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  • In Fig. 6.13, BA || ED and BC || EF. Show that ∠ ABC + ∠ DEF = 180°

In Fig. 6.13, BA \parallel ED  and BC \parallel EF. Show that \angle ABC +\angle DEF = 180^{\circ}

Answers (1)

Given
BA \parallel ED and BC \parallel EF
To show \angle ABC + \angle DEF = 180^{\circ}
Construction:
Extend EF to point P on AB

Proof: In the figure, BC \parallel EF, so BC \parallel PF
\because \angle EPB + \angle PBC = 180^{\circ} (Sum of co-interior angles is 180^{\circ}) …(i)
Now, AB \parallel ED PE is the transversal line.
\angle EPB = \angle DEF (Corresponding angle) …(ii)
From Equations (i) and (ii)
\angle DEF + \angle PBC = 180^{\circ}
\angle ABC + \angle DEF = 180^{\circ} (\because \angle PBC = \angle ABC)

Hence proved

Posted by

infoexpert26

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