# Difference between revisions of "Menelaus' Theorem"

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The mass at A is <math>m_{3}+m_{2}</math> | The mass at A is <math>m_{3}+m_{2}</math> | ||

<cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \Rightarrow \frac{AR}{RB} = \frac{m_{1}}{m_{3}+m_{2}} </cmath> | <cmath>AR\cdot (m_{3}+m_{2}) = RB \cdot m_{1} \Rightarrow \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> | + | Multiplying them together, <math>\fbox{\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 == | == Converse == |

## Revision as of 14:50, 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, $\fbox{\frac{BP}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = \frac{{m_{2}}}{{m_{1}}} \cdot \frac{{m_{3}+m_{2}}}{{m_{2}}} \cdot \frac{{m_{1}}}{{m_{3}+m_{2}}} = 1}$ (Error compiling LaTeX. ! Missing $ inserted.)

## 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.