From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to the diameter of the circle, show that triangle APB is equilateral.

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The given information can be represented using a figure as
 
⇒ ∠OPA = 30°
Similarly, it can be proved that ∠OPB = 30°.
Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°
In ∆PAB,
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°
∴ ∆PAB is an equilateral triangle.
 
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