The bisector of the angle $BAC$of the acute triangle $ABC$ ( $AC \ne AB$) intersects its circumscribed circle for the second time at the point $W$. Let $O$ be the center of the circumscribed circle $\Delta ABC$. The line $AW$ intersects for the second time the circumcribed circles of triangles $OWB$ and $OWC$ at the points $N$ and $M$, respectively. Prove that $BN + MC = AW$. (Mitrofanov V., Hilko D.)