Question

2 diameters of a circle intersect each other at 90o.Prove that the quadrilateral formed by joining their end points is a square.

Answer 1

Heya!!

Given : AB and CD are the diameter of the circle bisecting each other at O. ABCD is the quadrilateral formed by joining the end points of the diameter AB and CD.

To prove : ACBD is a square.

Proof :

AB and CD bisect each other at O.

∴ Quadrilateral ACBD is a parallelogram. (A quadrilateral is a parallelogram, if diagonals bisect each other)

⇒ AD = BC and AC = BD ...(1) (Opposite sides of parallelogram are equal)

In ΔBOD and ΔBOC,

OD = OC (Radius of the circle)

∠BOD = ∠BOC (Given)

OB = OB (Common)

∴ ΔBOD ΔBOC (SAS congruence axiom)

⇒ BD = BC (CPCT) ...(2)

Form (1) and (2), we get

AD = BC = BD = AC ...(3)

∠ACB =∠ ADB ...(4) (Opposite angles of parallelogram are equal)

ABCD is a cyclic,

∴ ∠ACB + ∠ADB = 180° (Sum of opposite angles of cyclic quadrilateral)

∴ 2∠ACB = 180° (Using (4))

⇒∠ACB = 90° ...(5)

From (3) and (5), we have

Parallelogram ACBD is a square.