In the figure,$BA \parallel ED$ and $BC \parallel EF.$ Show that $\angle ABC + \angle DEF = 180^{\circ}.$

(N/A) Produce $ED$ to meet $BC$ at point $P$.
Since $EF \parallel BC$ and $EP$ is the transversal,the interior angles on the same side of the transversal are supplementary.
Therefore,$\angle DEF + \angle EPC = 180^{\circ} .....(1)$
Again,since $AB \parallel EP$ and $BC$ is the transversal,the corresponding angles are equal.
Therefore,$\angle ABC = \angle EPC .....(2)$
Substituting the value of $\angle EPC$ from equation $(2)$ into equation $(1)$,we get:
$\angle DEF + \angle ABC = 180^{\circ}$
Hence,$\angle ABC + \angle DEF = 180^{\circ}.$
Thus,it is proved.