# Difference between revisions of "Menelaus' Theorem"

Math.is.gud (talk | contribs) m (→Statement: deleted (-) in front of PC) |
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QED | QED | ||

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+ | ===Proof with Mass Points=== | ||

+ | Let's First define some points' masses. | ||

+ | |||

+ | <math>B_{m_{1}}</math>, <math>C_{m_{2}}</math>, and <math>Q_{m_{3}}</math> | ||

+ | |||

+ | By Mass Points: | ||

+ | <cmath>BP\cdot m_{1}=PC\cdot m_{2} \Rightarrow \frac{BP}{CP}=\frac{m_{2}}{m_{1}}</cmath> | ||

+ | <cmath>\frac{QC}{QA}=\frac{AC+QA}{QA}=1+\frac{AC}{QA}=1+\frac{m_{3}}{m_{2}}=\frac{m_{2}}{m_{2}}+\frac{m_{3}}{m_{2}}=\frac{m_{3}+m_{2}}{m_{2}}</cmath> | ||

+ | 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> | ||

+ | 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> | ||

== Converse == | == Converse == |

## Revision as of 15:47, 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.