If the non-parallel sides of a trapezium are equal,prove that it is a cyclic quadrilateral.

(N/A) Let $ABCD$ be a trapezium such that $AB \parallel CD$ and the non-parallel sides $AD = BC$.
To prove that $ABCD$ is a cyclic quadrilateral,we need to show that the sum of opposite angles is $180^{\circ}$,i.e.,$\angle A + \angle C = 180^{\circ}$ or $\angle B + \angle D = 180^{\circ}$.
Draw $DE \perp AB$ and $CF \perp AB$.
In $\triangle ADE$ and $\triangle BCF$:
$AD = BC$ (Given)
$\angle AED = \angle BFC = 90^{\circ}$ (By construction)
$DE = CF$ (Distance between parallel lines is constant)
Therefore,$\triangle ADE \cong \triangle BCF$ by $RHS$ congruence rule.
This implies $\angle A = \angle B$ (by $CPCT$).
Since $AB \parallel CD$,the sum of consecutive interior angles is $180^{\circ}$,so $\angle A + \angle D = 180^{\circ}$.
Substituting $\angle B$ for $\angle A$,we get $\angle B + \angle D = 180^{\circ}$.
Since the sum of opposite angles is $180^{\circ}$,$ABCD$ is a cyclic quadrilateral.