If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$, then $\angle B+\angle D =$

$\text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ$
As a strategy, think of how $\angle B + \angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\angle B$ or $\angle D$ alone.  If you see $\triangle BFD$, then you can see that the problem is solved quickly after determining $\angle BFD$.
But start with $\triangle AGF$, since that's where most of our information is.  Looking at $\triangle AGF$, since $\angle F = \angle G$, and $\angle A = 20$, we can write:
$\angle A + \angle G + \angle F = 180$
$20 + 2\angle F = 180$
$\angle AFG = 80$
By noting that $\angle AFG$ and $\angle GFD$ make a straight line, we know
$\angle AFG + \angle GFD = 180$
$80 + \angle GFD = 180$
$\angle GFD = 100$
Ignoring all other parts of the figure and looking only at $\triangle BFD$, you see that $\angle B + \angle D + \angle F = 180$.  But $\angle F$ is the same as $\angle GFD$.  Therefore:
$\angle B + \angle D + \angle GFD = 180$
$\angle B + \angle D + 100 = 180$
$\angle B + \angle D = 80^\circ$, and the answer is thus $\boxed{D}$