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. | |
+ | |||
+ | == 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> | ||
+ | |||
+ | 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 20:27, 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.
Contents
[hide]Statement
If line intersecting
on
, where
is on
,
is on the extension of
, and
on the intersection of
and
, then
Alternatively, when written with directed segments, the theorem becomes .
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 through
to intersect
at
:
Multiplying the two equalities together to eliminate the
factor, we get:
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 the following coordinates:
Note that this says the following:
$$ (Error compiling LaTeX. Unknown error_msg)\frac{QA}{QC}=\frac{1-Q}{Q}
R
P
<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(Q-1)(R-1)(P-1) = QPR
\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.B_{m_{1}}
C_{m_{2}}
Q_{m_{3}}
m_{3}+m_{2}
{\;\; \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} = 1P, Q, R$ are collinear. The converse is useful in proving that three points are collinear.