In the figure,$BA \parallel ED$ and $BC \parallel EF.$ Show that $\angle ABC = \angle DEF.$

(N/A) Produce $DE$ to intersect $BC$ at $P$.
Since $EF \parallel BC$ and $DP$ is the transversal,
$\therefore \angle DEF = \angle DPC$ ....$(1)$ [Corresponding angles]
Now,$AB \parallel ED$ (which is the same line as $DP$),and $BC$ is the transversal,
$\therefore \angle DPC = \angle ABC$ ....$(2)$ [Corresponding angles]
From $(1)$ and $(2)$,we get
$\angle ABC = \angle DEF$
Hence,proved.