<s>[INST] find $c$ and $b$ in terms of $x$ and $a$ 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 $x$
 [/INST] 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.
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