Question

In the adjoining figure, ABCD is a quadrilateral in which AB || DC and AD || BC. Prove that ∠ADC = ∠ABC.

Answer 1

Let $AD\parallel BC$ and CD is the transversal. Then,
$\angle ADC+\angle DCB=180°...\left(i\right)\left[\text{Consecutive Interior Angles}\right]$
Also,
$AB\parallel CD$ and BC is the transversal. Then,
$\angle DCB+\angle ABC=180°...\left(ii\right)\left[\text{Consecutive Interior Angles}\right]$
From (i) and (ii), we get:
$\angle ADC+\angle DCB=\angle DCB+\angle ABC$
$\Rightarrow \angle ADC=\angle ABC$