# 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.

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?
• $(\frac{b-a}{2})^2 = \frac{a^2 - b^2}{2c} \cdot \frac{c(a-b)}{2(a+b)} = \frac{(a^2-b^2)c(a-b)}{4c(a+b)} = \frac{(a^2-b^2)(a-b)}{4(a+b)} = \frac{(a^2-b^2)(a-b)^2}{4(a+b)(a-b)} = \frac{(a^2-b^2)(a-b)^2}{4(a^2-b^2)} = \frac{(a-b)^2}{4}$ which is correct because of $(a-b)^2 = (b-a)^2$. Thank you! – Feuermagier Aug 29 '19 at 12:46