IMO Shortlist

Consider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\angle H K T = \angle H Q K$ and we are to prove that $\angle M K T = \angle C F K$. Thus it remains to show that $\angle H Q K = \angle C F K + \angle H K M$. Due to $\angle H Q K = 90^{\circ} - \angle Q' H A'$, and $\angle C F K = 90^{\circ} - \angle K F A$, this means the same as $\angle Q' H A' = \angle K F A - \angle H K M$. Now, since the triangles $K H E$ and $A H Q'$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\angle K F A = \angle H J A$, and analogously one may obtain $\angle H K M = \angle J Q H$. Thereby our task is reduced to verifying
$$\angle Q' H A' = \angle H J A - \angle J Q H .$$

Figure 1

Figure 2
To avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\angle Q' H A' = \angle J Q H + \angle H J Q$ and $\angle H J A = \angle Q J A + \angle H J Q$, we just have to show that $2 \angle J Q H = \angle Q J A$. To this end, it suffices to remark that $A Q A' Q'$ is a rectangle and that $J$, being defined to be the midpoint of $H Q'$, has to lie on the mid parallel of $Q A'$ and $Q' A$.

We define the points $A'$ and $E$ and prove that the ray $M H$ passes through $Q$ in the same way as in the first solution. Notice that the points $A'$ and $E$ can play analogous roles to the points $Q$ and $K$, respectively: point $A'$ is the second intersection of the line $M H$ with $\Gamma$, and $E$ is the point on $\Gamma$ with the property $\angle H E A' = 90^{\circ}$ (see Figure 3).

In the circles $K Q H$ and $E A' H$, the line segments $H Q$ and $H A'$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $M H$. Let $R$ be the radical center of the circles $A B C, K Q H$ and $E A' H$. Their pairwise radical axes are the lines $Q K$, $A' E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $H R$; by $\angle Q K H = \angle H E A' = 90^{\circ}$, the quadrilateral $H E R K$ is cyclic and its circumcenter is $S$; hence we have $S K = S E = S H$. The line $B C$, being the perpendicular bisector of $H E$, passes through $S$.
The circle $H M F$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $H M F$ we have
$$S M \cdot S F = S H^{2} = S K^{2} .$$
So, the power of $S$ with respect to the circles $K Q H$ and $K F M$ is $S K^{2}$. Therefore, the line segment $S K$ is tangent to both circles at $K$.

Figure 3