Let $G$ on $BC$ be the foot of the perpendicular from $I$ on $BC$, then
$$|IG| = |IH|$$
is the radius of the incircle of $\triangle ABC$. Because $AI$ is the angle bisector of the angle at $A$, we have $2\angle HAI = \angle BAC$.

Because $BH$ and $BG$ are tangent to the incircle of triangle $ABC$, we have
$$|BG| = |BH|$$
Because $DE$ is parallel to $IG$, the Intercept Theorem implies
$$\frac{|BE|}{|BG|} = \frac{|DE|}{|IG|}$$
Because $EH \parallel AF$, the Intercept Theorem or the similarity of $\triangle BEH$ and $\triangle BFA$ implies
$$\frac{|BE|}{|BH|} = \frac{|EF|}{|HA|}$$
Combining the above, we obtain
$$\frac{|DE|}{|IH|} = \frac{|DE|}{|IG|} = \frac{|BE|}{|BG|} = \frac{|BE|}{|BH|} = \frac{|EF|}{|HA|}$$
Because $\angle FED = 90^\circ = \angle AHI$, we see now that the triangles $FED$ and $AHI$ are similar. In particular, $\angle BAC = 2\angle HAI = 2\angle EFD$.