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

(N/A) Given: $ABCD$ is a trapezium in which $AD \parallel BC$ and its non-parallel sides $AB$ and $DC$ are equal,i.e.,$AB = DC$.
To prove: Trapezium $ABCD$ is cyclic.
Construction: Draw $AM \perp BC$ and $DN \perp BC$.
Proof: In right-angled triangles $\Delta AMB$ and $\Delta DNC$:
$\angle AMB = \angle DNC = 90^{\circ}$ (By construction)
$AB = DC$ (Given)
$AM = DN$ (Distance between two parallel lines is constant)
Therefore,$\Delta AMB \cong \Delta DNC$ (By $RHS$ congruence rule).
Thus,$\angle B = \angle C$ $(CPCT)$.
Also,$\angle BAM = \angle CDN$ $(CPCT)$.
Since $\angle AMB = 90^{\circ}$ and $\angle DNC = 90^{\circ}$,we have $\angle MAB = 90^{\circ} - \angle B$ and $\angle NDC = 90^{\circ} - \angle C$. Since $\angle B = \angle C$,then $\angle MAB = \angle NDC$.
Now,$\angle BAD = \angle BAM + \angle MAD = \angle BAM + 90^{\circ}$ and $\angle CDA = \angle CDN + \angle NDA = \angle CDN + 90^{\circ}$.
Since $\angle BAM = \angle CDN$,it follows that $\angle BAD = \angle CDA$.
In quadrilateral $ABCD$,$\angle B + \angle C + \angle CDA + \angle BAD = 360^{\circ}$.
Substituting $\angle C = \angle B$ and $\angle CDA = \angle BAD$,we get $2\angle B + 2\angle BAD = 360^{\circ}$,which implies $\angle B + \angle BAD = 180^{\circ}$.
Since the sum of a pair of opposite angles is $180^{\circ}$,the trapezium $ABCD$ is cyclic.