#### Question

In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal to radius OA.

Is AC produced and BD produced meet at point P; show that ∠APB = 60°

#### Solution

Given – In the figure, AB is the diameter of a circle with centre O. CD is the chord with length equal radius OA. AC and BD produced meet at point P To Prove : ∠APB = 60°

Construction – join OC and OD

Proof – We have CD = OC = OD [given] Therefore, ∆OCD is an equilateral triangle

∴ ∠OCD = ∠ODC = ∠COD = 60°

In ∆AOC, OA = OC [ radii of the same circle]

∴ ∠A = ∠1

Similarly, in ∆BOD, OB = OD [ radii of the same circle]

∴ ∠B = ∠2

Now, in cyclic quad ACBD Since,

∠ACD + ∠B = 180°

[Since sum of opposite angles of a cyclic quadrilateral are supplementary]

⇒ 60° + ∠1+ ∠B = 180°

⇒ ∠1 + ∠B = 180° - 60°

⇒ ∠1 + ∠B = 120°

But, ∠1 = ∠A

∴ ∠A + ∠B = 120° …..(1)

Now, in ∆APB,

∠P + ∠A + ∠B = 180° [sum of angles of a triangles]

⇒ ∠P +120° = 180°

⇒ ∠P = 180° -120° [from (1)]

⇒ ∠P = 60° or ∠APB = 60°