Question 39
Let $ABC$ be an acute triangle, $O$ its circumcenter, $\omega$ its circumcircle, $H$ its orthocenter, and $M$ the midpoint of $BC$. Line $AH$ meets $\omega$ again at $D \neq A$ and line $DM$ meets $\omega$ again at $E \neq D$. Finally, line $AE$ meets the circumcircle of triangle $OME$ again at $F \neq A$. Prove that $FH = FA$. (Michal Pecho)