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Consider trapezium ABCD such that AB is parallel to CD and $AD = BC$.

Consider two points M and N that can be considered as the foot of perpendicular drawn on AB from vertices D and C respectively.

Now, if we compare $\Delta DAM$and $\Delta CBN$, we have:

$AD = BC{\text{ }}...{\text{(Given)}}$

$\angle AMD = \angle BNC{\text{ }}...{\text{(Right angles i}}{\text{.e}}{\text{. 9}}{{\text{0}}^ \circ }{\text{)}}$

And $DM = CN{\text{ }}....{\text{(Distance between two parallel lines)}}$

From this we can say that both the triangles are congruent.

$\Delta DAM \cong \Delta CBN$

We know that corresponding parts of congruent triangles are equal. So we have:

$ \Rightarrow \angle A = \angle B .....(i)$

Also $\angle B + \angle C = {180^ \circ }{\text{ }}...{\text{(sum of co - interior angles)}}$

Substituting the value of $\angle B$ from equation first, we have:

$ \Rightarrow \angle A + \angle C = {180^ \circ } ....(ii)$

Equation $(ii)$ shows that the sum of the pair of opposite angles of trapezium ABCD is ${180^ \circ }$.

Therefore the trapezium is a cyclic quadrilateral.

Note: A quadrilateral is said to be cyclic quadrilateral if all of its 4 vertices lie on the same circle.

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