AMC 12 2001 Problem 24

In $\triangle ABC$, $\angle ABC=45^\circ$. Point $D$ is on $\overline{BC}$ so that $2\cdot BD=CD$ and $\angle DAB=15^\circ$. Find $\angle ACB$.

$\text{(A) }54^\circ \text{(B) }60^\circ \text{(C) }72^\circ \text{(D) }75^\circ \text{(E) }90^\circ$