We notice that $$\angle AND=360^{\circ}-\angle MNA-\angle MND=\angle MCB+\angle MBC=180^{\circ}-\angle BMC.$$Let $M'$ be the image of $M$ in the translation that takes $B$ to $A$ and $C$ to $D$. We see that $$\angle AM'D=\angle BMC=180^{\circ}-\angle AND,$$so $AM_1DN$ is a cyclic quadrilateral. We also find out $$180^{\circ}-\angle ANM=\angle MCB=\angle ADM_1=\angle ANM_1,$$$$180^{\circ}-\angle DNM=\angle MBC=\angle DAM_1=\angle DNM_1.$$Both imply that $M,M_1,N$ are collinear, and because $MM_1\|AB$, we have $MN\|AB$.