In ΔAED and ΔBFC,
AD = BC (Given)
DE = CF (Distance between parallel sides is same)
∠AED = ∠BFC = 90°
ΔAED ≅ ΔBFC (RHS Congruence criterion)
Hence ∠DAE = ∠CBF (CPCT) … (1)
Since AB||CD, AD is transversal
∠DAE + ∠ADC = 180° (Sum of adjacent interior angles is supplementary)
⇒ ∠CBF + ∠ADC = 180° [from (1)]
Since sum of opposite angles is supplementary in trapezium ABCD.
Thus ,ABCD is a cyclic trapezium .