Difference between revisions of "Menelaus' Theorem"

Revision as of 00:30, 29 May 2013

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Menelaus's 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:

$P: (0,P,1-P)$

$R: (R,1-R,0)$

$Q: (1-Q,0,Q)$

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:

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

QED

@@ Line 41: / Line 41: @@
 <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]]==
+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 <math>P, Q, R</math> the following coordinates:
+<math>P: (0,P,1-P)</math>
+<math>R: (R,1-R,0)</math>
+<math>Q: (1-Q,0,Q)</math>
+The line through <math>R</math> and <math>P</math> is given by:
+<center><math>\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0</math>
+</center>
+Which yields, after simplification,
+<center> <math>-X\cdot(R-1)(P-1)+Y\cdotR(1-P)-Z\cdotPR = 0</math>
+<math>Z\cdotPR = -X\cdot(R-1)(P-1)+Y\cdotR(1-P)</math>
+Plugging in the coordinates for <math>Q</math> yields:
+<math>(Q-1)(R-1)(P-1) = QPR</math>
+</center>
+QED
 == See also ==
 * [[Ceva's Theorem]]

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Difference between revisions of "Menelaus' Theorem"

Revision as of 00:30, 29 May 2013

Contents

Statement

Proof

Proof Using Barycentric Coordinates

See also