1. Given $\triangle A B C$ with the three angles $\angle A, \angle B, \angle C$ opposite to the sides of lengths $a, b, c$, and $\angle A=2 \angle B$. Then $\frac{a^{2}}{b(b+c)}=$ $\qquad$ .
1. 1.

As shown in Figure 4, it is easy to know,
$$
\begin{array}{l}
\triangle A B C \backsim \triangle D A C \\
\Rightarrow \frac{A B}{D A}=\frac{A C}{D C}=\frac{B C}{A C} .
\end{array}
$$

Let $B D=x, D C=y$. Then
$$
\begin{array}{l}
\frac{c}{x}=\frac{b}{y}=\frac{a}{b} \Rightarrow \frac{b+c}{x+y}=\frac{a}{b} \\
\Rightarrow b(b+c)=a^{2} \Rightarrow \frac{a^{2}}{b(b+c)}=1 .
\end{array}
$$