SELECTION and TRAINING SESSION

Without loss of generality assume that $P$ lies between $A$ and $Q$. Let the line passing through $M$ parallel to $AC$ intersect the sides $AB$ and $BC$ at the points $K$ and $L$ respectively. Since $M$ is the midpoint of the arc $PQ$, the line $KL$ is parallel to $AC$ whence $\omega$ is the tangency point of $B$-excircle of the triangle $BKL$. Thus the point $X=BM\cap AC$ is the tangency point of $B$-excircle of the triangle $BAC$. Therefore $$2AX=AC+BC-AB=AC+\sqrt{CP\cdot CQ}-\sqrt{AP\cdot AQ},$$ we used the fact that the difference $BC-AB$ equals to the difference between the lengths of the tangents from $C$ and $A$ to $\omega$. Note that $2AX$ doesn't depend on $\omega$, this circle must only pass through $P$ and $Q$. Therefore if $Y=DM\cap AC$ then $2AX=2AY$, i.e. $X=Y$ and $BM$, $DN$ and $AC$ are concurrent.