4. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$. If $\angle B=2 \angle A$, then which of the following relationships is true? ( ).
(A) $\frac{a}{b}>\frac{a+b}{a+b+c}$
(B) $\frac{a}{b}<\frac{a+b}{a+b+c}$
(C) $\frac{a}{b}=\frac{a+b}{a+b+c}$
(D) Cannot be determined
4.C.

As shown in Figure 5, extend $C B$ to $D$, such that $A B = B D$. Then
$$
\begin{array}{l}
\angle 1 = \angle 2. \\
Also, \angle C B A = 2 \angle C A B, \\
\angle C B A = \angle 1 + \angle 2 \\
= 2 \angle 1,
\end{array}
$$

Therefore, $\angle C A B = \angle 1 = \angle 2$.
Thus, $\triangle A C B \sim \triangle D C A$.
So, $\frac{A C}{C B} = \frac{C D}{A C}$, which means $\frac{b}{a} = \frac{a+c}{b}$.
Then $b^{2} = a^{2} + a c, b^{2} + a b = a^{2} + a c + a b$. Hence $\frac{a+b}{a+b+c} = \frac{a}{b}$.