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 $MAN$.