# 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.

**Menelaus's theorem** uses a very similar structure. Both theorems are very useful in Olympiad geometry.

Ceva's theorem is useful in proving the concurrence of cevians in triangles and is widely used in Olympiad geometry.

## Statement

Given a triangle \(\triangle ABC\) with a point \(P\) inside the triangle, continue lines \(AP, BP, CP\) to hit \(BC, CA, AB\) 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 \(\frac{AF}{FB}\cdot \frac{BD}{DC}\cdot \frac{CE}{EA} = 1\), 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,

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

Multiplying the previous three equations together, we get

\[\dfrac{AF}{FB}\cdot \dfrac{BD}{DC}\cdot \dfrac{CE}{EA} =\dfrac{[APC]}{[BPC]}\cdot \dfrac{[APB]}{[APC]}\cdot \dfrac{[BPC]}{[APB]} = 1.\ _\square\]

## Practice Problems

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

In this case, \(D,E,F\) are the midpoints of their respective sides. Therefore, \(AE=EC, CD=DB,\) and \(BF=FA,\) so it immediately satisfies Ceva's theorem, demonstrating the existence of the intersection, which is the centroid. \(_\square\)

Prove that cevians perpendicular to opposite sides are concurrent.

Let \(D, E, F\) be the feet of the altitudes.

Note that \(\triangle BFC \sim \triangle BDA\) and, similarly, \(\triangle AEB \sim \triangle AFC, \triangle CDA \sim \triangle CEB\). Therefore,

\[\frac{BF}{BD} = \frac{BC}{BA}, \frac{AE}{AF} = \frac{AB}{AC}, \frac{CD}{CE} = \frac{CA}{BC}.\]

Multiplying these three equations gives us

\[\frac{BF}{BD} \cdot \frac{AE}{AF} \cdot \frac{CD}{CE} = \frac{BC}{BA} \cdot \frac{AB}{AC} \cdot \frac{CA}{BC} = 1.\]

Rearranging the left-hand side gives us

\[\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.\]

Therefore, the three altitudes coincide at a single point, the orthocenter. \(_\square\)