Circles with centers $A$ and $B$ have radii $4$ and $9$, respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $AB$ and $CD$ intersect at $E$, and $AE=5$. What is $CD$?

1. A tangent line is perpendicular to the radius at the point of tangency, so $AC\perp CD$ and $BD\perp CD$. Hence $\triangle ACE$ and $\triangle BDE$ are similar right triangles (they share the angle at $E$), with $AC:BD=4:9$.
2. In the right triangle $ACE$, $AE=5$ and $AC=4$, so $CE=\sqrt{5^2-4^2}=3$.
3. By similarity, $DE:CE=BD:AC=9:4$, so $DE=\dfrac94\cdot3=\dfrac{27}4$.
4. $E$ lies between $C$ and $D$ (the tangent is an internal tangent), so $CD=CE+DE=3+\dfrac{27}4=\boxed{\dfrac{39}4}$.