Fall Mathematical Competition

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 $△ABC$. 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 $C{D}_1$, we see as above that the quadrilateral $MBC{N}_1$ is cyclic. Hence the points $M,B,C,N$ and $N_1$ are concyclic. However, this is impossible when $D\ne D_1$ since then $N_1$ lies on the midsegment of $△CDE$ through $N$, which means that $N_1$ is inside $△MCN$.