Menelaus's Theorem

The collinearity test for a transversal cutting a triangle's sides.

The idea

A straight line that cuts all three side-lines of triangle $ABC$ is called a transversal, where a side may need extending for the line to reach it.

Theorem (Menelaus's Theorem).

Let points $X$, $Y$, $Z$ lie on the side-lines $BC$, $CA$, $AB$ of triangle $ABC$. Then $X$, $Y$, $Z$ are collinear if and only if $\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB} = 1,$ each ratio taken as a ratio of unsigned lengths.

Each factor compares the two pieces into which one cut point divides its side, and the factors cycle through the vertices $B$, $C$, $A$, so every vertex appears once in a numerator and once in a denominator. The product is therefore a pure number, unchanged when the triangle is rescaled.

Ceva's Theorem multiplies the same three ratios to test whether three cevians pass through one point. The two criteria separate once lengths carry signs: count a ratio as negative when its cut point falls outside its segment. A transversal must leave the triangle, so it cuts an odd number of the three sides externally and its signed product is $-1$; three cevians through an interior point cut all three sides internally, giving $+1$.

Ways to work on it

• Walkthrough. The transversal ratio product and how it differs from Ceva.
• Practice. Solve for the missing side ratio on a transversal.
• Hardest. Locate a crossing, then use the converse as a collinearity test.

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