Question

In the figure, arms BA and BC of $\angle ABC$ are respectively parallel to arms ED and EF of $\angle DEF$. Prove that $\angle ABC$ = $\angle DEF$ = ${180}^0$

Answer 1

Given: In $\angle ABCand\angle DEF$
BA ||ED and BC || EF
To prove: $\angle ABC+\angle DEF={180}^0$
Construction : Produce BC to H intersecting ED at G
Proof: $\because AB||ED$
$\therefore \angle ABC=\angle EGH$ ...(i) (Corresponding angles)
$\because \angle EGH+\angle DEF={180}^0$ (Sum of co-interior angles)
$\Rightarrow \angle ABC+\angle DEF={180}^0$ [From (i)]
Hence proved.