Question

In figure OP is equal to diameter of the circle. AP and PB are two tangents to the circle. AB is a secant to the circle. Prove that triangle ABP 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         [lengths 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.