2. In $\triangle A B C$, $a$, $b$, $c$ are the lengths of the sides opposite to $\angle A$, $\angle B$, $\angle C$ respectively. If $\angle B=60^{\circ}$, then the value of $\frac{c}{a+b}+\frac{a}{c+b}$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$
2. C.

Draw $A D \perp C D$ at point $D$.
In Rt $\triangle B D A$, since $\angle B=60^{\circ}$, we have $D B=\frac{c}{2}, A D=\frac{\sqrt{3}}{2} c$.
In Rt $\triangle A D C$, we have
$$
\begin{array}{l}
D C^{2}=A C^{2}-A D^{2} \\
\Rightarrow\left(a-\frac{c}{2}\right)^{2}=b^{2}-\frac{3}{4} c^{2} \\
\Rightarrow a^{2}+c^{2}=b^{2}+a c .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } \frac{c}{a+b}+\frac{a}{c+b} \\
=\frac{c^{2}+c b+a^{2}+a b}{(a+b)(c+b)} \\
=\frac{a^{2}+c^{2}+a b+b c}{a c+a b+b c+b^{2}}=1 .
\end{array}
$$