Question

Prove that :

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

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

Answer 1

Here is the proof to the mentioned question.
(1)
Let ABCD be a parallelogram inscribed in a circle.

In parallelogram, opposite angles are equal
∴ ∠A = ∠C
It is known that, in a cyclic quadrilateral, opposite angles are supplementary.
∴ ∠A + ∠C = 180°
⇒2 ∠A = 180° [∠A = ∠C]
⇒∠A = 180° ¸ 2 = 90°
∴ ∠C = ∠A = 90°
Similarly, it can be shown that ∠B = ∠D = 90°.

Thus, ABCD be the parallelogram where ∠A = ∠B = ∠C = ∠D = 90°
Hence, ABCD be a rectangle.
(2)
Let PQRS be a rhombus inscribed in a circle.

In rhombus, opposite angles are equal.
∴ ∠P = ∠R
It is known that, in a cyclic quadrilateral, opposite angles are supplementary.
∴ ∠P + ∠R = 180°
⇒2 ∠P = 180° [∠A = ∠C]
⇒ ∠P = 180° ¸ 2 = 90°
∴ ∠R = ∠P = 90°
Similarly, it can be shown that ∠Q = ∠S = 90°.
In rhombus, all the sides are equal.

Thus, PQRS is a quadrilateral where ∠P = ∠Q = ∠R = ∠ S = 90° and PQ = QR = RS = SP
Hence, PQRS is a square.

C