Clockwise $\pi/3$ rotation wrt $C$ sends $A\to B$, $D\to E$ $\therefore$ the answer is $\pi/3$

Let $(CDE) \cap AD = F\neq A$. $\angle AFC = \pi - \angle CFD = 60^{\circ} = \angle ABC$.

Afo wrote: Let $(CDE) \cap AD = F\neq A$. $\angle AFC = \pi - \angle CFD = 60^{\circ} = \angle ABC$. I sketch it, everything is ok, but how we know that AD and BE intersect at F? How is that F point defined, please reply