Triangle $ABC$ has $AB=\frac{\sqrt{3}+1}{2}\cdot AC$. 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 $△CFE$ is equilateral. What is $\angle ACB$?