$ABCD$ is a cyclic quadrilateral. If $AB \parallel CD$,then prove that $BC = AD$.

(N/A) Given: $ABCD$ is a cyclic quadrilateral with $AB \parallel CD$.
To prove: $BC = AD$.
Proof:
$1$. Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 180^{\circ}$.
$2$. Since $AB \parallel CD$,the sum of consecutive interior angles is $180^{\circ}$. Thus,$\angle A + \angle D = 180^{\circ}$.
$3$. From $\angle A + \angle C = 180^{\circ}$ and $\angle A + \angle D = 180^{\circ}$,we get $\angle C = \angle D$.
$4$. Similarly,since $AB \parallel CD$,$\angle B + \angle C = 180^{\circ}$. Given $\angle B + \angle D = 180^{\circ}$,we get $\angle C = \angle D$.
$5$. In a cyclic quadrilateral,if one pair of opposite sides is parallel,it must be an isosceles trapezium. In an isosceles trapezium,the non-parallel sides are equal.
$6$. Therefore,$BC = AD$.