Triangle it Out!!!!

Drop an altitude from A to BC with foot F. Let AF equal x. This means $DF =x$ , $CF =x\sqrt3$ , $\rightarrow CD=x\sqrt3-x$ . Since D is the midpoint of CB, then $CB = x\sqrt3-x$ and $DF=2x-x\sqrt3$ . Therefore $\tan\angle DAF =2-\sqrt3$ . It can easily be derived from half-angle identities that $\angle BAF =15^\circ$ . So our desired $\angle BAD =45-15=\boxed{30^\circ}$

Mr Wang's solution is more elegant than mine, definitely.

This prints the angles of ABC to be 30, 45 and 105 respectively from which the rest follows easily.