In the given figure,$AB \parallel DE$ and $BC \parallel EF$. Prove that $\angle ABC + \angle DEF = 180^{\circ}$.

(N/A) Given: $AB \parallel DE$ and $BC \parallel EF$.
To prove: $\angle ABC + \angle DEF = 180^{\circ}$.
Construction: Extend $ED$ to meet $BC$ at point $G$.
Proof:
Since $AB \parallel DE$ and $ED$ is extended to $G$,we have $AB \parallel DG$.
Since $AB \parallel DG$ and $BG$ is a transversal,the interior angles on the same side are supplementary,but here $AB$ and $DG$ are parallel and $BC$ is a transversal,so $\angle ABC + \angle BGD = 180^{\circ}$ (Consecutive interior angles).
Now,consider $BC \parallel EF$. Since $BC \parallel EF$ and $EG$ is a transversal,$\angle BGD = \angle DEF$ (Corresponding angles).
Substituting $\angle BGD = \angle DEF$ in the first equation,we get $\angle ABC + \angle DEF = 180^{\circ}$.
Hence proved.