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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.

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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.

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