Question

If non-parallel sides of a trapezium are

equal, prove that it is cyclic.​

Answer 1

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°.