Question

OP is the diameter of circle with centre O. PA and PB are tangents. Prove that triangle APB is an equilateral triangle.

Answer 1

PA and PB are the tangents to the circle. 

∴ OA ⊥ PA 

⇒ ∠OAP = 90° 

In ΔOPA, 

 sin ∠OPA = OA OP  =  r 2r   (Given OP is the diameter of the circle)

⇒ sin ∠OPA = 1 2 =  sin  30 ⁰

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB         (length of tangents drawn from an external point to a circle are equal)

⇒∠PAB = ∠PBA ............(1)   (Equal sides have equal angles opposite to them)

∠PAB + ∠PBA + ∠APB = 180°    (Angle sum property)

⇒∠PAB + ∠PAB = 180° – 60° = 120°  (Using (1)

⇒2∠PAB = 120°

⇒∠PAB = 60°    .............(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ΔPAB is an equilateral triangle.