Mongolian Mathematical Olympiad

Let denote $\angle BAD=\alpha$. Then $\angle BCA=BAD=\alpha$.

Since $\angle BCA=90^{\circ }-\alpha =\angle AEM\Rightarrow \angle MED=\alpha$ and we conclude that $△EMD$ is. Similarly, it is easy to show $AM=EM=MD$. In other words, point $M$ is the midpoint of $AD$. Similarly, $N$ is the midpoint of $BC$. Denote $P$ as the midpoint of $AC$. It implies that $NP=\frac{1}{2}AB$ and $MP=\frac{1}{2}CD$.



Therefore the inequality $MN\le NP+MP=\frac{1}{2}(CD+AB)$ holds. If points $N$, $P$, $M$ lie on a line then the $MN=\frac{1}{2}(CD+AB)$ equality holds and $NP\parallel MP\Rightarrow NP\parallel AB$ and $MP\parallel CD\Rightarrow CD\parallel AB$. It means $ABCD$ is a trapezoid.