Difference between revisions of "Menelaus' Theorem"

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'''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]].
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#REDIRECT[[Menelaus' theorem]]
It is named for 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>
 
 
 
<center><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></center>
 
 
 
Alternatively, when written with [[directed segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math>.
 
 
 
== Proofs ==
 
 
 
===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>:
 
<center><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></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{CP}{KP}</math>
 
 
 
Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get:
 
 
 
<math>\frac{AR}{RB}\cdot\frac{QC}{QA}=\frac{CP}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{CP}=1</math>
 
 
 
===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:
 
 
 
<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,
 
 
 
<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>
 
 
 
QED
 
 
 
===Proof with Mass Points===
 
Let's First define some points' 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} \Rightarrow \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} \Rightarrow \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' Statement 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.
 
 
 
<center><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></center>
 
 
 
== 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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