Question

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

Answer 1

Given AB || DE, BC || EF

To prove: `∠`ABC + `∠`DEF =180°

Construction: produce BC to intersect DE at M

Proof: Since AB || EM and BL is the transversal

`∠`ABC = `∠`EML          [Corresponding angle]            ……(1)

Also,  

EF || ML and EM is the transversal

By the property of co-interior angles are supplementary

`∠`DEF + `∠`EML = 180°

From (i) and (ii) we have

∴`∠`DEF + `∠`ABC = 180°