View source for 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 after Menelaus of Alexandria.

== Statement ==

If line <math>PQ</math> intersecting <math>AB</math> on <math>\triangle ABC</math>, where <math>P</math> is on <math>BC</math>, <math>Q</math> is on the extension of <math>AC</math>, and <math>R</math> on the intersection of <math>PQ</math> and <math>AB</math>, then
<cmath>\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.</cmath>

<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 segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR = CP\cdot QA\cdot RB</math>.
Also, the theorem works with all three points on the extension of their respective sides.

==Proof==

===Proof with Areas===

<cmath>\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = \frac{[BQR]}{[QRC]} \cdot \frac{[CQR]}{[QRA]} \cdot \frac{[QRA]}{[QRB]} = -1.</cmath>

=== Proof with Similar Triangles ===

Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>:
<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>
<cmath>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</cmath>
<cmath>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{CP}{KP}</cmath>
Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get:
<cmath>\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{CP}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{CP}=1</cmath>

=== 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 <math>P, Q, R</math> the following coordinates:

<cmath>P: (0, P, 1-P)</cmath>
<cmath>R: (R , 1-R, 0)</cmath>
<cmath>Q: (1-Q ,0 , Q)</cmath>

Note that this says the following:

<cmath>\frac{CP}{PB}=\frac{1-P}{P}</cmath>
<cmath>\frac{BR}{AR}=\frac{1-R}{R}</cmath>
<cmath>\frac{QA}{QC}=\frac{1-Q}{Q}</cmath>

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, 

<cmath>-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0</cmath>
<cmath>Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).</cmath>

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


Substituting these values yields <cmath>(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}</cmath> which simplifies to <math>QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.</math>

=== Proof with [[Mass points]] ===
First let's define some masses.  

<math>B_{m_{1}}</math>, <math>C_{m_{2}}</math>, and <math>Q_{m_{3}}</math>

By Mass Points: 
<cmath>BP\cdot m_{1}=PC\cdot m_{2} \implies \frac{BP}{CP}=\frac{m_{2}}{m_{1}}</cmath>
<cmath>\frac{QC}{QA}=\frac{AC+QA}{QA}=1+\frac{AC}{QA}=1+\frac{m_{3}}{m_{2}}=\frac{m_{2}}{m_{2}}+\frac{m_{3}}{m_{2}}=\frac{m_{3}+m_{2}}{m_{2}}</cmath>
The mass at A is <math>m_{3}+m_{2}</math>
<cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \implies \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}} </cmath>
Multiplying them together,<math>{\;\; \frac{BP}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = \frac{{m_{2}}}{{m_{1}}} \cdot \frac{{m_{3}+m_{2}}}{{m_{2}}} \cdot \frac{{m_{1}}}{{m_{3}+m_{2}}} = 1}</math>

== Converse ==

The converse of Menelaus' theorem is also true.  If <math>\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = 1</math> in the below diagram, then <math>P, Q, R</math> 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 ==

* [[Ceva's Theorem]]
* [[Stewart's Theorem]]

[[Category:Theorems]]
[[Category:Geometry]]