+704 The vertices of triangle $ABC$ lie on the sides of equilateral triangle $DEF$, as shown. If $CD = 5$, $CE = BD = 4$, and $\angle ACB = 60^\circ$, then find $AE$.
+5 To solve this problem, consider extending $\overline{BC}$ and $\overline{AB}$ until they intersect at point $G$, forming an equilateral triangle $ABCG$.
Now, $\angle ACB = \angle ACB = \angle GCB = 60^\circ$ because all three angles of equilateral triangle $DEF$ are $60^\circ$.
Triangle $GBC$ is isosceles with $GB = GC$ because $G$ is the intersection of the equilateral triangle's sides extended. Thus, $\angle GBC = \angle GCB = 60^\circ$.
Now, $\angle BGC = 60^\circ + 60^\circ = 120^\circ$. Since $\angle ACB = 60^\circ$, the exterior angle at $B$ of triangle $ABC$ is $120^\circ - 60^\circ = 60^\circ$.
Therefore, $\triangle ABC$ is an equilateral triangle, and $AE = \boxed{4}$.
