From $|A B|=|C D|$ and $A B \| C D$ it follows that $A B$ is a midline in triangle $C D E$. Therefore, $A$ is the midpoint of $D E$. Since $\angle D B E=90^{\circ}$, according to Thales, $A$ is the center of the circle through $D, B$, and $E$. Thus, $|A D|=|A E|=|A B|$, and we already knew that $|A B|=|D M|=|M C|$. In the same way, using Thales, we show that $|B M|=|C M|=|D M|$, so in quadrilateral $A B M D$ all sides are of equal length. Therefore, it is a rhombus (a). The diagonals of a rhombus bisect each other, so $O$ is the midpoint of $B D$. Since $A$ is also the midpoint of $D E$, $N$ is the centroid of triangle $B D E$. Thus, $D N$ passes through the midpoint of $B E$ (b).