0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ When you write, for example, $\sqrt58$, do you mean $\sqrt{58}$? $\endgroup$ – Gerry Myerson Dec 5 '16 at 8:35
  • $\begingroup$ Yes. Sorry. I don't know how to write it correctly. $\endgroup$ – jonyoung2002 Dec 5 '16 at 8:44
  • $\begingroup$ Like this: \sqrt{58}. Only, enclosed in dollar signs. $\endgroup$ – Gerry Myerson Dec 5 '16 at 8:53
  • $\begingroup$ I tried this problem again and I still can't do it... Please help... $\endgroup$ – jonyoung2002 Dec 5 '16 at 8:58
  • $\begingroup$ Maybe there are some facts about the orthocenter that would come in handy. en.wikipedia.org/wiki/Altitude_(triangle)#Orthocenter $\endgroup$ – Gerry Myerson Dec 5 '16 at 12:03
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.