Two tangents PA and PB are drawn from an external point P to a circle with the centre O. Prove that OAPB is a cyclic quadrilateral.
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Answered by
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by theorem
angle PAO = angle PBO = 90°
therefore
angle PAO+angle PBO =90°+90°=180°
since angle between the tangent and ange between the radius are supplementary
therefore
angleAOB+angleAPB=180°
SINCE OPPOSITE ANGLE ARE EQUAL TO 180° HENCE OAPB IS A CYCLIC QUADRILATERAL
angle PAO = angle PBO = 90°
therefore
angle PAO+angle PBO =90°+90°=180°
since angle between the tangent and ange between the radius are supplementary
therefore
angleAOB+angleAPB=180°
SINCE OPPOSITE ANGLE ARE EQUAL TO 180° HENCE OAPB IS A CYCLIC QUADRILATERAL
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I can do this one too but it is a HOTS type question and you cannot understand easily that's why I am saying no to this question
sorry not this one the last one in which I said I can't do
and this one I cannot find
and also ABCD is a quadrilateral. A circle centred at O is inscribed in the quadrilateral. ............ And also point P, two tangents PA and PB are drawn to a circle C ( 0, r ). .....
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