B. If $a, b, c$ are positive numbers, and satisfy
$$
\begin{array}{c}
a+b+c=9, \\
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{10}{9}, \\
\text { then } \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=
\end{array}
$$
B. 7 .

From the given information, we have
$$
\begin{array}{l}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\
=\frac{9-b-c}{b+c}+\frac{9-c-a}{c+a}+\frac{9-a-b}{a+b} \\
=\frac{9}{b+c}+\frac{9}{c+a}+\frac{9}{a+b}-3 \\
=9 \times \frac{10}{9}-3=7 .
\end{array}
$$