Solution:

The triangle $ABC$ is a right triangle with a right angle at $A.$ This makes $D$ the circumcenter of the triangle since it is the midpoint of the hypotenuse.

Therefore, $$AD = BD = DC = 5.$$ Also, the area of $ABC$ is $$\dfrac{6\cdot 8}2 = 24.$$

Since $BD$ and $DC$ have the same altitude and base, the triangles $ABD$ and $ACD$ have the same area of $12.$

Then, for each triangle, we have $A = rs$ where $A$ is the area, $r$ is the inradius, and $s$ is the semiperimeter. This means $12 = \frac12 {rP}$ for each triangle, where $P$ is the perimeter. Thus, we know that $$r=\dfrac{24}P.$$ We apply this fact for $ABD,$ to see that $r=\dfrac{24}{5+5+6} = \dfrac 32.$ Similarly, for $ACD,$ it $$r=\dfrac{24}{5+5+8} = \dfrac 43.$$

Their sum is $\dfrac 32 + \dfrac 43 = \dfrac{17}6 .$

Thus, the correct answer is D .