69th Belarusian Mathematical Olympiad

Answer: 60°. We will count the angle $AFB$ as the sum of angles $AFE$ and $BFE$. Note that $\angle AFE=\angle ACE$ since they share the arc $AE$ in ${\Gamma }_B$. And $\angle BFE=\angle BEF$ in the isosceles triangle $BEF$. The angle $BEF$ equals to the half of the arc $BD$, which equals to the angle $BCD$ in the circle ${\Gamma }_C$, whence $\angle BFE=0.5\angle BCD$. The diagonal $AC$ of the rhombus $ABCD$ bisects the angle $DCB$, hence $\angle BFE=0.5\angle BCD=\angle ACB$. Thus



$$\angle AFB=\angle AFE+\angle EFB=\angle ECA+\angle ACB=\angle ECB.$$ Since the triangle $ECB$ is equilateral, $\angle ECB=60^{\circ }$, and therefore $\angle AFB=60^{\circ }$.