The triangle $ABC$ with $AB> AC$ is inscribed in a circle, the angle bisector of $\angle BAC$ intersects the side $BC$ of the triangle at the point $K$, and the circumscribed circle at the point $M$. The midline of $\Delta ABC$, which is parallel to the side $AB$, intersects $AM$ at the point $O$, the line $CO$ intersects the line $AB$ at the point $N$. Prove that a circle can be circumscribed around the quadrilateral $BNKM$. (Nagel Igor)