NL-NWO 2020 Problem 3

Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$ . The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$ . Point $O$ is the centre of the circle through $M, C$ , and $N$ . Prove that $\angle OBC = \angle ODC$ .