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प्रश्न
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that : ∠PAQ = 2∠OPQ
उत्तर
In quadrilateral OPAQ,
∠OPA = ∠OQA = 90°
(∵ OP ⊥ PA and OQ ⊥ QA)
∴ `∠`POQ + `∠`PAQ + 90° + 90° = 360°
`=>` ∠POQ + ∠PAQ = 360° – 180° = 180° ...(i)
In triangle OPQ,
OP = OQ ...(Radii of the same circle)
∴ OPQ = ∠OQP
But ∠POQ + ∠OPQ + ∠OQP = 180°
`=>` ∠POQ + ∠OPQ + ∠OPQ = 180°
`=>` ∠POQ + 2∠OPQ = 180° ...(ii)
From (i) and (ii)
∠POQ + ∠PAQ = ∠POQ + 2∠OPQ
`=>` ∠PAQ = 2∠OPQ
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