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.



Let us consider a circle centred at point O.


Let P be an external point from which two tangents PA and PB are drawn to the circle which touches the 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 the circle.


It can be observed that


OA PA (radius of circle is always perpendicular to tangent)


Therefore, OAP = 90°


Similarly, OB PB


OBP = 90°


In quadrilateral OAPB,


Sum of all interior angles = 360°


OAP + APB+ PBO + BOA = 360°


90° + APB + 90° + BOA = 360°


APB + BOA = 180°


Hence, it can be observed 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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