If $ABCD$ is a cyclic quadrilateral in which $AD \| BC$. Prove that $\angle B = \angle C$.
"
Given:
$ABCD$ is a cyclic quadrilateral in which $AD \| BC$.
To do:
We have to prove that $\angle B = \angle C$.
Solution:
$AD \| BC$
This implies,
$\angle A + \angle B = 180^o$ (Sum of cointerior angles)
$\angle A + \angle C = 180^o$ (Opposite angles of the cyclic quadrilateral)
This implies,
$\angle A + \angle B = \angle A + \angle C$
$\Rightarrow \angle B = \angle C$
Hence proved.
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