2. Let the three sides of the right triangle $\triangle ABC$ be $a$, $b$, and $c$, and $a < b < c$. If $\frac{b}{c+a} + \frac{a}{c+b} = \frac{13}{15}$, then $a: b: c=$ $\qquad$
2. 5: 12: 13.

Since $c^{2}-a^{2}=b^{2}, c^{2}-b^{2}=a^{2}$, we have
$$
\begin{array}{l}
\frac{13}{15}=\frac{b}{c+a}+\frac{a}{c+b}=\frac{c-a}{b}+\frac{c-b}{a} \\
=\frac{c(a+b-c)}{a b}=\frac{2 c(a+b-c)}{(a+b)^{2}-\left(a^{2}+b^{2}\right)} \\
=\frac{2 c}{a+b+c}
\end{array}
$$

Thus, $13(a+b)=17 c$.
Squaring both sides and simplifying, we get
$$
\begin{array}{l}
60 a^{2}-169 a b+60 b^{2}=0 \\
\Rightarrow(5 a-12 b)(12 a-5 b)=0 \\
\Rightarrow \frac{a}{b}=\frac{5}{12} \text { or } \frac{12}{5} .
\end{array}
$$

Since $5^{2}+12^{2}=13^{2}$, and $a<b<c$, then $a: b: c=5: 12: 13$.