Menelaus' Theorem

Revision as of 17:03, 30 November 2019 by Rockmanex3 (talk | contribs) (Made the theorem more friendly for solvers who have no experience with directed segments.)

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

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 $BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB$.

Proofs

Proof with Transversals

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:

$\frac{CP}{PB}=\frac{1-P}{P}$

$\frac{BR}{AR}=\frac{1-R}{R}$

$\frac{QA}{QC}=\frac{1-Q}{Q}$

The line through $R$ and $P$ is given by: $\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0$


which yields, after simplification,

\[-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0\]

\[Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).\]

Plugging in the coordinates for $Q$ yields $(Q-1)(R-1)(P-1) = QPR$. From $\frac{CP}{PB}=\frac{1-P}{P},$ we have \[P=\frac{(1-P)\cdot PB}{CP}.\] Likewise, \[R=\frac{(1-R)\cdot AR}{BR}\] and \[Q=\frac{(1-Q)\cdot QC}{QA}.\]


Substituting these values yields \[(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdot CP\cdot BR}\] which simplifies to $QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.$

QED

Converse

The converse of Menelaus' Statement is also true. If $\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = 1$ 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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