1. Let $a, b, c$ be positive real numbers, satisfying $a^{2}=b(b+c), b^{2}=c(c+a)$.
Then ( ).
(A) $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$
(B) $\frac{1}{b}=\frac{1}{a}+\frac{1}{c}$
(C) $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$
(D) $\frac{2}{c}=\frac{1}{a}+\frac{1}{b}$
- 1. C.

Multiplying the two equations given in the problem, we get
$$
a^{2}(b+c)(b-c)=a b c(b+c) \text {. }
$$

Since $a>0, b>0, c>0$, it follows that $b+c>0$.
Thus, $a(b-c)=b c \Rightarrow a b=b c+a c$.
Therefore, $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$.