Example 6 Let $a, b, c$ be the lengths of the sides of $\triangle ABC$, and $\frac{a}{b}=\frac{a+b}{a+b+c}$. Then the relationship between the interior angles $\angle A, \angle B$ is ( ).
(A) $\angle B>2 \angle A$
(B) $\angle B=2 \angle A$
(C) $\angle B<2 \angle A$
(D) Uncertain
(2000, National Junior High School Mathematics Competition)
The given proportional equation can be simplified as
$$
b^{2}=a(a+c) \Rightarrow \frac{a}{b}=\frac{b}{a+c} \text {. }
$$

Therefore, as shown in Figure 11, we can
construct similar triangles with $a, b$ and $b, a+c$
as corresponding sides. It is easy to see that
$$
\angle B=2 \angle A \text {. }
$$