Let $A B C$ be an obtuse triangle with circumcenter $O$ such that $\angle A B C=15^{\circ}$ and $\angle B A C>90^{\circ}$. Suppose that $A O$ meets $B C$ at $D$, and that $O D^{2}+O C \cdot D C=O C^{2}$. Find $\angle C$.
Let the radius of the circumcircle of $\triangle A B C$ be $r$. \[ \begin{gathered} O D^{2}+O C \cdot C D=O C^{2} \\ O C \cdot C D=O C^{2}-O D^{2} \\ O C \cdot C D=(O C+O D)(O C-O D) \\ O C \cdot C D=(r+O D)(r-O D) \end{gathered} \] By the power of the point at $\mathrm{D}$, \[ \begin{gathered} O C \cdot C D=B D \cdot D C \\ r=B D \end{gathered} \] Then, $\triangle O B D$ and $\triangle O A B$ and $\triangle A O C$ are isosceles triangles. Let $\angle D O B=\alpha . \angle B A O=90-\frac{\alpha}{2}$. In $\triangle A B D, 15+90-\frac{\alpha}{2}=\alpha$. This means that $\alpha=70$. Furthermore, $\angle A C B$ intercepts minor arc $A B$, thus $\angle A C B=\frac{\angle A O B}{2}=\frac{70}{2}=35$ $\fbox{35}$.