A tetrahedron $ABCD$ with acute-angled faces is inscribed in a sphere with center $O$. A line passing through $O$ perpendicular to plane $ABC$ crosses the sphere at point $D'$ that lies on the opposide side of plane $ABC$ than point $D$. Line $DD'$ crosses plane $ABC$ in point $P$ that lies inside the triangle $ABC$. Prove, that if $\angle APB=2\angle ACB$, then $\angle ADD'=\angle BDD'$.