Circles with centers $A$ and $B$ have radius 3 and 8, respectively. A [ common internal tangent](https://artofproblemsolving.com/wiki/index.php/Common_internal_tangent_line) intersects the circles at $C$ and $D$, respectively. Lines $AB$ and $CD$ intersect at $E$, and $AE=5$. What is $CD$?

$\textbf{(A) } 13\qquad\textbf{(B) } \frac{44}{3}\qquad\textbf{(C) } \sqrt{221}\qquad\textbf{(D) } \sqrt{255}\qquad\textbf{(E) } \frac{55}{3}\qquad$
[2006 AMC12A-16a.png](https://artofproblemsolving.com/wiki/index.php/File:2006_AMC12A-16a.png)
$\angle AED$ 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](https://artofproblemsolving.com/wiki/index.php/Tangent_line) at a point on a circle are always perpendicular). Thus, $\triangle ACE \sim \triangle BDE$.
By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/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 $CD = CE + DE = 4 + \frac{32}{3} = \boxed{\textbf{(B) }\frac{44}{3}}$.