Let $ABCD$ be a cyclic quadrilateral. Since $\angle ABD = \angle CBD$ it follows that $AD = CD$. Denote by $S$ the midpoint of $DE$. Then $SM = \frac{AD}{2} = \frac{CD}{2} = CN$ and $SN \parallel AC$. Hence $MCNS$ is an isosceles trapezoid and therefore it is cyclic. On the other hand, we have $\angle MSB = \angle ADB = \angle ACB$ and we conclude that the quadrilateral $MBCS$ is inscribed. Thus the points $M, B, C, N$ and $S$ are concyclic, i.e. the quadrilateral $MBCN$ is cyclic.

Conversely, let $MBCN$ be a cyclic quadrilateral. Let us denote by $D_1$ the intersection point of the line $BD$ and the circumcircle of $\triangle ABC$. We shall prove that $D_1 \equiv D$. Let $D_1$ lie between $B$ and $D$ (the case, when $D$ is between $B$ and $D_1$, is analogous). If $N_1$ is the midpoint of $CD_1$, we see as above that the quadrilateral $MBCN_1$ is cyclic. Hence the points $M, B, C, N$ and $N_1$ are concyclic. However, this is impossible when $D \neq D_1$ since then $N_1$ lies on the midsegment of $\triangle CDE$ through $N$, which means that $N_1$ is inside $\triangle MCN$.