First we prove that $E$, $M$ and $F$ are collinear. Let $H'$ be the reflection of $H$ with respect to $M$. It is known that $AH'$ is diameter of circumcircle of triangle $ABC$, and so $\angle H'CA = 90^\circ$.

Let $E'$ be the intersection point of perpendicular line to $HH'$ through $M$ with $AC$. The goal is to show that $E' \equiv E$.
$\angle H'ME' + \angle H'CE' = 180^\circ$, thus $H'ME'C$ is a cyclic quadrilateral and thus
$$\angle C = \angle EH'M = \angle E'HM \implies E' \equiv E$$
Using the same argument, it is proved that $\angle HMF = 90^\circ$. So $EF$ is the perpendicular bisector of $HH'$. Hence $E$, $M$ and $F$ are collinear. Now we show that $DAH'K$ is cyclic and points $E$ and $F$ lie on the circumcircle passing through these points. Notice that
$$\frac{HM}{HH'} = \frac{HN}{HA} = \frac{1}{2},$$
therefore $AH' \parallel MN$ and $\angle MDK = \angle MNK = \angle H'AK$ so $DAH'K$ is cyclic. It suffices to prove that $E$ lies on circumcircle of $DAH'$.
$$\left. \begin{array}{l} \angle EH'D = \angle C \\ \angle DAE = \angle A + \angle B \end{array} \right\} \implies \angle EH'D + \angle DAE = 180^\circ \\ \implies DAEH' \text{ is cyclic.}$$
Hence the claim of the problem.