Prove that:

(i) the parallelogram, inscribed in a circle, is a rectangle.

(ii) the rhombus, inscribed in a circle, is a square.

(i) Let's assume that ABCD is a parallelogram which is inscribed in a circle.

⇒ ∠A = ∠C and ∠B = ∠D [Opposite angles of a parallelogram are equal]

⇒ ∠A + ∠C = 180° and ∠B + ∠D = 180° [Pair of opposite angles in a cyclic quadrilateral are supplementary]

⇒ 2∠A = 180° and 2∠B = 180°

⇒ ∠A = 90° and ∠B = 90°

⇒ ∠A = ∠C = 90° and ∠B = ∠D = 90°.

Also,

AB = CD and AD = BC [Opposite sides of parallelogram are equal]

Hence, proved that ABCD is a rectangle.

(ii) Let's assume that ABCD is a rhombus which is inscribed in a circle.

So, we have

∠BAD = ∠BCD [Opposite angles of a rhombus are equal]

From figure,

∠BAD + ∠BCD = 180° [Pair of opposite angles in a cyclic quadrilateral are supplementary]

⇒ 2∠BAD = 180°

Thus, ∠BAD = ∠BCD = 90°.

Also,

∠ABC = ∠ADC [Opposite angles of a rhombus are equal]

From figure,

∠ABC + ∠ADC = 180° [Pair of opposite angles in a cyclic quadrilateral are supplementary]

⇒ 2∠ABC = 180°

Thus, ∠ABC = ∠ADC = 90°.

Since, each side of rhombus are equal.

Thus,

Each angle of quadrilateral = 90° and all sides are equal.

Hence, proved that ABCD is a square.