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| − | '''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]].
| + | #REDIRECT[[Menelaus' theorem]] |
| − | It is named for Menelaus of Alexandria.
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| − | == Statement ==
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| − | 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
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| − | <center><math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math></center>
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| − | where all segments in the formula are [[directed segment]]s.
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| − | <center><asy>
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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></center>
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| − | == Proof ==
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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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| − | <center><asy>
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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></center>
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| − | <math>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</math>
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| − | <math>\triangle QCP \sim \triangle ACK \implies \frac{QC}{QA}=\frac{PC}{PK}</math>
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| − | | |
| − | Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get:
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| − | <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>
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| − | | |
| − | ==Proof Using [[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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| − | <math>P: (0, P, 1-P)</math>
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| − | <math>R: (R , 1-R, 0)</math>
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| − | <math>Q: (1-Q ,0 , Q)</math>
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| − | Note that this says the following:
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| − | <math>\frac{CP}{PB}=\frac{1-P}{P}</math>
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| − | <math>\frac{BR}{AR}=\frac{1-R}{R}</math>
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| − | <math>\frac{QA}{QC}=\frac{1-Q}{Q}</math>
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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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| − | QED
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| − | == See also ==
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| − | * [[Ceva's Theorem]]
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| − | * [[Stewart's Theorem]]
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| − | [[Category:Theorems]]
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| − | [[Category:Geometry]]
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