Problems

The bisector of the outer corner at the vertex $C$ of the triangle $ABC$ intersects the circumscribed circle at the point $D$. Prove that $AD=BD$.

In the triangle $ABC$, the height $AH$ is drawn; $O$ is the center of the circumscribed circle. Prove that $\mathrm{\angle }OAH=|\mathrm{\angle }B-\mathrm{\angle }C$|.