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If a trapezium is cyclic then
Let $$ABCD$$ be the trapezium with $$AB \parallel CD$$. Then $$\angle A + \angle D = 180^\circ$$, and since it is cyclic, $$\angle A + \angle C = 180^\circ$$, so $$\angle C = \angle D$$. Hence the trapezium is isosceles, so its non parallel sides are equal (its diagonals are also equal, and it need not be a rectangle).
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