Difference between revisions of "Menelaus' Theorem"

Revision as of 23:42, 5 December 2015

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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:

$\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=1$

Proof Using 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

@@ Line 39: / Line 39: @@
 Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get:
-<math>\frac{AR}{RB}\cdot\frac{QC}{QA}=-\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=-1</math>
+<math>\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=1</math>
 ==Proof Using [[Barycentric coordinates]]==

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Difference between revisions of "Menelaus' Theorem"

Revision as of 23:42, 5 December 2015

Contents

Statement

Proof

Proof Using Barycentric coordinates

See also