Example 3 In the right triangle $ABC$, $\angle C=90^{\circ}$, $BC=a$, $CA=b$, $AB=c$. If $\frac{b}{c+a}+\frac{a}{c+b}=\frac{13}{15}$, find: $a: b: c$.
Explanation: Using the Pythagorean theorem to transform the two ratios on the left side of the given equation, we have
$$
\begin{array}{l}
\frac{b}{c+a}=\frac{c-a}{b}, \frac{a}{c+b}=\frac{c-b}{a}. \\
\text { Therefore, } \frac{13}{15}=\frac{b}{c+a}+\frac{a}{c+b}=\frac{c-a}{b}+\frac{c-b}{a} \\
=\frac{a c-a^{2}+b c-b^{2}}{a b}=\frac{a c+b c-c^{2}}{a b} \\
=\frac{2 c(a+b-c)}{2 a b}=\frac{2 c(a+b-c)}{(a+b+c)(a+b-c)} \\
=\frac{2 c}{a+b+c}.
\end{array}
$$

From this, it is easy to get
$$
13(a+b)=17 c \text {, }
$$

or $169(a+b)^{2}=289 c^{2}=289\left(a^{2}+b^{2}\right)$.
Simplifying, organizing, and transforming, we get
$$
60\left(\frac{a}{b}\right)^{2}-169\left(\frac{a}{b}\right)+60=0.
$$

Solving this, we get $\frac{a}{b}=\frac{5}{12}$ or $\frac{12}{5}$.
Thus, $a: b: c=5: 12: 13$ or $12: 5: 13$.