The altitudes $AD$ and $BE$ of an acute triangle $ABC$ intersect at point $H$. Let $F$ be the intersection of the line $AB$ and the line that is parallel to the side BC and goes through the circumcenter of $ABC$. Let $M$ be the midpoint of the segment $AH$. Prove that $\angle CMF = 90^o$
Problem
Source: 2018 Grand Duchy of Lithuania, Mathematical Contest p3 (Baltic Way TST)
Tags: geometry, right angle, Circumcenter