4. In an acute $\triangle A B C$, $a$, $b$, and $c$ are the sides opposite to $\angle A$, $\angle B$, and $\angle C$ respectively, and $\angle A=60^{\circ}$. Then $\frac{c}{a+b}+\frac{b}{a+c}=(\quad)$.
(A) 1
(B) $\frac{1}{2}$
(C) $\frac{\sqrt{3}}{2}$
(D) None of the above answers
Figure 5
4. A.

As shown in Figure 5, construct $C D \perp$
$A B$ at point $D$. Then
$$
\begin{array}{c}
A D=\frac{1}{2} b \\
C D=\frac{\sqrt{3}}{2} b \\
B D=c-\frac{1}{2} b \\
\Rightarrow \frac{a^{2}}{}=\left(\frac{\sqrt{3}}{2} b\right)^{2}+\left(c-\frac{1}{2} b\right)^{2}=b^{2}+c^{2}-b c \\
\Rightarrow \frac{c}{a+b}+\frac{b}{a+c}=\frac{c(a+c)+b(a+b)}{(a+b)(a+c)} \\
=\frac{b^{2}+c^{2}+a b+a c}{a^{2}+a b+a c+b c}=1 .
\end{array}
$$