Mongolian Mathematical Olympiad

Let us denote $\angle PNM=\alpha$. Then $\angle PMC=\angle PNB=180^{\circ }-\alpha$ and $\angle PAC=\angle PMB$. Hence $△APM\sim △MPN$. As we have $AM=MC$, $MN=NB$ it yields $△APC\sim △MPB$. It implies $\angle APC=\angle MPB$ and $\angle MPC=\angle NPB$. Therefore, $\angle BPC=90^{\circ }$. Since $\angle PBM=\angle PCM$, quadrilateral $BCMP$ is cyclic. Hence $\angle MPC=\angle MBC$. Also, $\angle PMC=\angle BNA=\angle BNC$, which means $△MPC\sim △NBC$. --- Hence $\angle BCN=\angle PCA$ is true. It implies that $\angle NCA=\angle PCB$. Also, we have $\angle APM=\angle BPC=90^{\circ }$. By AAA property, we have $△APM\sim △CPB$.