$I$ is the incenter of $\triangle ABC$. The perpendicular bisector of $CI$ cuts $AI$, $BI$ and $CA$ respectively at $D$, $E$ and $F$. The line that passes through the midpoint of $IF$ and is perpendicular to $BC$ intersects the line that passes through $E$ and is perpendicular to $AI$ at $H$. Prove that $IH \perp BD$.
This is supposed to be an easy problem yet I still can't think of a way to solve this. But I have a theory that $H$ is the centre of cyclic quadrilateral $AIFE$, if I could have proved it.
If you can solve the problem, thanks so much for that!