# Difference between revisions of "Menelaus' Theorem"

Rockmanex3 (talk | contribs) (Made the theorem more friendly for solvers who have no experience with directed segments.) |
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'''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]]. | '''Menelaus' Theorem''' deals with the [[collinearity]] of points on each of the three sides (extended when necessary) of a [[triangle]]. | ||

It is named for Menelaus of Alexandria. | It is named for Menelaus of Alexandria. | ||

− | |||

− | |||

− | + | == Statement == | |

− | where | + | 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> | <center><asy> | ||

Line 19: | Line 18: | ||

</asy></center> | </asy></center> | ||

− | == Proof | + | 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 Transversals=== | ||

+ | |||

Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>: | Draw a line parallel to <math>QP</math> through <math>A</math> to intersect <math>BC</math> at <math>K</math>: | ||

<center><asy> | <center><asy> | ||

Line 35: | Line 39: | ||

<math>\triangle RBP \sim \triangle ABK \implies \frac{AR}{RB}=\frac{KP}{PB}</math> | <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{ | + | <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: | Multiplying the two equalities together to eliminate the <math>PK</math> factor, we get: | ||

− | <math>\frac{AR}{RB}\cdot\frac{QC}{QA}= | + | <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. | 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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QED | QED | ||

− | == See | + | == 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]] | * [[Ceva's Theorem]] | ||

* [[Stewart's Theorem]] | * [[Stewart's Theorem]] |

## Revision as of 17:03, 30 November 2019

**Menelaus' Theorem** deals with the collinearity of points on each of the three sides (extended when necessary) of a triangle.
It is named for Menelaus of Alexandria.

## Contents

## 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 .

## Proofs

### Proof with Transversals

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:

The line through and is given by:

which yields, after simplification,

Plugging in the coordinates for yields . From we have Likewise, and

Substituting these values yields which simplifies to

QED

## Converse

The converse of Menelaus' Statement is also true. If in the below diagram, then are collinear. The converse is useful in proving that three points are collinear.