# Prove that triangle APB is an equilateral triangle :

27.1k views
From a point P two tangents PA and PB are drawn to a circle with centre O. If OP is equal to the length of the diameter of the circle  then prove that triangle APB is an equilateral triangle.

by (9.9k points)
selected by

From the question we have: 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.