30. [20] Let $A B C$ be a triangle with circumcenter $O$, incenter $I, \angle B=45^{\circ}$, and $O I \| B C$. Find $\cos \angle C$.
Answer: $1-\frac{\sqrt{2}}{2}$ Let $M$ be the midpoint of $B C$, and $D$ the foot of the perpendicular of $I$ with $B C$. Because $O I \| B C$, we have $O M=I D$. Since $\angle B O C=2 \angle A$, the length of $O M$ is $O A \cos \angle B O M=$ $O A \cos A=R \cos A$, and the length of $I D$ is $r$, where $R$ and $r$ are the circumradius and inradius of $\triangle A B C$, respectively.
Thus, $r=R \cos A$, so $1+\cos A=(R+r) / R$. By Carnot's theorem, $(R+r) / R=\cos A+\cos B+\cos C$, so we have $\cos B+\cos C=1$. Since $\cos B=\frac{\sqrt{2}}{2}$, we have $\cos C=1-\frac{\sqrt{2}}{2}$.