jmc

$△ABC$ is a right triangle, since $9^2+12^2=15^2$. Thus, $\angle ABC=90^{\circ }$.



By the angle bisector theorem, $BD/AD=BC/AC$, so $$BD=\frac{BC}{BC+AC}\cdot AB=\frac{4}{9}\cdot 9=4.$$ Then by use of the Pythagorean Theorem on right triangle $BCD$, $CD=\sqrt{B{C}^2+B{D}^2}=\sqrt{12^2+4^2}=\sqrt{160}=\boxed{4\sqrt{10}}$.