# Difference between revisions of "2010 AMC 10A Problems/Problem 14"

## Problem

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

$[asy] pair A,B,C,D,E,F,G,H; G=(0,10); A=(0,3.464); B=(6,0); C=(0,0); draw(A--B--C--cycle); F=(1,1.73); E=(2,0); draw(C--F--E); D=(1.5,2.6); draw(C--D); label("A",A,W); label("B",B,S); label("C",C,S); label("F",F,N); label("D",D,NE); label("E",E,S); draw(A--E); draw(anglemark(E,A,B)); draw(anglemark(D,C,A)); [/asy]$

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

\begin{align*}\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\end{align*}

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