smc

We start by finding the length of $AD$ and $BD$ as in solution 1. Using the angle bisector theorem, we see that $AD=\frac{15}{7}$ and $BD=\frac{20}{7}$. Using Stewart's Theorem gives us the equation $5d^2+\frac{1500}{49}=\frac{240}{7}+\frac{180}{7}$, where $d$ is the length of $CD$. Solving gives us $d=\frac{12\sqrt{2}}{7}$, so $CD=\frac{12\sqrt{2}}{7}$. Call the incenters of triangles $ACD$ and $BCD$ $O_a$ and $O_b$ respectively. Since $O_a$ is an incenter, it lies on the angle bisector of $\angle ACD$. Similarly, $O_b$ lies on the angle bisector of $\angle BCD$. Call the point on $CD$ tangent to $O_a$ $M$, and the point tangent to $O_b$ $N$. Since $△C{O}_{a}M$ and $△C{O}_{b}N$ are right, and $\angle O_{a}CM=\angle O_{b}CN$, $△C{O}_{a}M\sim △C{O}_{b}N$. Then, $\frac{r_a}{r_b}=\frac{CM}{CN}$. We now use common tangents to find the length of $CM$ and $CN$. Let $CM=m$, and the length of the other tangents be $n$ and $p$. Since common tangents are equal, we can write that $m+n=\frac{12\sqrt{2}}{7}$, $n+p=\frac{15}{7}$ and $m+p=3$. Solving gives us that $CM=m=\frac{6\sqrt{2}+3}{7}$. Similarly, $CN=\frac{6\sqrt{2}+4}{7}$. We see now that $\frac{r_a}{r_b}=\frac{\frac{6\sqrt{2}+3}{7}}{\frac{6\sqrt{2}+4}{7}}=\frac{6\sqrt{2}+3}{6\sqrt{2}+4}=\frac{60-6\sqrt{2}}{56}=\frac{3}{28}(10-\sqrt{2})\Rightarrow \boxed{\frac{3}{28}(10-\sqrt{2})}$