Q:

Prove the following equation involving angle bisector and altitude?

In a triangle $ABC$ with $a > b$ is $M$ the midpoint of $c$, $CW$ the angle bisector of $\gamma$ and $CL$ the altitude on $c$. I have to prove that
$(\frac{b-a}{2})^2 = MW \cdot ML$. How can I do this? I tried Pythagoras' theorem in the triangles $CMW$ and $CML$, but that leads to way to long and complex equations.

A:

In the standard notation 
$$ML=\frac{c}{2}-b\cos\alpha=\frac{c}{2}-b\cdot\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-b^2}{2c}.$$
In another hand,
$$MW=BW-BM=\frac{ac}{a+b}-\frac{c}{2}=\frac{c(a-b)}{2(a+b)}.$$
Can you end it now?

