# 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.

Title proof of Menelaus’ theorem ProofOfMenelausTheorem 2013-03-22 12:46:46 2013-03-22 12:46:46 mathwizard (128) mathwizard (128) 4 mathwizard (128) Proof msc 51A05