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Tangents AP and AQ are drawn to a circle, with center O, from an exterior point A. Prove that :

∠PAQ = 2∠OPQ

Circles

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Answer

We know that,

The tangent at any point of a circle and the radius through this point are perpendicular to each other.

Tangents AP and AQ are drawn to a circle, with center O, from an exterior point A. Prove that ∠PAQ = 2∠OPQ. Tangents and Intersecting Chords, Concise Mathematics Solutions ICSE Class 10.

In quadrilateral OPAQ,

∠OPA = ∠OQA = 90°.

∠OPA + ∠OQA + ∠POQ + ∠PAQ = 360° [∵ Sum of angles in quadrilateral = 360°]

90° + 90° + ∠POQ + ∠PAQ = 360°

∠POQ + ∠PAQ = 360° - 180°

∠POQ + ∠PAQ = 180° ……….(1)

In △OPQ,

OP = OQ [Radius of same circle]

∴ ∠OPQ = ∠OQP [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OPQ + ∠OQP + ∠POQ = 180°

⇒ ∠OPQ + ∠OPQ + ∠POQ = 180°

⇒ 2∠OPQ + ∠POQ = 180° ………(2)

From (1) and (2) we get,

∠POQ + ∠PAQ = 2∠OPQ + ∠POQ

⇒ ∠PAQ = 2∠OPQ.

Hence, proved that ∠PAQ = 2∠OPQ.

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