smc

$\angle AED$ and $\angle BEC$ are vertical angles so they are congruent, as are angles $\angle ADE$ and $\angle BCE$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $△ACE\sim △BDE$. By the Pythagorean Theorem, line segment $DE=4$. The sides are proportional, so $\frac{DE}{AD}=\frac{CE}{BC}\Rightarrow \frac{4}{3}=\frac{CE}{8}$. This makes $CE=\frac{32}{3}$ and $CD=CE+DE=4+\frac{32}{3}=\boxed{\frac{44}{3}}$.