Mongolian Mathematical Olympiad

Note that $AF$, $FK$ are sim medians of triangles $\Delta ABF$, $\Delta BKF$ respectively. Therefore we get $\angle CMK=\angle CMP$. Since $LK$ angle bisector and quadrilateral $APLK$ is inscribed in a circle, $\angle LPM=\angle LKM$ and from this follows $PM=MK$. Since $\angle MAC\sim \angle MCK$, we get $AM/CM=MC/MK\Rightarrow AM\cdot PM=AM\cdot MK=M{C}^2=a^2/4$. On the other hand $AM\cdot AP=A{M}^2-AM\cdot PM=(1/4)(2c^2+2b^2-a^2)-(1/4)a^2=bc\cos \alpha$. The fact that quadrilateral $IHNB$ implies $AH\cdot AN=AI\cdot AB=cb\cos \alpha =AM\cdot PM$. Hence quadrilateral $HNMP$ can be inscribed in a circle and it implies $\angle APM=90^{\circ }\Rightarrow \angle HPM=\angle APH=90^{\circ }$. The proof completed.