Ceva's Theorem

When three cevians of a triangle meet at a single point.

The idea

A cevian of a triangle is a segment from a vertex to a point on the opposite side. Ceva's Theorem decides when three cevians, one from each vertex, pass through a single point.

Theorem (Ceva's Theorem).

In triangle $ABC$, let $D$ lie on side $BC$, $E$ on $CA$, and $F$ on $AB$. The cevians $AD$, $BE$, $CF$ pass through one common point if and only if $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = 1.$

Three lines in the plane generally meet pairwise in three separate points, so concurrence is a genuine condition, and the theorem measures it exactly: choose $D$ and $E$ freely, and concurrence forces the position of $F$, since the product must equal $1$.

The factors follow the cycle $B$, $C$, $A$: each compares the two pieces into which one cevian divides its side. Every vertex appears once in a numerator and once in a denominator, so the product is a pure number, unchanged when the triangle is rescaled.

Ways to work on it

• Walkthrough. The concurrency criterion and the median case.
• Practice. Find the third ratio that forces concurrence.
• Hardest. Test concurrence from explicit side divisions.

Not sure where to start? Take the ten-question placement test.