# Difference between revisions of "Menelaus' Theorem"

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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{PC}{PB}\implies \frac{AR}{RB}\cdot\frac{QC}{QA}\cdot\frac{PB}{PC}=1</math> |

==Proof Using [[Barycentric coordinates]]== | ==Proof Using [[Barycentric coordinates]]== |

## Revision as of 00:42, 6 December 2015

*This article is a stub. Help us out by expanding it.*

**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.

## Statement

A necessary and sufficient condition for points on the respective sides (or their extensions) of a triangle to be collinear is that

where all segments in the formula are directed segments.

## Proof

Draw a line parallel to through to intersect at :

Multiplying the two equalities together to eliminate the factor, we get:

## Proof Using 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