Iranian Mathematical Olympiad

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



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