# Difference between revisions of "Menelaus' Theorem"

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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 | 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 | ||

− | <center><math>BP\cdot CQ\cdot AR = | + | <center><math>BP\cdot CQ\cdot AR = PC\cdot QA\cdot RB</math></center> |

where all segments in the formula are [[directed segment]]s. | where all segments in the formula are [[directed segment]]s. |

## Revision as of 22:55, 26 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