Let $ABC$ be an acute triangle ($AB < AC$) which is circumscribed by a circle with center $O$. $BE$ and $CF$ are two altitudes and $H$ is the orthocenter of the triangle. Let $M$ be the intersection of $EF$ and $BC$.
Now draw the diameter $AK$ of the circle and let I be the intersection of $KH$ and the circle.
Prove $A,I,M$ are in the same line, i.e, collinear.
I have an idea that if we can prove $MH$ to be perpendicular to $AD$ where $D$ is the midpoint of $BC$, then $H$ would be the orthocenter of triangle $AMD$, which means $DH$ is perpendicular to $AM$. However, I still cannot prove that $MH$ is perpendicular to $AD$
I include here the figure of this problem.
Point $M$ is the radical center of the circle with diameter $AH$ (passing through $E$, $F$, $I$), the circle with diameter $BC$ (passing through $E$, $F$) and the circumcircle of $\triangle ABC$. So done.