Let $I_a$ be the point of the center of an ex-circle of the triangle $ABC$, which touches the side $BC$ . Let $W$ be the intersection point of the bisector of the angle $\angle A$ of the triangle $ABC$ with the circumcircle of the triangle $ABC$. Perpendicular from the point $W$ on the straight line $AB$, intersects the circumcircle of $ABC$ at the point $P$. Prove, that if the points $B, P, I_a$ lie on the same line, then the triangle $ABC$ is isosceles. (Mykola Moroz)