Difference between revisions of "Menelaus' Theorem"

Revision as of 05:40, 16 May 2014

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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,

$-X\cdot(R-1)(P-1)+Y\cdotR(1-P)-Z\cdotPR = 0$ (Error compiling LaTeX. Unknown error_msg)

$Z\cdotPR = -X\cdot(R-1)(P-1)+Y\cdotR(1-P)$ (Error compiling LaTeX. Unknown error_msg)

Plugging in the coordinates for $Q$ yields:

$(Q-1)(R-1)(P-1) = QPR$

We know that, since

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

$P=\frac{(1-P)\cdot PB}{CP}$ . Likewise,

$R=\frac{(1-R)\cdot AR}{BR}$ , and

$Q=\frac{(1-Q)\cdot QC}{QA}$ .

Substituting these values yields:

$(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdotCP\cdotBR}$ (Error compiling LaTeX. Unknown error_msg)

Which simplifies to:

$QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB$

QED

@@ Line 6: / Line 6: @@
 A necessary and sufficient condition for points <math>P, Q, R</math> on the respective sides <math>BC, CA, AB</math> (or their extensions) of a triangle <math>ABC</math> to be collinear is that
-<center><math>BP\cdot CQ\cdot AR = -PC\cdot QA\cdot RB</math></center>
+<center><math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math></center>
 where all segments in the formula are [[directed segment]]s.

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Difference between revisions of "Menelaus' Theorem"

Revision as of 05:40, 16 May 2014

Contents

Statement

Proof

Proof Using Barycentric coordinates

See also