What is D?

wording has been corrected

Let $P$ be the midpoint of $\overarc{BAC}$, and $J$ be the intersection of $\overrightarrow{AH}$ and $(ABC)$. $\angle DFC=90^o-\angle BCA=\angle JAC=\angle JPC$. $\therefore D, C, F, P$ are concyclic. Similarly, $D, B, E, P$ are concyclic. $\angle EPF=\angle EPD+\angle DPF=180^o-\angle EBD+180^o-\angle DCF=360^o-2\angle ABC-2\angle BCA=2\angle BAC=\angle EAF$. $\therefore A, P, F, E$ are concyclic.

This problem feels similar to 2012 EGMO/7, though not concretely. Let $M$ be the midpoint of major arc $\widehat{BC}$. It suffices to show $E, M, A, F$ are concyclic. Observe that $\angle EAF = 2\angle A$. Claim. [Key observation] Let $Q$ be the reflection of $H$ over $\overline{AB}$ and define $P$ similarly. Then $E, Q, M$ and $A, P, F$ are collinear. Notice that $HPFD$ and $EQHD$ are isosceles trapezoids. Then $$\angle HPF = \angle PHD = 90^\circ + \frac A2$$while $$\angle MPB = \frac 12 \widehat{MB} = 90^\circ - \frac A2.$$The other collinearity follows similarly. $\blacksquare$ Then $$\angle EMF = \angle QMP = 180^\circ - \angle QBP = 2\angle A = \angle EAF,$$done.