Prove that a cyclic parallelogram is a rectangle.

Answer: Each angle of the parallelogram is 90°, so it is a rectangle (proved below).

[Figure: parallelogram ABCD with all four vertices on a circle.]

Given: \(\square ABCD\) is a parallelogram and it is cyclic. To prove: \(\square ABCD\) is a rectangle.

Proof: Opposite angles of a parallelogram are congruent: \(\angle A = \angle C\) ...... (I)

Opposite angles of a cyclic quadrilateral are supplementary: \(\angle A + \angle C = 180^\circ\) ...... (II)

From (I) and (II), \(2\angle A = 180^\circ\) \(\therefore \angle A = 90^\circ\).

A parallelogram with one right angle has all angles right angles, so \(\square ABCD\) is a rectangle. Hence proved.