# 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. $_\square$