Main thing in this problem is to observe that $H$, orthocenter of $IEF$ lies on $BC$. Also $HI||AD$.

Consider the degenerate cases when $D=B,C$ and then you find that $H$ lies on $BC$. Then use linearity.

The degenerate cases where $D=B,C$ give $H$ lies on $BC$. This can be proved by noticing that $E$ is the intersection of the $B$-angle bisector and the perp bisector of $AD$ which gives us $E$ is the midpoint of $\overarc{AD}$ not containing $B$ in $\odot(ABD)$, a similar conclusion can be deduced for $F$ and then trivial angle chase finishes $\blacksquare$