Given:ABCD is a trapezium where non-parallel sides AD and BC are equal.
Construction:DM and CN are perpendicular drawn on AB from D and C, respectively.
To prove:ABCD is cyclic trapezium.Proof:
In △DAM and △CBN,AD=BC       ...[Given]∠AMD=∠BNC      ...[Right angles]DM=CN     ...[Distance between the parallel lines]Therefore, △DAM≅△CBN  by RHS congruence condition.
Now, ∠A=∠B      ...[by CPCT]Also, ∠B+∠C=180°    ....[Sum of the co-interior angles]⇒∠A+∠C=180°.
Thus, ABCD is a cyclic quadrilateral as the sum of the pair of opposite angles is 180°.