TP and TQ are tangents. TP=TQ. Angle PTQ=2angle OPQ.

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Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.To Prove: {tex}\\angle{/tex}PTQ = 2{tex}\\angle{/tex}OPQProof: Let {tex}\\angle{/tex}PTQ = {tex}\\theta{/tex}Since TP, TQ are tangents drawn from point T to the circle.TP = TQ{tex}\\therefore{/tex}\xa0TPQ is an isoscles triangle{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}TPQ = {tex}\\angle{/tex}TQP = {tex}\\frac12{/tex} (180o – {tex}\\theta{/tex}) = 90o – {tex}\\fracθ2{/tex}Since, TP is a tangent to the circle at point of contact P{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OPT = 90o{tex}\\therefore{/tex}\xa0{tex}\\angle{/tex}OPQ = {tex}\\angle{/tex}OPT – {tex}\\angle{/tex}TPQ = 90o – (90o -\xa0{tex}\\frac12{/tex}{tex}\\theta{/tex}) = {tex}\\fracθ2{/tex}= {tex}\\frac12{/tex}{tex}\\angle{/tex}PTQThus, {tex}\\angle{/tex}PTQ = 2{tex}\\angle{/tex}OPQ