Prove that cyclic parallelogram is always a rectangle.

 Let ABCD be a cyclic quadrilateral such that its diagonals AC and BD are the diameters of the circle through the vertices A, B, C, and D.

Since AC is a diameter and angle in a semi-circle is a right angle,

angle ADC = 90⁰ and angle ABC = 90 ⁰

Similarly, BD is a diameter.

Therefore, angle BCD = 90 ⁰ and angle BAD = 90 ⁰

Thus, ABCD is a rectangle

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Let ABCD be a cyclic parallelogram.

Angle A = Angle C (opposite angles of a parallelogram are equal)

Angle A + Angle C = 180 (opposite angles of a cyclic quadrilateral are supplementary)

i.e., 2 * Angle A = 180

Angle A = 180/2 = 90

Angle C = Angle A = 90

Similarly, we find that Angle B = Angle D = 90

Since all four angles are 90, ABCD is a rectangle

i.e a cyclic parallelogram is a rectangle.

Hope this is clear.