Question

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 $\angle ACB.$

$54^\circ$

$60^\circ$

$72^\circ$

$75^\circ$

$90^\circ$

Solution:

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.

2001 AMC 12 Problem 24

Solution:

Problem 24 in Other Years