Mongolian Mathematical Olympiad

Denote the incircles of $BMF$, $CME$ by ${\omega }_1$, ${\omega }_2$, respectively, with their centers denoted by $I_1,I_2$. Points $P$ and $Q$ are the midpoints of $BF$ and $CE$, respectively. Point $N$ is the intersection of external common tangents of ${\omega }_1$ and ${\omega }_2$. The orthocenter of triangle $ABC$ is denoted as $H$. Given that $BC$ is external tangent, $N$ lies on $BC$. Knowing that $AD$ is the diameter of circumcenter, we have $BD\parallel CF$ and $BE\parallel CD$. Therefore, $$\angle CBD=\angle CAD=\angle BCF,$$ and $$\angle BCD=\angle BAD=90^{\circ }-\angle C=\angle CBE.$$ Thus, $BHCD$ is parallelogram, implying that points $D$, $M$, $H$ are collinear. It is also given that $$MB=MF=ME=MC.$$ Consequently, $I_1\in MP$, $I_2\in MQ$ and $M{I}_1$ is passing through the midpoint of $BH$. Since $M{I}_2\parallel BH$, $M(B{I}_{1}H{I}_2)$ is pencil. Thus, intersection points with $I_1{I}_2$ are harmonic. On the other hand, $(N{I}_{1}K{I}_2)$ is harmonic. It follows that $K$ is in $MH$.