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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]]. It is named after Menelaus of Alexandria.
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== Statement ==
 
== Statement ==
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''(awaiting image)''
 
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A necessary and sufficient condition for points D, E, F on the respective side lines BC, CA, AB of a triangle ABC to be collinear is that
 
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<br><center><math>BD*CE*AF = -DC*EA*FB</math></center><br>
 
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where all segments in the formula are directed segments.
 
  
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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
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<cmath>\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.</cmath>
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<asy>
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unitsize(16);
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defaultpen(fontsize(8));
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pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R;
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draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);
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draw((7,6)--(6,8)--(4,0));
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R=intersectionpoint(A--B,Q--P);
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dot(A^^B^^C^^P^^Q^^R);
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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));
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</asy>
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Alternatively, when written with [[directed segment|directed segments]], the theorem becomes <math>BP\cdot CQ\cdot AR = CP\cdot QA\cdot RB</math>.
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Also, the theorem works with all three points on the extension of their respective sides.
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 +
==Proof==
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===Proof with Areas===
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<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>
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=== Proof with Similar Triangles ===
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Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>:
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<asy>
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unitsize(16);
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defaultpen(fontsize(8));
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pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R, K=(5.5,0);
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draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);
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draw((7,6)--(6,8)--(4,0));
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draw(A--K, dashed);
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R=intersectionpoint(A--B,Q--P);
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dot(A^^B^^C^^P^^Q^^R^^K);
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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));
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label("K",K,(0,-1));
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</asy>
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<cmath>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</cmath>
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<cmath>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{CP}{KP}</cmath>
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Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get:
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<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>
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=== Proof with [[Barycentric coordinates]] ===
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 +
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.
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 +
Suppose we give the points <math>P, Q, R</math> the following coordinates:
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<cmath>P: (0, P, 1-P)</cmath>
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<cmath>R: (R , 1-R, 0)</cmath>
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<cmath>Q: (1-Q ,0 , Q)</cmath>
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Note that this says the following:
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<cmath>\frac{CP}{PB}=\frac{1-P}{P}</cmath>
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<cmath>\frac{BR}{AR}=\frac{1-R}{R}</cmath>
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<cmath>\frac{QA}{QC}=\frac{1-Q}{Q}</cmath>
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The line through <math>R</math> and <math>P</math> is given by:
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<math>\begin{vmatrix} X & 0 & R \\ Y & P & 1-R\\ Z & 1-P & 0 \end{vmatrix} = 0</math>
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which yields, after simplification,
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<cmath>-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0</cmath>
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<cmath>Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).</cmath>
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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>
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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>
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=== Proof with [[Mass points]] ===
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First let's define some masses. 
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<math>B_{m_{1}}</math>, <math>C_{m_{2}}</math>, and <math>Q_{m_{3}}</math>
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By Mass Points:
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<cmath>BP\cdot m_{1}=PC\cdot m_{2} \implies \frac{BP}{CP}=\frac{m_{2}}{m_{1}}</cmath>
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<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>
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The mass at A is <math>m_{3}+m_{2}</math>
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<cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \implies \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}} </cmath>
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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>
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== Converse ==
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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.
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<asy>
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unitsize(16);
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defaultpen(fontsize(8));
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pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R;
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draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);
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draw((7,6)--(6,8)--(4,0));
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R=intersectionpoint(A--B,Q--P);
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dot(A^^B^^C^^P^^Q^^R);
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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));
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</asy>
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== See Also ==
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* [[Ceva's Theorem]]
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* [[Stewart's Theorem]]
  
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== Example ==
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[[Category:Theorems]]
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''Does anyone have a good example? ''
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[[Category:Geometry]]

Latest revision as of 13:18, 29 July 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 Areas

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

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:

\[\frac{CP}{PB}=\frac{1-P}{P}\] \[\frac{BR}{AR}=\frac{1-R}{R}\] \[\frac{QA}{QC}=\frac{1-Q}{Q}\]

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,

\[-X\cdot (R-1)(P-1)+Y\cdot R(1-P)-Z\cdot PR = 0\] \[Z\cdot PR = -X\cdot (R-1)(P-1)+Y\cdot R(1-P).\]

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


Substituting these values yields \[(Q-1)(R-1)(P-1) = \frac{(1-Q)\cdot QC \cdot (1-P) \cdot PB \cdot (1-R) \cdot AR}{QA\cdot CP\cdot BR}\] which simplifies to $QA\cdot CP \cdot BR = -QC \cdot AR \cdot PB.$

Proof with Mass points

First let's define some masses.

$B_{m_{1}}$, $C_{m_{2}}$, and $Q_{m_{3}}$

By Mass Points: \[BP\cdot m_{1}=PC\cdot m_{2} \implies \frac{BP}{CP}=\frac{m_{2}}{m_{1}}\] \[\frac{QC}{QA}=\frac{AC+QA}{QA}=1+\frac{AC}{QA}=1+\frac{m_{3}}{m_{2}}=\frac{m_{2}}{m_{2}}+\frac{m_{3}}{m_{2}}=\frac{m_{3}+m_{2}}{m_{2}}\] The mass at A is $m_{3}+m_{2}$ \[AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \implies \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}}\] 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 $\frac{BP}{PC} \cdot \frac{CQ}{QA} \cdot \frac{AR}{RB} = 1$ 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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