Problem

Source: 2003 Peru Cono Sur TST P3

Tags: geometry



Let $M$ and $N$ be points on the side $BC$ of a triangle $ABC$ such that $BM = CN$ ($M$ lies between $B$ and $N$). Points $P$ and $Q$ lie on $AN$ and $AM$ respectively, so that $\angle PMC =\angle MAB$ and $\angle QNB = \angle NAC$. Prove that $\angle QBC = \angle PCB$.