AB is a diameter of a circle with centre O. Chord CD is equal to radius OC. AC and BD produced intersect at P. Prove that : ∠APB = 60°.

From figure,

∠ADB = 90° [Angle subtended by diameter is a right angle.]

In △COD,

OC = OD [Radius of same circle]

∴ OC = OD = CD.

So, △COD is an equilateral triangle.

∴ ∠COD = 60°.

We know that,

Angle subtended by chord in the center is double the angle in the circumference.

⇒ ∠COD = 2∠CAD

⇒ ∠CAD = $\dfrac{1}{2}$∠COD

⇒ ∠CAD = $\dfrac{1}{2} \times 60°$ = 30°.

In △APD,

⇒ ∠PAD = ∠CAD = 30°.

⇒ ∠ADP + ∠ADB = 180° [Linear pair]

⇒ ∠ADP = 180° - ∠ADB = 180° - 90° = 90°.

By angle sum property.

⇒ ∠ADP + ∠APD + ∠PAD = 180°

⇒ 90° + ∠APD + 30° = 180°

⇒ ∠APD + 120° = 180°

⇒ ∠APD = 180° - 120°

⇒ ∠APD = 60°.

From figure,

⇒ ∠APB = ∠APD = 60°.

Hence, proved that ∠APB = 60°.