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Homeproof of Menelaus' theorem

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# proof of Menelaus’ theorem

First we note that there are two different cases: Either the line connecting $X$, $Y$ and $Z$ intersects two sides of the triangle or none of them. So in the first case that it intersects two of the triangle’s sides we get the following picture:

From this we follow ($h_{1}$, $h_{2}$ and $h_{3}$ being undircted):

$\displaystyle\frac{AZ}{ZB}$ | $\displaystyle=$ | $\displaystyle-\frac{h_{1}}{h_{2}}$ | ||

$\displaystyle\frac{BY}{YC}$ | $\displaystyle=$ | $\displaystyle\frac{h_{2}}{h_{3}}$ | ||

$\displaystyle\frac{CX}{XA}$ | $\displaystyle=$ | $\displaystyle\frac{h_{3}}{h_{1}}.$ |

Mulitplying all this we get:

$\frac{AZ}{ZB}\cdot\frac{BY}{YC}\cdot\frac{CX}{XA}=-\frac{h_{1}h_{2}h_{3}}{h_{2% }h_{3}h_{1}}=-1.$ |

The second case is that the line connecting $X$, $Y$ and $Z$ does not intersect any of the triangle’s sides:

In this case we get:

$\displaystyle\frac{AZ}{ZB}$ | $\displaystyle=$ | $\displaystyle-\frac{h_{1}}{h_{2}}$ | ||

$\displaystyle\frac{BY}{YC}$ | $\displaystyle=$ | $\displaystyle-\frac{h_{2}}{h_{3}}$ | ||

$\displaystyle\frac{CX}{XA}$ | $\displaystyle=$ | $\displaystyle-\frac{h_{3}}{h_{1}}.$ |

So multiplication again yields Menelaus’ theorem.

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## Versions

(v4) by mathwizard 2013-03-22