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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.

Asked by Topperlearning User 4th June 2014, 1:23 PM
Answered by Expert
Answer:

Let us consider a circle centred at point O. Let P be a external point from which two tangents PA and PB are drawn to circle which are touching circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends AOB at centre O of circle.

Now we may observe that:

OA (radius) PA (tangent)

So, OAP = 90o

Similarly, OB (radius) PB (tangent)

OBP = 90o

Now in quadrilateral OAPB, sum of all interior angles = 360o

OAP + APB + PBO +BOA = 360o

90o + APB + 90o + BOA = 360o

APB +BOA = 180o

Hence, 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.

Answered by Expert 4th June 2014, 3:23 PM
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