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):

${\displaystyle \frac{AZ}{ZB}}$ $=$ $-{\displaystyle \frac{h_1}{h_2}}$

${\displaystyle \frac{BY}{YC}}$ $=$ ${\displaystyle \frac{h_2}{h_3}}$

${\displaystyle \frac{CX}{XA}}$ $=$ ${\displaystyle \frac{h_3}{h_1}}.$

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:

${\displaystyle \frac{AZ}{ZB}}$ $=$ $-{\displaystyle \frac{h_1}{h_2}}$

${\displaystyle \frac{BY}{YC}}$ $=$ $-{\displaystyle \frac{h_2}{h_3}}$

${\displaystyle \frac{CX}{XA}}$ $=$ $-{\displaystyle \frac{h_3}{h_1}}.$

So multiplication again yields Menelaus’ theorem.

Title	proof of Menelaus’ theorem
Canonical name	ProofOfMenelausTheorem
Date of creation	2013-03-22 12:46:46
Last modified on	2013-03-22 12:46:46
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	4
Author	mathwizard (128)
Entry type	Proof
Classification	msc 51A05