jmc

Since $\angle AFG=\angle AGF$ and $\angle GAF+\angle AFG+\angle AGF=180^{\circ },$ we have $20^{\circ }+2(\angle AFG)=180^{\circ }.$ So $\angle AFG=80^{\circ }.$ Also, $\angle AFG+\angle BFD=180^{\circ },$ so $\angle BFD=100^{\circ }.$ The sum of the angles of $△BFD$ is $180^{\circ },$ so $\angle B+\angle D=\boxed{80^{\circ }}.$

Note: In $△AFG,$ $\angle AFG=\angle B+\angle D.$ In general, an exterior angle of a triangle equals the sum of its remote interior angles. For example, in $△GAF,$ $\angle AFE=\angle GAF+\angle AGF.$