# 2001 AMC 12 Problems/Problem 24

## Problem

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 $\angle ACB$.

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

## Solution

$[asy] unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(4,0), D=intersectionpoint( A -- (dir(15)*100), B -- (B+100*dir(135)) ), C=B+3*(D-B); pair ortho=rotate(-90)*(D-A); pair E=intersectionpoint(A--D, C--(C+10*ortho)); draw(A--B--C--cycle); draw(A--D); draw(C--E, dashed); draw(B--E, dashed); draw( rightanglemark(C,E,D) ); draw( scale(2)*anglemark(B,A,D) ); draw( anglemark(D,B,A) ); label("15^\circ",(0.6,0),5*ENE); label("45^\circ",B,5*WNW,UnFill); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,NE); label("E",E,S); [/asy]$

We start with the observation that $\angle ADB = 180^\circ - 15^\circ - 45^\circ = 120^\circ$, and $\angle ADC = 15^\circ + 45^\circ = 60^\circ$.

We can draw the height $CE$ from $C$ onto $AD$. In the triangle $CED$, we have $\frac {ED}{CD} = \cos EDC = \cos 60^\circ = \frac 12$. Hence $ED = CD/2$.

By the definition of $D$, we also have $BD=CD/2$, therefore $BD=DE$. This means that the triangle $BDE$ is isosceles, and as $\angle BDE=120^\circ$, we must have $\angle BED = \angle EBD = 30^\circ$.

Then we compute $\angle ABE = 45^\circ - 30^\circ = 15^\circ$, thus $\angle ABE = \angle BAE$ and the triangle $ABE$ is isosceles as well. Hence $AE=BE$.

Now we can note that $\angle DCE = 180^\circ - 90^\circ - 60^\circ = 30^\circ$, hence also the triangle $EBC$ is isosceles and we have $BE=CE$.

Combining the previous two observations we get that $AE=EC$, and as $\angle AEC=90^\circ$, this means that $\angle CAE = \angle ACE = 45^\circ$.

Finally, we get $\angle ACB = \angle ACE + \angle ECD = 45^\circ + 30^\circ = \boxed{75^\circ}$.