AHSME 1983 Problem 19

Point $D$ is on side $CB$ of triangle $ABC$. If $\angle{CAD} = \angle{DAB} = 60^\circ\text{, }AC = 3\text{ and }AB = 6$, then the length of $AD$ is $\textbf{(A)} \ 2 \textbf{(B)} \ 2.5 \textbf{(C)} \ 3 \textbf{(D)} \ 3.5 \textbf{(E)} \ 4$