Point $M$ is the midpoint of base $AD$ of an isosceles trapezoid $ABCD$ with circumcircle $\omega$. The angle bisector of $ABD$ intersects $\omega$ at $K$. Line $CM$ meets $\omega$ again at $N$. From point $B$, tangents $BP, BQ$ are drawn to $(KMN)$. Prove that $BK, MN, PQ$ are concurrent. A. Kuznetsov