# Menelaus' Theorem

Menelaus' Theorem deals with the collinearity of points on each of the three sides (extended when necessary) of a triangle. It is named for Menelaus of Alexandria.

## Statement

A necessary and sufficient condition for points $P, Q, R$ on the respective sides $BC, CA, AB$ (or their extensions) of a triangle $ABC$ to be collinear is that $BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB$

where all segments in the formula are directed segments. $[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); [/asy]$

## Proof

Draw a line parallel to $QP$ through $A$ to intersect $BC$ at $K$: $[asy] defaultpen(fontsize(8)); pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0); draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75); draw((7,6)--(6,8)--(4,0)); draw(A--K, dashed); R=intersectionpoint(A--B,Q--P); dot(A^^B^^C^^P^^Q^^R^^K); label("A",A,(1,1));label("B",B,(-1,0));label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1)); label("K",K,(0,-1)); [/asy]$ $\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}$ $\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}$

Multiplying the two equalities together to eliminate the $PK$ factor, we get: $\frac{AR}{RB}\cdot\frac{QC}{QA}=-\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=-1$