# Ceva's Theorem

Ceva's theorem is a theorem about triangles in Euclidean plane geometry. It regards the ratio of the side lengths of a triangle divided by cevians.

## Definition

Given a triangle \(\triangle ABC\) with a point \(P\) inside the triangle, continue lines \(AP\), \(BP\), and \(CP\) to hit \(BC\), \(CA\), and \(AB\) at \(D, E, F\) respectively.

Ceva's Theoremstates that \[\dfrac{AF}{FB}\cdot \dfrac{BD}{DC}\cdot \dfrac{CE}{EA} = 1\]The converse of Ceva's Theorem is also true: If \(D, E, F\) are on sides \(BC, CA, AB\) respectively such that \(\dfrac{AF}{FB}\cdot \dfrac{BD}{DC}\cdot \dfrac{CE}{EA} = 1\), then lines \(AD, BE, CF\) are concurrent at a point \(P\).

## Proof of Ceva's Theorem

There are various proofs for Ceva's Theorem. In this wiki, we're going to prove it by using triangle's area.

Note that \[\dfrac{AF}{FB}=\dfrac{[AFP]}{[FBP]}=\dfrac{[AFC]}{[FBC]}\] because \(\triangle AFP\) and \(\triangle FBP\) have the same altitudes (and ditto for the last two triangles).

By subtracting the triangle areas of the second equality with the first equality, we get \[\dfrac{AF}{FB}=\dfrac{[APC]}{[BPC]}\]

Similarly, \[\dfrac{BD}{DC}=\dfrac{[APB]}{[APC]}\] \[\dfrac{CE}{EA}=\dfrac{[BPC]}{[APB]}\]

Multiplying the previous three equations together, we get \[\dfrac{AF}{FB}\cdot \dfrac{BD}{DC}\cdot \dfrac{CE}{EA} =\dfrac{[APC]}{[BPC]}\cdot \dfrac{[APB]}{[APC]}\cdot \dfrac{[BPC]}{[APB]} = 1\]

## Practice Problems

\(1\). Prove that if \(X, Y, Z\) are midpoints of the sides, the three cevians are concurrent.

\(2\). Prove that cevians perpendicular to the opposite sides are concurrent.