In △ABC, \angle ABC \equal{} 45^\circ. Point D is on BC so that 2 \cdot BD \equal{} CD and \angle DAB \equal{} 15^\circ. Find ∠ACB.
[asy]
pair A, B, C, D;
A = origin;
real Bcoord = 3*sqrt(2) + sqrt(6);
B = Bcoord/2*dir(180);
C = sqrt(6)*dir(120);
draw(A--B--C--cycle);
D = (C-B)/2.4 + B;
draw(A--D);label("A", A, dir(0));
label("B", B, dir(180));
label("C", C, dir(110));
label("D", D, dir(130));
[/asy]
<spanclass=′latex−bold′>(A)</span>54∘<spanclass=′latex−bold′>(B)</span>60∘<spanclass=′latex−bold′>(C)</span>72∘<spanclass=′latex−bold′>(D)</span>75∘<spanclass=′latex−bold′>(E)</span>90∘