# Two tangent TP and TQ are drawn to a circle with centre O from an external point T. Prove that ?PTQ=2?OPQ

$We\;know\;that\;length\;of\;tangents\;drawn\;from\;an\;external\;point\;to\; a\;circle\;are\;equal\\* \therefore TP=TQ..eq(1)\\* \therefore \angle TQP= \angle TPQ\;\;\;\;(angles\;of\;equal\;sides\;are\;equal) ...eq(2)\\* Now,\;PT\;is\;tangent\;and\;OP\;is\;radius.\\* \therefore OP \perp TP\;(Tangent\;at\;any\;point\;of\;circle\;is\;perpendicular\;to\;the\;radius\;through\;point\;of\;contact)\\* \angle OPT=90^{\circ}\\*or\; \angle OPQ + \angle TPQ=90^{\circ}\\*or,\; \angle TPQ=90^{\circ} - \angle OPQ ...eq(3)\\*In\;\triangle PTQ\\* \angle TPQ+ \angle PQT+ \angle QTP= 180^{\circ}\;\;\;\;(\therefore Sum\;of\;angles\;of\;triangle\;is\;180^{\circ})\\* or,\;90^{\circ}-\angle OPQ+\angle TPQ+\angle QTP=180^{\circ}\\* or,\;2(90^{\circ}-\angle OPQ)+\angle QTP=180^{\circ}\;\;\;\;\;(from\;(2)\;and\;(3))\\* or,\;180^{\circ}-2\angle OPQ+\angle PTQ=180^{\circ}\\* \therefore 2\angle OPQ=\angle PTQ\;\;\;\;Hence,\;proved$

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