Triangle $ABC$ has $AC=3$, $BC=4$, and $AB=5$. Point $D$ is on $\overline{AB}$, and $\overline{CD}$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b$, respectively. What is $r_a/r_b$?
$\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \frac{3}{56}\left(10-\sqrt{2}\right)\qquad\mathrm{(C)}\ \frac{1}{14}\left(10-\sqrt{2}\right)\qquad\mathrm{(D)}\ \frac{5}{56}\left(10-\sqrt{2}\right)\\\mathrm{(E)}\ \frac{3}{28}\left(10-\sqrt{2}\right)$
By the [Angle Bisector Theorem](https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem), 
\[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\]
By [Law of Sines](https://artofproblemsolving.com/wiki/index.php/Law_of_Sines) on $\triangle BCD$, 
\[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\]
Since the area of a triangle satisfies $[\triangle]=rs$, where $r =$ the [inradius](https://artofproblemsolving.com/wiki/index.php/Inradius) and $s =$ the [semiperimeter](https://artofproblemsolving.com/wiki/index.php/Semiperimeter), we have 
\[\frac{r_A}{r_B} = \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A}\]
Since $\triangle ACD$ and $\triangle BCD$ share the [altitude](https://artofproblemsolving.com/wiki/index.php/Altitude) (to $\overline{AB}$), their areas are the ratio of their bases, or \[\frac{[ACD]}{[BCD]} = \frac{AD}{BD} = \frac{3}{4}\]
The semiperimeters are $s_A = \left(3 + \frac{15}{7} + \frac{12\sqrt{2}}{7}\right)\left/\right.2 = \frac{18+6\sqrt{2}}{7}$ and $s_B = \frac{24+ 6\sqrt{2}}{7}$. Thus,
\begin{align*} \frac{r_A}{r_B} &= \frac{[ACD] \cdot s_B}{[BCD] \cdot s_A} = \frac{3}{4} \cdot \frac{(24+ 6\sqrt{2})/7}{(18+6\sqrt{2})/7} \\ &= \frac{3(4+\sqrt{2})}{4(3+\sqrt{2})} \cdot \left(\frac{3-\sqrt{2}}{3-\sqrt{2}}\right) = \frac{3}{28}(10-\sqrt{2}) \Rightarrow \mathrm{(E)}\qquad \blacksquare \end{align*}