proof of Ceva’s theorem


sign of

As in the article on Ceva’s Theorem, we will consider directed line segments.

Let X, Y and Z be points on BC, CA and AB, respectively, such that AX, BY and CZ are concurrentMathworldPlanetmath, and let P be the point where AX, BY and CZ meet. Draw a parallelMathworldPlanetmathPlanetmath to AB through the point C. Extend AX until it intersects the parallel at a point A. Construct B in a similarMathworldPlanetmath way extending BY.

The triangles ABX and ACX are similar, and so are ABY and CBY. Then the following equalities hold:


and thus


Notice that if directed segments are being used then AB and BA have opposite signs, and therefore when cancelled change the sign of the expression. That’s why we changed CA to AC.

Now we turn to consider the following similarities: AZPACP and BZPBCP. From them we get the equalities


which lead to


Multiplying the last expression with (1) gives


and we conclude the proof.

To prove the converse, suppose that X,Y,Z are points on BC,CA,AB respectively and satisfying


Let Q be the intersection point of AX with BY, and let Z be the intersection of CQ with AB. Since then AX,BY,CZ are concurrent, we have


and thus


which implies Z=Z, and therefore AX,BY,CZ are concurrent.

Title proof of Ceva’s theorem
Canonical name ProofOfCevasTheorem
Date of creation 2013-03-22 12:38:54
Last modified on 2013-03-22 12:38:54
Owner yark (2760)
Last modified by yark (2760)
Numerical id 11
Author yark (2760)
Entry type Proof
Classification msc 51A05