AMC 12 2008a Problem 24

Triangle $ABC$ has $\angle C = 60^{\circ}$ and $BC = 4$. Point $D$ is the midpoint of $BC$. What is the largest possible value of $\tan{\angle BAD}$?

$\mathrm{(A)}\ \frac{\sqrt{3}}{6} \mathrm{(B)}\ \frac{\sqrt{3}}{3} \mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}} \mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3} \mathrm{(E)}\ 1$