7. 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 $\angle C$.
Answer: 35 Let the radius of the circumcircle of $\triangle A B C$ be $r$.
$$
\begin{array}{c}
O D^{2}+O C \cdot C D=O C^{2} \\
O C \cdot C D=O C^{2}-O D^{2} \\
O C \cdot C D=(O C+O D)(O C-O D) \\
O C \cdot C D=(r+O D)(r-O D)
\end{array}
$$

By the power of the point at $\mathrm{D}$,
$$
\begin{array}{c}
O C \cdot C D=B D \cdot D C \\
r=B D
\end{array}
$$

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 $\angle A C B=\frac{\angle A O B}{2}=\frac{70}{2}=35$