Cute angle chasing problem. Well clearly $\triangle BCE$ is equilateral. Observe that: $$ \angle AFE = \angle ACE =\alpha \quad \text{and} \quad \angle ABE = 2\alpha $$Note that $\angle ABC =60^{\circ} - 2\alpha $. Now this implies that $\angle BCD = 120^{\circ} +2\alpha \implies DCE = 60^{\circ}+2\alpha$. Now to finish the problem observe that: $$ \angle CDE = 60^{\circ} - \alpha \implies \angle ADE = \alpha $$Consequently we have that $AD=AF = AB = BF$ and as a result $\angle AFB=60^{\circ}$