Question

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 $\angle ACB?$

$60^\circ$

$75^\circ$

$90^\circ$

$105^\circ$

$120^\circ$

Solution:

Let $\angle BAE = \angle ACD = x.$ Note that $\angle CFE = 60^{\circ}$ since $\triangle CFE$ is equilateral.

We then have that $\angle AFC = 180^{\circ} - \angle CFE = 120^{\circ}.$

Then: $\begin{align*} \angle FAC &= 180^{\circ} - 120^{\circ} - x\\ &=60^{\circ} - x \\ &= \angle EAC.\end{align*}$

We then get that $\begin{align*} \angle BAC &= \angle BAE + \angle EAC\\ &= x + 60^{\circ} - x \\&= 60^{\circ}. \end{align*}$

Since $AB = 2 \cdot AC$ and $\angle BAC = 60^{\circ},$ we have that $\triangle ABC$ is a $30-60-90$ triangle.

Thus, C is the correct answer.

2010 AMC 10A Problem 14

Solution:

Problem 14 in Other Years