74th Romanian Mathematical Olympiad

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.