XXIX Rioplatense Mathematical Olympiad

Let $N^{\prime }$ be the symmetric of $N$ relative to $M$. Since the diagonals cut in half, $A{N}^{\prime }CN$ and $D{N}^{\prime }BN$ are parallelograms. Hence, $\angle D{N}^{\prime }A=\angle BNC$ and $\angle B{N}^{\prime }C=\angle DNA$. As $\angle DNA=\angle BNC$, then the quadrilaterals $D{N}^{\prime }NA$ and $BN{N}^{\prime }C$ are cyclic. Thus, $\angle MNC=\angle N^{\prime }NC=\angle A{N}^{\prime }N$ ($A{N}^{\prime }\parallel NC$) and $\angle A{N}^{\prime }N=\angle NDA$ (cyclic $D{N}^{\prime }NA$), hence $\angle MNC=\angle NDA$. Similarly, $\angle MND=\angle NCB$.