6. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$, respectively. Given that $a^{2}=b(b+c)$ and $\angle C$ is an obtuse angle, 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) $a=2 b<c$
6. A.
$$
\because a+b>c, \therefore a^{2}=b(b+c)<b(2b+a),(a-2b) \text {. }
$$
$(a+b)<0, \therefore a<2b$. Also $\because \angle C$ is an obtuse angle, $\therefore a^{2}+b^{2}<c^{2}$, i.e., $b(b+c)+b^{2}<c^{2},(2b-c)(b+c)<0, \therefore 2b<c$, then $a$ $<2b<c$.