$FH \perp CE \perp AB$, so $FH \parallel AB$. In this case, we have $\angle KFH = \angle AEK =\angle HDK$, the last is because $B, C, D, E, K$ are lies on one circle. But now we have that $\angle KFH = \angle HDK$, so $H, D, K, F$ are concyclic. So, $\angle HKE = \angle HDA = 90^{\circ}$ and we are done.

$FH^2=FD.FC=FK.FE\implies HK\perp EF$

$CE \perp AB \perp FH \implies FH \parallel AB$ As $EKDC$ is cyclic $$\angle DCE = \angle FKD = 90 - \angle A$$but $$\angle DFH = \angle CAB = \angle A \implies \angle FHD = 90 - \angle A$$ hence we get $FDHK$ cycic which give us $HK \perp EF$ $\blacksquare$

Clearly $\angle{ACE}=90^{\circ}-A$ and $\angle{DHC}=A \implies \angle{FHD}=90^{\circ}-A$ Since $K \in \odot{EDC} \implies EKDC$ is cyclic $\implies \angle{EKD}=90^{\circ}+A \implies \angle{FKD}=90^{\circ}-A \implies \angle{FKD}=\angle{FHD} \implies FDHK$ is cyclic. Hence $\angle{HKD}=\angle{HFD}=A \implies \angle{HKE}=90^{\circ} \implies HK \perp EF$. $\square$