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

From figure,

CO = DO = CD [Radius of circle]

∴ COD is an equilateral triangle.

∴ ∠DOC = ∠ODC = ∠DCO = 60°.

Let ∠A = x and ∠B = y

OA = OB = OC = OD [Radius of circle]

As angles opposite to equal sides are equal.

∴ ∠ODA = ∠OAD = x

and

∠OCB = ∠OBC = y.

In △OAD,

⇒ ∠ODA + ∠OAD + ∠AOD = 180°

⇒ x + x + ∠AOD = 180°

⇒ ∠AOD = 180° - 2x.

In △BOC,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + y + ∠BOC = 180°

⇒ ∠BOC = 180° - 2y.

Since, AOB is a straight line

⇒ ∠AOD + ∠BOC + ∠DOC = 180°

⇒ 180° - 2x + 180° - 2y + 60° = 180°

⇒ 2x + 2y = 420° - 180°

⇒ 2x + 2y = 240°

⇒ x + y = 120°.

In △ABP,

⇒ ∠A + ∠B + ∠P = 180°

⇒ x + y + ∠P = 180°

⇒ 120° + ∠P = 180°

⇒ ∠P = 180° - 120° = 60°.

From figure,

∠APB = ∠P = 60°.

Hence, ∠APB = 60°.