find $c$ and $b$ in terms of $x$ and $a$

Question

I have this geometry problem.

Supose any $\triangle{ABC}$, where $\overline{CE} \perp \overline{AB}$; $\overline{CM}$ is median; $n$ is $proy_{\overline{CM}}\overline{AB}$; $\angle{CMA}$ is obtuse.

Find:

$c^2;b^2;c^2+b^2$ in terms of $a$ and $x$

Answer 1

By Pythagoras' Theorem, from triangle $CME$, we have that $EM=n=\sqrt{x^2-h^2}$.
Similarly, by Pythagoras' Theorem, from triangle $CBE$, we have that $BE=\sqrt{a^2-h^2}$.
Since $CM$ is a median, $BM=\frac{c}{2}$ and also we have $BM=BE+EM$.
Hence, we get $$\frac{c}{2}=\sqrt{a^2-h^2}+\sqrt{x^2-h^2}$$

See if this helps you even a little bit.

Answer 2

For $CME$, $EM=\sqrt{x^2-h^2}=n$

for $CBE$, we have that $BE=\sqrt{a^2-h^2}$

We have,$BM=BE+EM$, Therefore,$BM=\sqrt{a^2-h^2}+\sqrt{x^2-h^2}$

As CM is median,we can say, $c=2BM= 2[\sqrt{a^2-h^2}+\sqrt{x^2-h^2}]$

Also, $b^2 = AC^2 = CE^2+AE^2=h^2+(AM+ME)^2 = h^2+(c/2+n)^2$

therefore, $b^2=h^2+[(\sqrt{a^2-h^2}+\sqrt{x^2-h^2})+\sqrt{x^2-h^2}]^2$

$b^2= h^2+(\sqrt{a^2-h^2}+2\sqrt{x^2-h^2})^2$

You get $b,c$ so you can calculate $b^2+c^2$ also.