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

Given: BA || ED and BC || EF.To Prove: ∠ ABC = ∠ DEFConstruction: Produce ED to meet BC at P.Proof:AB || ED and BPC is a transversal.   ∠ 1 = ∠ 2 (a pair of alternate angles)    …………..(i)BC and EP' are two lines intersecting at P.  ∴ ∠ 2 = ∠ 3 (Vertically opposite angles)  …………..(ii)and ∠ 4 = ∠ 5(Vertically opposite angles) …………..(iii)From (i) and (ii), ∠ 1 = ∠ 3………………(iv)BC || EF (Given)∠ 3 = ∠ 6 (Corresponding angles on the same side of the transversal)But ∠ 3 = ∠ 2 and ∠ 2 = ∠ 1  ∴ ∠ 1 = ∠ 6But ∠ 1 = ∠ ABC and ∠ 6 = ∠ DEF∴ ∠ ABC = ∠ DEF.