Prove cyclic trapezium is isosceles and its diagonals are equal to each other.

In a cyclic trapezium,

∠BAD + ∠BCD = 180°

∠BAD + ∠ABC = 180°

∴ ∠ABC = ∠BCD

In ∆ABC and ∆BDC,

∠ABC = ∠BCD

∠BAC = ∠CDB (angles in the same segment)

BC = BC (common side)

∴ ∆ABC ≅ ∆BDC

∴ AB = CD and AC = BD

Hence, cyclic trapezium ABCD is isosceles and diagonals are equal to each other.