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Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. |
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Answer» Required to prove -: ∠PTQ = 2∠OPQ Let ∠PTQ = θ Now by the theorem TP = TQ. So, TPQ is an isosceles triangle ∠TPQ = ∠TQP = \(\frac{1}{2}\) (180° - θ) = 90° - \(\frac{1}{2}\) θ ∠OPT = 90° ∠OPQ = ∠OPT - ∠TPQ = 90° - (90° - \(\frac{1}{2}\)θ) = \(\frac{1}{2}\)θ = \(\frac{1}{2}\) ∠PTQ ∠PTQ = 2∠OPQ |
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