Acute triangle $ABC$ has orthocenter H. The foot of the altitude from $C$ to $AB$ is point $F$ and the foot of the altitude from $A$ to $BC$ is $D$. Let $M$ be the midpoint of $AC$ and let $N$ be that midpoint of $HC$. If $AH=10$ , $HD=6$, and $CN= \sqrt{58}$, compute $MF^2$.
So I know I can pythag to find $AC=\sqrt{452}$ and $HC=\sqrt{232}$.
I find that $FM$ is the perpendicular bisector of $AD$.
Now, I don't know how to continue.
$MF$ is the median of right triangle $CFA$ and is thus half the hypotenuse $AC$.