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Menelaus' Theorem: Difference between revisions

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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 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 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>
<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>
 
=== 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$ 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]]

Revision as of 19:27, 28 April 2025

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 $PQ$ intersecting $AB$ on $\triangle ABC$, where $P$ is on $BC$, $Q$ is on the extension of $AC$, and $R$ on the intersection of $PQ$ and $AB$, then \[\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.\]

[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 segments, the theorem becomes $BP\cdot CQ\cdot AR = CP\cdot QA\cdot RB$. Also, the theorem works with all three points on the extension of their respective sides.

Proof

Proof with Similar Triangles

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

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 $P, Q, R$ the following coordinates:

\[P: (0, P, 1-P)\] \[R: (R , 1-R, 0)\] \[Q: (1-Q ,0 , Q)\]

Note that this says the following:

\[\frac{CP}{PB}=\frac{1-P}{P}\] \[\frac{BR}{AR}=\frac{1-R}{R}\] $$ (Error compiling LaTeX. Unknown error_msg)\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,

<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$ (Error compiling LaTeX. Unknown error_msg)Q$yields$(Q-1)(R-1)(P-1) = QPR$. From$\frac{CP}{PB}=\frac{1-P}{P},$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$ (Error compiling LaTeX. Unknown error_msg)QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.$=== Proof with [[Mass points]] === First let's define some masses.$B_{m_{1}}$,$C_{m_{2}}$, and$Q_{m_{3}}$By Mass Points: <cmath>BP\cdot m_{1}=PC\cdot m_{2} \implies \frac{BP}{CP}=\frac{m_{2}}{m_{1}}</cmath> <cmath>\frac{QC}{QA}=\frac{AC+QA}{QA}=1+\frac{AC}{QA}=1+\frac{m_{3}}{m_{2}}=\frac{m_{2}}{m_{2}}+\frac{m_{3}}{m_{2}}=\frac{m_{3}+m_{2}}{m_{2}}</cmath> The mass at A is$m_{3}+m_{2}$<cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \implies \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}} </cmath> Multiplying them together,${\;\; \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}$== Converse ==

The converse of Menelaus' theorem is also true. If$ (Error compiling LaTeX. Unknown error_msg)\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = 1$in the below diagram, then$P, Q, R$ 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

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