# Find the value of $MF^2$ with the given information

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.

• When you write, for example, $\sqrt58$, do you mean $\sqrt{58}$? – Gerry Myerson Dec 5 '16 at 8:35
• Yes. Sorry. I don't know how to write it correctly. – jonyoung2002 Dec 5 '16 at 8:44
• Like this: \sqrt{58}. Only, enclosed in dollar signs. – Gerry Myerson Dec 5 '16 at 8:53
• I tried this problem again and I still can't do it... Please help... – jonyoung2002 Dec 5 '16 at 8:58
• Maybe there are some facts about the orthocenter that would come in handy. en.wikipedia.org/wiki/Altitude_(triangle)#Orthocenter – Gerry Myerson Dec 5 '16 at 12:03

$MF$ is the median of right triangle $CFA$ and is thus half the hypotenuse $AC$.