# Difference between revisions of "Menelaus' Theorem"

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## Revision as of 14:48, 13 December 2020

**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 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:

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

### Proof with Mass points

Let's First define some points' masses.

, , and

By Mass Points: The mass at A is Multiplying them together,

## 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.