The angle $POQ$ is given ($OP$ and $OQ$ are rays). Let $M$ and $N$ be points inside the angle $POQ$ such that $\angle POM = \angle QON$ and $\angle POM < \angle PON$. Consider two circles: one touches the rays $OP$ and $ON$, the other touches the rays $OM$ and $OQ$. Denote by $B$ and $C$ the points of their intersection. Prove that $\angle POC = \angle QOB$.