9/20/2011

Ceva’s Theorem

MA 341 – Topics in Geometry

Lecture 11

Ceva’s Theorem
The three lines containing the vertices A, B, and
C of ᇞABC and intersecting opposite sides at
points L, M, and N, respectively, are concurrent
if and only if

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Ceva’s Theorem

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Ceva’s Theorem

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Ceva’s Theorem

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Ceva’s Theorem

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Ceva’s Theorem

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Ceva’s Theorem
Now assume that

Let BM and AL
intersect at P and
construct CP
intersecting AB at
N’, N’ different
from N.

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Ceva’s Theorem
Then AL, BM, and CN’ are concurrent and

From our hypothesis it follows that

So N and N’ must coincide.

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Medians
In ΔABC, let M, N, and P be midpoints of AB,
BC, AC.
Medians: CM, AN, BP
Theorem: In any triangle the three medians
meet in a single point, called the centroid.

M – midpoint  AM=BM, N - midpoint  BN=CN

P - midpoint  AP=CP

By Ceva’s Theorem they are concurrent.

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Orthocenter
Let ΔABC be a triangle and let P, Q, and R
be the feet of A, B, and C on the opposite
sides.
AP, BQ, and CR are the altitudes of ΔABC.

Theorem: The altitudes of a triangle

ΔABC meet in a single point, called the
orthocenter, H.

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Orthocenter

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Orthocenter
By AA
ΔBRC~ΔBPA (a right angle and B)
 BR/BP=BC/BA
ΔAQB~ΔARC (a right angle and A)
 AQ/AR=AB/AC
ΔCPA~ΔCQB (a right angle and C)
 CP/CQ=AC/BC

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Orthocenter
By Ceva’s Theorem, the altitudes meet at
a single point.

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Orthocenter
Traditional route:
BQ intersects AP. B
Now construct CH and let
it intersect AB at R. R
Prove ΔARC~ΔAQB
making R=90. H
P
A
C
Q

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Incenter
Let ΔABC be a triangle and let AP, BQ,
and CR be the angle bisectors of A, B,
and C.
Angle Bisector Theorem: If AD is the
angle bisector of A with D on BC, then

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Incenter
Proof: Want to use similarity.
Where is similarity?

Construct line through

C parallel to AB

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Incenter
Proof: Want to use similarity.
Where is similarity?

Construct line through

C parallel to AB
Extend AD to meet parallel line
through C at point E.

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Incenter
BAE  CEA – Alt Int Angles
BDA  CDE – vertical angles
ΔBAD ~ ΔCDE – AA
Therefore

Note that CEA  BAE  CAE

 ΔACE isosceles  CE = AC and
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Incenter
Let ΔABC be a triangle and let AP, BQ,
and CR be the angle bisectors of A, B,
and C.
Theorem: The angle bisectors of a triangle
ΔABC meet in a single point, called the
incenter, I.

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Incenter
Proof: Angle bisector means:

By Ceva’s Theorem we need to find the

product:

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Incenter

Thus by Ceva’s Theorem the

angle bisectors are
concurrent.

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Circumcenter & Perp Bisectors

Does Ceva’s Theorem apply to
perpendicular bisectors?

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Circumcenter & Perp Bisectors

How can we get Ceva’s Theorem to apply
to perpendicular bisectors?

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Circumcenter & Perp Bisectors

Draw in
midsegments

EF||BC 
perpendicular
bisector of BC is
perpendicular to
EF  is an
altitude of ΔDEF
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Circumcenter & Perp Bisectors

Perpendicular bisectors of
AB, BC and AC are
altitudes of ΔDEF.

Altitudes meet in a single

point  perpendicular
bisectors are concurrent.

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Circumcircle
Theorem: There is exactly one circle through
any three non-collinear points.

The circle = the circumcircle

The center = the circumcenter, O.
The radius = the circumradius, R.
Theorem: The circumcenter is the point of
intersection of the three perpendicular
bisectors.

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Question
Where do the perpendicular bisectors of
the sides intersect the circumcircle?

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Question
Where do the perpendicular bisectors of
the sides intersect the circumcircle?
At one end is point of intersection of
angle bisector with circumcircle
The other end is point of intersection of
exterior angle bisector with circumcircle.

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Extended Law of Sines

Theorem: Given ΔABC with circumradius R, let
a, b, and c denote the lengths of the sides
opposite angles A, B, and C, respectively.
Then

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Proof
Three cases:

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Proof
Case I: A < 90º
BP = diameter
ΔBCP right triangle
BP = 2R
 sin P = a/2R
A = P
 2R = a/sin A

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Proof
Case II: A > 90º
BP = diameter
ΔBCP right triangle
BP = 2R
 sin P = a/2R
A = P
 2R = a/sin A

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Proof
Case III: A = 90º
BP = a = diameter
BP = 2R
2R = a = a/sin A

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Circumradius and Area

Theorem: Let R be the circumradius and K be
the area of ΔABC and let a, b, and c denote the
lengths of the sides as usual. Then 4KR=abc

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Proof

K = ½ ab sin C
2K = ab sin C
c/sin C = 2R
sin C = c/2R
2K = abc/2R
4KR = abc

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