Solution.

Construct the parallel to $AB$ through $N$ and denote by $P$ its intersection with the line $BE$.

Using the fundamental theorem of similarity in the triangle $EAB$:
$$\frac{NP}{AB} = \frac{EN}{EA} = \frac{CM}{CD}.$$
From here we obtain $NP = CM$ and, since $NP \parallel MC$, it follows that $MNPC$ is a parallelogram, so $MN \parallel CP$.

Given that $MN \perp NB$, it follows that $CP \perp NB$.

In the triangle $BNC$, $BE$ and $CP$ are the lines supporting the altitudes, so the point $P$ is the orthocenter.

Therefore $NP \perp BC$ and, since $NP \parallel CD$, it follows that $BC \perp CD$, thus $ABCD$ is a rectangle.