1. Given that $\angle A = 20^\circ$ and $\angle AFG = \angle AGF$, we need to find the sum of $\angle B$ and $\angle D$.

2. Since $\angle AFG = \angle AGF$, let $\angle AFG = \angle AGF = x$.

3. In $\triangle AFG$, the sum of the angles is $180^\circ$. Therefore, we have:
$$\angle A + \angle AFG + \angle AGF = 180^\circ$$
Substituting the given values:
$$20^\circ + x + x = 180^\circ$$
Simplifying, we get:
$$20^\circ + 2x = 180^\circ$$
Solving for $x$:
$$2x = 160^\circ \implies x = 80^\circ$$

4. Now, we know that $\angle AFG = 80^\circ$ and $\angle AGF = 80^\circ$.

5. Next, consider $\triangle BDF$. We need to find $\angle BFD$. Since $\angle AFG = \angle AGF = 80^\circ$, the remaining angle $\angle FAG$ in $\triangle AFG$ is:
$$\angle FAG = 180^\circ - \angle AFG - \angle AGF = 180^\circ - 80^\circ - 80^\circ = 20^\circ$$

6. In $\triangle BDF$, the sum of the angles is $180^\circ$. We know that:
$$\angle BFD + \angle B + \angle D = 180^\circ$$

7. Since $\angle BFD$ is the exterior angle for $\triangle AFG$, it is equal to the sum of the remote interior angles:
$$\angle BFD = \angle AFG + \angle AGF = 80^\circ + 80^\circ = 160^\circ$$

8. Therefore, in $\triangle BDF$:
$$\angle B + \angle D = 180^\circ - \angle BFD = 180^\circ - 160^\circ = 20^\circ$$

9. However, this contradicts the earlier steps. Let's re-evaluate the problem. The correct approach is to consider the sum of the angles around point F. Since $\angle AFG = \angle AGF = 80^\circ$, the remaining angle $\angle BFD$ should be:
$$\angle BFD = 180^\circ - \angle AFG = 180^\circ - 80^\circ = 100^\circ$$

10. Therefore, the correct sum of $\angle B$ and $\angle D$ is:
$$\angle B + \angle D = 180^\circ - \angle BFD = 180^\circ - 100^\circ = 80^\circ$$

Conclusion:
$$\boxed{80^\circ}$$