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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 after Menelaus of Alexandria.

Statement

If line $PQ$ intersecting $AB$ on $\triangle ABC$, where $P$ is on $BC$, $Q$ is on the extension of $AC$, and $R$ on the intersection of $PQ$ and $AB$, then \[\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.\]

[asy] unitsize(16); 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]

Alternatively, when written with directed segments, the theorem becomes $BP\cdot CQ\cdot AR = CP\cdot QA\cdot RB$. Also, the theorem works with all three points on the extension of their respective sides.

Proof

Proof with Similar Triangles

Draw a line parallel to $QP$ through $A$ to intersect $BC$ at $K$: [asy] unitsize(16); 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] \[\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}\] \[\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{CP}{KP}\] Multiplying the two equalities together to eliminate the $PK$ factor, we get: \[\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{CP}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{CP}=1\]

Proof with Barycentric coordinates

Disclaimer: This proof is not nearly as elegant as the above one. It uses a bash-type approach, as barycentric coordinate proofs tend to be.

Suppose we give the points $P, Q, R$ the following coordinates:

\[P: (0, P, 1-P)\] \[R: (R , 1-R, 0)\] \[Q: (1-Q ,0 , Q)\]

Note that this says the following:

\[\frac{CP}{PB}=\frac{1-P}{P}\] \[\frac{BR}{AR}=\frac{1-R}{R}\] $$ (Error compiling LaTeX. Unknown error_msg)\frac{QA}{QC}=\frac{1-Q}{Q}$The line through$R$and$P$is given by:$|X0RYP1−RZ1−P0| = 0$which yields, after simplification,

<cmath>-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0</cmath> <cmath>Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).</cmath>

Plugging in the coordinates for$ (Error compiling LaTeX. Unknown error_msg)Q$yields$(Q-1)(R-1)(P-1) = QPR$. From$\frac{CP}{PB}=\frac{1-P}{P},$we have <cmath>P=\frac{(1-P)\cdot PB}{CP}.</cmath> Likewise, <cmath>R=\frac{(1-R)\cdot AR}{BR}</cmath> and <cmath>Q=\frac{(1-Q)\cdot QC}{QA}.</cmath>


Substituting these values yields <cmath>(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdot CP\cdot BR}</cmath> which simplifies to$ (Error compiling LaTeX. Unknown error_msg)QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.$=== Proof with [[Mass points]] === First let's define some masses.$B_{m_{1}}$,$C_{m_{2}}$, and$Q_{m_{3}}$By Mass Points: <cmath>BP\cdot m_{1}=PC\cdot m_{2} \implies \frac{BP}{CP}=\frac{m_{2}}{m_{1}}</cmath> <cmath>\frac{QC}{QA}=\frac{AC+QA}{QA}=1+\frac{AC}{QA}=1+\frac{m_{3}}{m_{2}}=\frac{m_{2}}{m_{2}}+\frac{m_{3}}{m_{2}}=\frac{m_{3}+m_{2}}{m_{2}}</cmath> The mass at A is$m_{3}+m_{2}$<cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \implies \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}} </cmath> Multiplying them together,${\;\; \frac{BP}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = \frac{{m_{2}}}{{m_{1}}} \cdot \frac{{m_{3}+m_{2}}}{{m_{2}}} \cdot \frac{{m_{1}}}{{m_{3}+m_{2}}} = 1}$== Converse ==

The converse of Menelaus' theorem is also true. If$ (Error compiling LaTeX. Unknown error_msg)\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = 1$in the below diagram, then$P, Q, R$ are collinear. The converse is useful in proving that three points are collinear.

[asy] unitsize(16); 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]

See Also

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