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[EMHL] New Proof of Ceva's Theorem
Posted:
Mar 10, 2003 10:45 AM


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



