Triangle $ABC$ has $AC=5,BC=12,AB=13$. Point $D$ is on $\overline{AB}$, and $\overline{CD}$ bisects the right angle. The incircles of $\triangle ADC,\triangle BCD$ have radii $r_a,r_b$. What is $r_a/r_b$?

1. By the angle bisector theorem, $AD=\frac{65}{17},BD=\frac{156}{17}$, and $CD=\frac{60\sqrt2}{17}$ (from $CD=\frac{ab\sqrt2}{a+b}$ for legs $a,b$).
2. $[ACD]:[BCD]=AD:BD=5:12$; computing semiperimeters $s_a=\frac{75+30\sqrt2}{17}, s_b=\frac{180+30\sqrt2}{17}$ and $r=\text{Area}/s$.
3. $r_a/r_b=\dfrac5{12}\cdot\dfrac{s_b}{s_a}=\boxed{\dfrac5{102}\left(26-7\sqrt2\right)}$.