#### 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°

#### Solution

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

∠OCB = ∠OBC = y and

∴ ∠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°