Difference between revisions of "Menelaus' Theorem"

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#REDIRECT[[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 for Menelaus of Alexandria.
 
== Statement ==
 
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>
 
 
 
where all segments in the formula are [[directed segment]]s.
 
 
 
<center><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></center>
 
 
 
== Proof ==
 
Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>:
 
<center><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></center>
 
<math>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</math>
 
 
 
<math>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}</math>
 
 
 
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>
 
 
 
==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>
 
 
 
Note that this says the following:
 
 
 
<math>\frac{CP}{PB}=\frac{1-P}{P}</math>
 
 
 
<math>\frac{BR}{AR}=\frac{1-R}{R}</math>
 
 
 
<math>\frac{QA}{QC}=\frac{1-Q}{Q}</math>
 
 
 
The line through <math>R</math> and <math>P</math> is given by:
 
<math>\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0</math>
 
 
 
 
 
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>
 
 
 
 
 
We know that, since
 
 
 
 
 
<math>\frac{CP}{PB}=\frac{1-P}{P}</math>,
 
 
 
 
 
<math>P=\frac{(1-P)\cdot PB}{CP}</math>. Likewise,
 
 
 
 
 
<math>R=\frac{(1-R)\cdot AR}{BR}</math>, and
 
 
 
 
 
<math>Q=\frac{(1-Q)\cdot QC}{QA}</math>.
 
 
 
 
 
Substituting these values yields:
 
 
 
 
 
<math>(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdotCP\cdotBR}</math>
 
 
 
Which simplifies to:
 
 
 
 
 
<math>QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB</math>
 
 
 
QED
 
 
 
== See also ==
 
* [[Ceva's Theorem]]
 
* [[Stewart's Theorem]]
 
 
 
[[Category:Theorems]]
 
 
 
[[Category:Geometry]]
 

Latest revision as of 16:20, 9 May 2021

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