jmc

Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$. Then, since $\angle BAD=\angle ADC$ and $\angle ABD=\angle DCE$, we know that $△ABD\sim △DCE$. Hence $\angle ADB=\angle DEC$, and $△BDE$ is isosceles. Then $BD=BE=10$. Using the similarity, we have: $$\frac{AB}{BD}=\frac{8}{10}=\frac{CD}{CE}=\frac{CD}{16}⟹CD=\frac{64}{5}$$ The answer is $m+n=\boxed{69}$.