2001 AMC 12 第 24 题

The exterior angle gives $\angle ADC$ $= \angle DAB + \angle ABD$ $= 15^\circ + 45^\circ$ $= 60^\circ$

Drop a perpendicular from $C$ to line $AD$ at $E;$ then $\triangle CDE$ is $30$-$60$-$90$ with $DE = \tfrac{1}{2}CD = BD$

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.