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Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. |
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Answer» let the centre of the circle be O and the two tangents forming an angle at point P from points A & B respectively now, `/_ APB is supplementary to /_AOB ` we have to prove that, `/_APB+ /_AOB = 180^@` now draw 2 `_|_` OA & OB TO AP & BP respectively now, `/_OAP = /_OBP= 90^@` now, in quadilateral AOBP, sum of 4 angles = `360^@` `/_AOB + /_OBP + /_APB+ /_OAP = 360^@` as we know that, `/_ OBP = 90^@ & /_OAP=90^@` so, `/_AOB + /_APB = 180^@` |
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