imc

We note that by the converse of the Pythagorean Theorem, $△ABC$ is a right triangle with a right angle at $A$. Therefore, $AD=BD=CD=5$, and $[ADB]=[ADC]=12$. Since $A=rs,$ we have $r=\frac{A}{s}$, so the inradius of $△ADB$ is $\frac{12}{(5+5+6)/2}=\frac{3}{2}$, and the inradius of $△ADC$ is $\frac{12}{(5+5+8)/2}=\frac{4}{3}$. Adding the two together, we have $\boxed{\frac{17}{6}}$.