2. Given $\triangle A B C$, the angle bisector of $\angle B$ intersects side $A C$ at $P$, and the angle bisector of $\angle A$ intersects side $B C$ at $Q$. If the circle passing through points $P, Q, C$ also passes through the incenter $R$ of $\triangle A B C$, and $P Q=1$, find the lengths of $P R$ and $R Q$.
Extend $C R$ and let it intersect side $A B$ at $T$, then
$$
\begin{aligned}
\angle P R Q= & \angle P R C+\angle C R Q \\
= & \angle B R T+\angle A R T \\
& =\frac{1}{2} \angle B+\frac{1}{2} \angle C+\frac{1}{2} \angle A \\
& +\frac{1}{2} \angle C \\
& =\frac{1}{2}(\angle A+\angle B)+\angle C
\end{aligned}
$$

Given that points $P, C, Q, R$ are concyclic, so
$$
\begin{array}{l}
\frac{1}{2}(\angle A+\angle B)+2 \angle C=180^{\circ}, \\
\frac{1}{2}(\angle A+\angle B+\angle C)+\frac{3}{2} \angle C=180^{\circ}
\end{array}
$$

Thus, $\angle C=60^{\circ}$, and hence $\angle P C R=\angle Q C R$ $=30^{\circ}$. This implies $P R=R Q$.

Let $P R=R Q=x$, and apply the Law of Cosines in $\triangle P Q R$:
$$
1^{2}=x^{2}+x^{2}-2 x^{2} \cos 120^{\circ} \text {, }
$$

Therefore, $P R=R Q=x=\frac{\sqrt{3}}{3}$.