In figure, arms BA and BC of ∠ ABC are respectively parallel to arms ED and EF of ∠ DEF. Prove that ∠ ABC + ∠ DEF = 180°.

Given: BA || ED and BC|| EF.
To Prove: ∠ABC + ∠DEF = 180° .
Construction: Produce FE to meet AB at E'.
Proof: BA || ED and FEE' is the transversal.
∠BE'E = ∠DEF……………..(i)
BC || E'EF and AB is the transversal.
∠ BE'E = E'BC………………(ii)
BC || E'EF and AB is the transversal.
∠E'BC + ∠BE'E = 180° [ Angles on the same side of the transversal are supplementary]
⇒ ∠ABC + ∠BE'E = 180° [∠E'BC = ∠ABC]
⇒ ∠ABC + ∠DEF = 180° [From (i) ∠BE'E  = ∠DEF]
\\ ∠ABC + ∠DEF = 180°