12. In $\triangle A B C, A B=\sqrt{5}, B C=1$ and $A C=2 . I$ is the incentre of $\triangle A B C$ and the circumcircle of $\triangle I B C$ cuts $A B$ at $P$. Find $B P$.
在 $\triangle A B C$ 中, $A B=\sqrt{5} 、 B C=1 、 A C=2 \circ I$ 是 $\triangle A B C$ 的內心，且 $\triangle I B C$ 的外接圆交 $A B$ 於 $P$ 。求 $B P$ 。
12. $\sqrt{5}-2$
12. Let $A, B, C$ denote the corresponding interior angles of $\triangle A B C$. Noting that $B, P, I, C$ are concyclic, we have
$$
\begin{array}{l}
\angle A P C=180^{\circ}-\angle C P B=180^{\circ}-\angle C I B=\frac{B}{2}+\frac{C}{2} \\
\angle A C P=180^{\circ}-A-\angle A P C=\frac{B}{2}+\frac{C}{2}
\end{array}
$$

It follows that $A P=A C=2$ and hence $B P=\sqrt{5}-2$.