Let ABCD is a trapezium. Now take two points E and F on AB and draw the perpendicular from D and C on it.Now from ΔAED and ΔBFC,AD = BC  (Given)DE = CF  (Distance between parallel sides are same)∠AED = ∠BFC = 90from congruence criterian,ΔAED ≅ ΔBFCfrom CPCT,So ∠DAE = ∠CBF ..............1Again since AB || CD∠DAE + ∠ADC = 180  (since sum of adjecent interior angles is supplymentory)∠CBF + ∠ADC = 180  (from equation 1)Since sum of opposite angles is suplymentory in trapezium ABCD.Hense ABCD is a cyclic trapezium.

Proof:

To prove that a trapezium with non-parallel sides equal is cyclic, we need to show that its opposite angles are supplementary, i.e., their sum is 180 degrees.

Let's consider a trapezium ABCD, where AB is parallel to CD. Given that AD = BC, we need to prove that ∠A + ∠B = 180 degrees.

Proof:
1. Draw perpendiculars from A and B to CD, meeting CD at E and F, respectively.
2. Since AB is parallel to CD, the opposite angles ∠AED and ∠BFC are equal due to the alternate interior angles theorem.
3. Also, the opposite angles ∠A and ∠B are equal since the non-parallel sides of the trapezium are equal.
4. We can observe that ∠AED + ∠BFC = 180 degrees since they form a straight line.
5. Therefore, ∠A + ∠B = 180 degrees, as required.
6. Hence, the trapezium ABCD is cyclic.

Visualization:
To better understand the proof, let's visualize the trapezium and the steps involved:
- Draw a trapezium ABCD with AB parallel to CD.
- Label the points of intersection of perpendiculars from A and B to CD as E and F, respectively.
- Mark the equal sides AD and BC.
- Observe that ∠AED and ∠BFC are equal due to the alternate interior angles theorem.
- Notice that ∠A and ∠B are equal since the non-parallel sides of the trapezium are equal.
- Understand that the sum of ∠AED and ∠BFC is 180 degrees since they form a straight line.
- Conclude that ∠A + ∠B = 180 degrees, indicating that the trapezium is cyclic.

Conclusion:
By establishing that the opposite angles of a trapezium with equal non-parallel sides are supplementary, we have proved that it is a cyclic trapezium.