In a parallelogram $ABCD$, the bisector of $\angle A$ intersects $BC$ at $M$ and the extension of $DC$ at $N$. Let $O$ be the circumcircle of the triangle $MCN$. Prove that $\angle OBC = \angle ODC$
Source: SMO Senior 2019 Q1
Tags: geometry, angle bisector, parallelogram, circumcircle
In a parallelogram $ABCD$, the bisector of $\angle A$ intersects $BC$ at $M$ and the extension of $DC$ at $N$. Let $O$ be the circumcircle of the triangle $MCN$. Prove that $\angle OBC = \angle ODC$