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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.

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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?

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  • $\begingroup$ $(\frac{b-a}{2})^2 = \frac{a^2 - b^2}{2c} \cdot \frac{c(a-b)}{2(a+b)} = \frac{(a^2-b^2)c(a-b)}{4c(a+b)} = \frac{(a^2-b^2)(a-b)}{4(a+b)} = \frac{(a^2-b^2)(a-b)^2}{4(a+b)(a-b)} = \frac{(a^2-b^2)(a-b)^2}{4(a^2-b^2)} = \frac{(a-b)^2}{4}$ which is correct because of $(a-b)^2 = (b-a)^2$. Thank you! $\endgroup$ – Feuermagier Aug 29 '19 at 12:46
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Aug 29 '19 at 14:08

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