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
Draw $AE\bot BC$. Since $\angle ADC=\angle B+\angle BAD=60^\circ$, triangle $ADE$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, and since $\angle B=45^\circ$, $BE=AE$.
Let $BD=1$, $CD=2$, $DE=x$; then $AE=\sqrt3 x$ and $BE=1+x$, so $1+x=\sqrt3x$ and $x=\frac{\sqrt3+1}2$.
