2. Given rhombus $A B C D, \Gamma_{B}$ and $\Gamma_{C}$ are circles centered at $B$ and passing through $C$, and centered at $C$ and passing through $B$, respectively. $E$ is one of the intersection points of circles $\Gamma_{B}$ and $\Gamma_{C}$, and the line $E D$ intersects circle $\Gamma_{B}$ at a second point $F$. Find the size of $\angle A F B$.
2. As shown in Figure 3.

Obviously, $\angle A F B=\angle A F E+\angle B F E$.
Notice that, $\angle A F E=\angle A C E$.
Since $B E=B F$, we have,
$$
\angle B F E=\angle B E F .
$$

And $\angle B E F$ is equal to half the degree of arc $\overparen{B D}$ of circle $\Gamma_{c}$, thus,
$$
\angle B E F=\frac{1}{2} \angle B C D \text {. }
$$

By the diagonals of rhombus $A B C D$ bisecting $\angle D C B$
$$
\begin{array}{c}
\Rightarrow \angle B F E=\frac{1}{2} \angle B C D=\angle A C B \\
\Rightarrow \angle A F B=\angle A F E+\angle E F B \\
=\angle E C A+\angle A C B=\angle E C B .
\end{array}
$$

Since $\triangle E C B$ is an equilateral triangle
$$
\begin{array}{l}
\Rightarrow \angle E C B=60^{\circ} \\
\Rightarrow \angle A F B=60^{\circ} .
\end{array}
$$