Belarus — Selection and Training Session

On the prolongation of the segment $CD$ over $D$ we mark the point $K$ such that $DK = BN$. Then the triangles $KDA$ and $NBA$ are equal since $\angle KDA = \angle NBA$, $KD = BN$, $DA = AB$. Hence $KA = NA$, $KM = DK + DM = BN + DM = NM$, so that the triangles $KMA$ and $NMA$ are equal. It easily follows that $\angle MAN = 0.5\angle DAB$. Let $O$, $C$ be the intersection points of $AC$ and the circumcircle of the triangle $MCN$.

Since $\angle MCO = \angle DCA = \angle BCA = \angle NCO$, we have $MO = NO$. Besides,
$$\angle MON = 180^\{\circ\} - \angle MCN = 180^\{\circ\} - \angle DCB = \angle DAB = 2\angle MAN.$$
This equality along with $MO = NO$ means that $O$ is the circumcenter of the triangle $MAN$.