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Topic: [EMHL] New Proof of Ceva's Theorem
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Darij Grinberg

Posts: 150
Registered: 12/3/04
[EMHL] New Proof of Ceva's Theorem
Posted: Mar 10, 2003 10:45 AM
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In a comment of Peter Yff concerning the equal
parallelians point, I have found a nice idea for a
new proof of Ceva's theorem.

We phrase Ceva's Theorem (without the converse)
as follows:

If a triangle ABC is given, and points A', B', C'
on BC, CA, AB respectively are constructed so that
the lines AA', BB', CC' concur at a point P, then

AC' BA' CB'
--- * --- * --- = 1.
C'B A'C B'A

C
/\
/ \ \
/ \ \
B'/ \ \
/-- \ \
/ -- \ \ A'
/ -- \ /\
/ -- \ / \
/ /P\-- \
/ / \ -- \
/ / \ -- \
/ / \ -- \
/ / \ -- \
/ / \ -- \
/ / \ -- \
A/-----------------------------\---------------\B
C'

For the proof, let the parallel to BC through P
intersect CA in Ba and AB in Ca, and analogously
define the points Cb, Ab, Ac, Bc.

C
/\
B / \
a / \
/ \ \
B'- \ \ A
/ \ \ b
/ \ / -A'
B / \ / \ A
c /---------------P\-------\ c
/ / \ \
/ / \ \
/ / \ \
/ / \ \
/ / \ \
/ / \ \
A/---------------/-------------|------C -------\B
C C' a
b

From parallel segments, we have

AC' BcP
--- = ---.
C'B PAc

On the other hand,

BcP PB' PAc PA'
--- = --- and --- = ---,
AB BB' AB AA'

therefore

BcP PB' / PA'
--- = --- / ---.
PAc BB' / AA'

So we have

AC' PB' / PA'
--- = --- / ---.
C'B BB' / AA'

Analogously,

BA' PC' / PB'
--- = --- / --- and
A'C CC' / BB'

CB' PA' / PC'
--- = --- / ---.
B'A AA' / CC'

Multiplying these equations yields

AC' BA' CB'
--- * --- * --- = 1,
C'B A'C B'A

qed.

This proves, although not as simple as others
using parallels, has the maximal symmetry.

Darij Grinberg





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