Question

OA & OB a radii of this circle O & there are two tangents drawn from P namely PA & PB with OP equal to the diameter of circle. Prove ABP is an equilateral triangle

Answer 1

Solution :- ∠OAP = 90°(PA and PB are the tangents to the circle.)

In ΔOPA,
sin ∠OPA = OA/OP  =  r/2r   [OP is the diameter = 2*radius]
sin ∠OPA = 1 /2 =  sin  30 ⁰

∠OPA = 30°

Similarly,

∠OPB = 30°.

∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB    (tangents from an external point to the circle)

⇒∠PAB = ∠PBA ............(1)   (angles opp.to equal sides are equal)

∠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°

all angles are equal in an equilateral triangle.( 60 degrees)

ΔPAB is an equilateral triangle