The angle bisector of BAD meet DC on K. 1) $ CM \perp BN \ \Longrightarrow \ K \equiv N$: clearly $ AN \perp BN$, so we must have $ AK \parallel CM$ and then $ \angle BMC = \frac{\angle DAB}{2}= 180 - \angle MBA - \angle BMC = \angle BCM$ and then $ \overline{AB}=2 \overline{BC}$. Call J the midpoint of CD, so BCJM and MJDA are rombus, so the angle bisector of $ \angle MAC$ and of $ \angle MAD$ pass through J (i.e. $ K \equiv N \equiv J$) 2) $ K \equiv N \ \Longrightarrow \ CM \perp BN$: we get $ \angle NBC = \angle BNC$ and $ \angle NAD = \angle AND$, from which N is the midpoint of DC, from which BCNM and MNDA are rombus, from which $ CM \perp BN$