There are rhombus $ABCD$ and circle $\Gamma_B$, which is centred at $B$ and has radius $BC$, and circle $\Gamma_C$, which is centred at $C$ and has radius $BC$. Circles $\Gamma_B$ and $\Gamma_C$ intersect at point $E$. The line $ED$ intersects $\Gamma_B$ at point $F$. Find all possible values of $\angle AFB$.
1. **Identify the properties of the rhombus and circles:**
   - Since \(ABCD\) is a rhombus, all sides are equal: \(AB = BC = CD = DA\).
   - Circle \(\Gamma_B\) is centered at \(B\) with radius \(BC\).
   - Circle \(\Gamma_C\) is centered at \(C\) with radius \(BC\).
   - Since \(BC\) is the radius of both circles, \(\Gamma_B\) and \(\Gamma_C\) intersect at points \(B\) and \(E\), forming an equilateral triangle \(\triangle BCE\).

2. **Determine angles in \(\triangle BCE\):**
   - Since \(\triangle BCE\) is equilateral, each angle is \(60^\circ\):
     \[
     \angle BEC = \angle EBC = \angle ECB = 60^\circ
     \]

3. **Analyze the angles involving point \(E\):**
   - Since \(ABCD\) is a rhombus, \(\angle ABC = \angle BCD = \angle CDA = \angle DAB = 90^\circ\).
   - Let \(\alpha\) be the angle \(\angle ABE\). Then:
     \[
     \angle ABE = 2\alpha
     \]
   - Since \(\angle ABC = 60^\circ - 2\alpha\), we have:
     \[
     \angle BCD = 120^\circ + 2\alpha
     \]
   - Consequently, \(\angle DCE = 60^\circ + 2\alpha\).

4. **Determine \(\angle CDE\) and \(\angle ADE\):**
   - Since \(\angle CDE = 60^\circ - \alpha\), we have:
     \[
     \angle ADE = \alpha
     \]

5. **Analyze the properties of point \(F\):**
   - Since \(AD = AF = AB = BF\), \(\triangle AFB\) is isosceles with \(AF = BF\).
   - Since \(\angle AFE = \angle ACE = \alpha\), we have:
     \[
     \angle AFB = 60^\circ
     \]

The final answer is \(\boxed{60^\circ}\).