Question

ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AD and BC produced meet at P, show that APB = 60°.

Answer 1

In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle

To prove – ∠APB = 60°

Construction – Join OC and OD

Proof – Since chord CD = CO = DO  ...[Radii of the circle]

∴ ΔDOC is an equilateral triangle

∴ ∠DOC = ∠ODC = ∠DCO = 60°

Let ∠A = x and ∠B = y

Since OA = OB = OC = OD    ...[Radii of the same circle]

∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC = y

∴ ∠AOD = 180° – 2x and ∠BOC = 180° – 2y

But AOB is a straight line

∴ ∠AOD + ∠BOC + ∠COD = 180°

`=>` 180° – 2x + 180° – 2y + 60° = 180°

`=>` 2x + 2y = 240°

`=>` x + y = 120°

But ∠A + ∠B + ∠P = 180°  ...[Angles of a triangle]

`=>` 120° + ∠P = 180°

`=>` ∠P = 180° – 120°

`=>` ∠P = 60°

Hence ∠APB = 60°