2001 AMC 12 Problem 24

Let $E$ be the foot of the perpendicular from $C$ to line $AD.$ The exterior angle of $\triangle ADB$ gives $\angle ADC = 15^\circ + 45^\circ = 60^\circ,$ so $\triangle CDE$ is a $30$-$60$-$90$ triangle with $DE = \tfrac{1}{2}CD = BD.$ Then $\triangle BDE$ is isosceles with $\angle EBD = \angle BED = 30^\circ,$ and since $\angle ECB = 30^\circ$ too, $\triangle BEC$ is isosceles with $BE = EC.$ Also $\angle ABE$ $= 45^\circ - 30^\circ$ $= 15^\circ$ $= \angle EAB,$ so $\triangle ABE$ is isosceles with $AE = BE.$ Hence $AE = BE = EC,$ making right triangle $AEC$ isosceles with $\angle ECA = 45^\circ.$ Therefore $\angle ACB$ $= \angle ECA + \angle ECD$ $= 45^\circ + 30^\circ$ $= 75^\circ.$ Thus, the correct answer is D.