Question

Question 6
In the figure, BA || ED and BC || EF. Show that $\angle ABC+\angle DEF={180}^{\circ }$.

Answer 1

Given BA || ED and BC || EF.
To show $\angle ABC+\angle DEF={180}^{\circ }$,
Construction:
Draw a ray PE opposite to ray EF.

Proof:
In the figure, BC || EF.
$\therefore \angle EPB+\angle PBC={180}^{\circ }$ [sum of cointerior angles is ${180}^{\circ }$]…(i)
Now, AB || ED and PE is a transversal line.
$\therefore \angle EPB=\angle DEF$ [corresponding angles] …(ii)
From equations (i) and (ii), $\angle DEF+\angle PBC={180}^{\circ }$
$\Rightarrow \angle ABC+\angle DEF={180}^{\circ }[\because \angle ABC=\angle PBC]$