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

(N/A) We have a trapezium $ABCD$ such that $AB || CD$ and $AD = BC$.
Let us draw $BE || AD$ such that $ABED$ is a parallelogram.
Since the opposite angles of a parallelogram are equal,
$\angle BAD = \angle BED$ ..... $(1)$
and $AD = BE$ [Opposite sides of a parallelogram] ...... $(2)$
But $AD = BC$ [Given] ......... $(3)$
From $(2)$ and $(3)$,we have
$BE = BC$
$\Rightarrow \angle BEC = \angle BCE$ [Angles opposite to equal sides are equal]
Now,$\angle BED + \angle BEC = 180^{\circ}$ [Linear pair]
Since $\angle BED = \angle BAD$ and $\angle BEC = \angle BCE$,we have
$\angle BAD + \angle BCE = 180^{\circ}$
In quadrilateral $ABCD$,$\angle BAD + \angle BCD = \angle BAD + (\angle BCE + \angle ECD)$. Since $ABED$ is a parallelogram,$\angle ECD = \angle BAD$. Thus,the sum of opposite angles is $180^{\circ}$.
Therefore,$ABCD$ is a cyclic quadrilateral.