. 3.

3:50
i.
6

ft.

3-30 sq. in. 336 sq. in. 198 sq. 42 sq.

9. 5.

12, 2-6 in.

ft.

10,000 sq. m. 10. 110 sq. 11. cm.

5

m.
48

14, 900 sq. yds. yds. 4'8". 15. 11700 sq.

ft. i
;

17,
ft. ft. 10

6".
ft.

yds. 600 sq. 1152 sq.

1

16. cm. 18. 19.

=
1

ft.

ft.

20. 100 sq. 21. 156 sq. 22. 110 sq.

ft.

ft.

23. 288 sq. 24. 72 sq. 25. 75 sq.

Exercises. Page 105.

cm.; (ii) 36".
22
(i)

5. 2,

3-4 sq. in. 574-5 sq. in.

4, 1.

3.

1°5". 193", 75°.

 ANSWERS. º hiſ

Exercises. Page 107.

ft.
(i) 15 (i) (i)
180 sq. (ii) 8.4 sq. in. hectare.

1
;

;
13:44 sq. om. (ii) 15:40 sq. cm.; (iii) 20:50 sq. cm.

;
:

6. 4,
sq. om. 6-3 sq. in.

-
8”; (ii) cm. 13 3-36 sq. in.

Exercises. Page 110.

1. 11400 sq. yds. 6312 sq. m.

4. 2.
2-4 cm.; 5:1 cm. 2.04"; 2:20".

Angle 30° 60° 90° 120° 150° 180°

||
0
|

|
Area sq. cm. 7-5 13-0 15-0 13-0 7-5
in

||

||

||

||

||
0

0
|
Exercises. Page 111.

m.
84

66 sq. ft.

6. 8,
sq. yds. 126 sq.
5. 2.

ft.

132 sq. cm. 180 sq. 306 sq. m.

Exercises. Page 113.

l
m.
ft.

sq. in. 170 sq. 615 sq. 8-4 sq. in,

2.

8.

7. 4.
31 6

sq. cm. 5:20 sq. in. 24 sq. cm,

6.
2

Exercises. Page 115.

1 25.5 sq. cm.;
(i) (i)

(ii) 15-6 sq. cm.

8-95 sq. in. (ii) 9-5 sq. in. 12500 sq. m.
3.
;

Exercises. Page 116.

5.

3.3 sq. in. 7-5 cm. 3-6 sq. in.

6.

Bxercises. Page 121.

cm.; cm.; l'6";
(i)

(i)

(ii) 6-5 (iii) 3.7". (ii) 2.8 cm.

2.
5

41 ft.
8, 4.

9. 5.

65 miles.
ft.

3. 6-1 km.
16
6,

48 m. 25 miles. T3 m. 10. 62 ft.

-

Exercises. Page 123.

4:24 cm.;
18

10. and (iii). 11. 2'83". 12, sq. cm

(i)

p=6'93 cm.
m.

13. 70-71 sq. 14,

20 cm.; (ii) 40 cm.; 39 cm.
15
(i)

16. cm.
35 cm.; cm.; 306 sq. cm.
12

17.
sq. in, (ii) 90 sq. ft.; (iii) 126 sq. cm.; (iv) 240 sq. yds.
36

18.
(i)

19. 5°l cm. nearly.

 GEOMETRY.

Exercises. Page 127.

7-1 cm. 4. 4-0 cm. 5. I-6". 6. 3-1 cm.; 15.6 sq. cm.

Exercises. Page 130.

23.90 sq. cm. 2. 8:40 sq. in.
27.52 sq. cm. 4. 129800 sq. m.

Exercises. Page 134.

(8, 5); (ii) (10, 10).
(i) (i)

(4, 5); (ii) (4, 5); (iii) (–4, -5); (iv) (–4, -5).
(6, 5), (12, 10). (5, 8).

.
6.
17; (ii) 17; (iii) 2.5"; 2-5".
(i) (i)

-
and (ii) 5; (iii) and (iv) 17; (v) and (vi) 37.

9,
10.
(0, 0). (7, 5). 15. 13; (9, 6).
straight line passing through the points (4,0), (0,
A

4).

in.
117 units
in

square. sq. in. sq.

of

area each case. 18.

1
Each =70 units 20. units 31°, 71°, 78°.

of
area.
of

area.

9
96; (ii) 80; (iii) 120;

r
(i) (i)

(iv) 104.
50; (ii) 60; (iii) 120; (iv) 132.
13; area 63. 27; (ii) 21; (iii) 30; (iv) 27°5.
5,

24,
(i)

Sides
(ii) 65-5; (iii)
50
(i)

21

(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.
=
5,
5,

10, 13, Area=60. 30. 160,000 sq. yds. 1000 yds. 320 yds.
3.

Side 15:23; area=232 units

of

area.
=

Exercises. Page 145.

2,

cm. 24”. 0-6", O'8".

5

7. 3.

N7=2.6 cm.
4.

ft. 0-6 sq. in.

1

6.

0-8".

- Exercises. Page 149. -

-
1.7" 3M2=42 cm.
6. 2.

3.

28/3=3-5 cm.
17". cm.
5

Exercises. Page 151.

cm.
7.

1.3%.
4

Exercises. Page 153.

'85'. (2.1", 2.1"); 2.97°.
3.

1-62”. 0.85";
5.
I
 ANSWERS. w

Exercises. Page 155.

51”, 1-6"; 1'5", 0.6".

6.
Exercises. Page 157.
(8) 11). 17; 10; (0, -8).

5.
Exercises. Page 161.
2,
74°, 148°, 16°, 115°, 230°. 55°, 8°, 47°.

3.
.

Exercises. Page 177.

2.

8-0 cm. 0-6". 12", 67°. 2.5".

3.

8-7 cm.

4.

5
Exercises. Page 179.
17

cm. and cm.

3

Exercises. Page 181.

&

72°, 108°, 108°.

Exercises. Page 187.

4,

1-6". 198", 1-6".

3.

17”.

Exercises. Page 198.

7. 3.

2-3 cm., 4-6 cm., 6.9 cm. 39".

1

6-9 cm.; 20-78 sq. cm. 3-2 cm.

Exercises. Page 199.

*
sq.

5.

2.0°.
4,
in.

2:12", 450 8.5 cm.

Exercises. Page 200.

128}”; 173".

Exercises. Page 201.

3:46"; 4:00". 259.8 sq. cm.
2.

41-57 sq. cm.; (ii) 77°25 sq. cm.

(i)

Exercises. Page 205.

28-3 cm.; (ii) 628-3 cm. 16-62 sq. in.; (ii) 352-99 sq.
(i)

in
56 (i)

11:31 cm.; 10:18 sq. cm. sq. cm. 43-98 sq. in.
5.
4.

l:
8.

30.5 sq. cm. 8'9". 4”; 3". 10. 12:57 sq. in.
l'54 sq. in., 3-14 sq.
in.

Circumferences, 4'4", 6'3". Areas,

 .***
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