Let $ABCD$ be a cyclic quadrilateral, $BL$ and $CN$ be the internal angle bisectors in triangles $ABD$ and $ACD$ respectively. The circumcircles of triangles $ABL$ and $CDN$ meet at points $P$ and $Q$. Prove that the line $PQ$ passes through the midpoint of the arc $AD$ not containing $B$.
My solution: Let $M$ be the midpoint of arc $AD$ not containing $B$. Then by Shooting Lemma, $ML \cdot MB = MN \cdot MC$. So $M$ lies on radical axis of $\odot (ABL)$ and $\odot (CDN)$, i.e. $PQ$.