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2010 AMC 12A Problems/Problem 8

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Problem 8

Triangle $ABC$ has $AB=2 \cdot AC$ . Let $D$ and $E$ be on $\overline{AB}$ and $\overline{BC}$ , respectively, such that $\angle BAE = \angle ACD$ . Let $F$ be the intersection of segments $AE$ and $CD$ , and suppose that $\triangle CFE$ is equilateral. What is $\angle ACB$ ?

$\textbf{(A)}\ 60^\circ \qquad \textbf{(B)}\ 75^\circ \qquad \textbf{(C)}\ 90^\circ \qquad \textbf{(D)}\ 105^\circ \qquad \textbf{(E)}\ 120^\circ$

Solution

Let $\angle BAE = \angle ACD = x$ .

$\angle BCD = \angle AEC = 60^\circ$

$\angle EAC + \angle FCA + \angle ECF + \angle AEC = \angle EAC + x + 60^\circ + 60^\circ = 180^\circ$

$\angle EAC = 60^\circ - x$

$\angle BAC = \angle EAC + \angle BAE = 60^\circ - x + x = 60^\circ$

Since $\frac{AC}{AB} = \frac{1}{2}$ , $\angle BCA = \boxed{90^\circ\ \textbf{(C)}}$

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