IrMO 2018 Problem 2

The triangle $ABC$ is right-angled at $A$ . Its incentre is $I$ , and $H$ is the foot of the perpendicular from $I$ on $AB$ . The perpendicular from $H$ on $BC$ meets $BC$ at $E$ , and it meets the bisector of $\angle ABC$ at $D$ . The perpendicular from $A$ on $BC$ meets $BC$ at $F$ . Prove that $\angle EFD = 45^o$