$17 \cdot 115$ In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$, respectively. It is known that $a^{2}=$ $b(b+c)$, and $\angle C$ is an obtuse angle, then the size relationship of $a, 2 b, c$ is
(A) $a<2 b<c$.
(B) $a<c<2 b$.
(C) $2 b<a<c$.
(D) $a=2 b<c$.
(Huanggang Region, Hubei Province, China Junior High School Mathematics Competition, 1996)
[Solution] $\because a+b>c$,
$$
\begin{array}{l}
\therefore \quad a^{2}=b(b+c)<b(2 b+a),(a-2 b) \cdot(a+b)<0, \\
\therefore \quad a<2 b .
\end{array}
$$

Also, $\because \angle C$ is an obtuse angle, $\therefore a^{2}+b^{2}<c^{2}$,
which means $b(b+c)+b^{2}<c^{2},(2 b-c)(b+c)<0$,
$\therefore 2 b<c$, then $a<2 b<c$.
Therefore, the answer is $(A)$.