A trapezium will be cyclic if its non-parallel sides are equal.
 

Given: ABCD is a trapezium in which AB || CD and AD = BC
To Prove: ABCD is cyclic
i.e. ∠B + ∠D = ∠A + ∠C = 180° 
Proof: In right-triangles ΔADE and ΔBCF, AD = BC and AE = BF.......(Given)
.^.. ΔADE ≅ ΔBCF
.^.. ∠1 = ∠2 and ∠D = ∠C  ...... (C.P.C.T.)
⇒ ∠1 + 90° = ∠2 + 90° 
⇒ ∠DAB = ∠ABC     ................. (i)
Now ∠BAD + ∠ADC = 180°  ........ (ii)   [Interior angles of the parellel sides] 
∠ABC + ∠ADC = 180°  [From Equations (i) and (ii)] 
Hence ABCD is cyclic.