Prove that the angle between 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.

This figure will surely help u to understand easily ......

** To prove==>** Angle BPA + Angle BOA = 180 ......

__Solution==>____In quadrilateral AOBP__

Angle PBO = 90^{0} [As angle between the radius and the tangent to the circle is 90^{0}]

Similarly angle PAO = 90.......

Angles (PAO + PBO + BOA + BPA) = 360^{0 ( sum of angles of a quadrilateral)}

-----> 90^{0} + 90^{0} + angle BOA + angle BPA = 360^{0} ^{(<PAO=90 and <PBO=90 proved)}

-----> Angle BOA + Angle BPA = 360^{0} - 180^{0}....

-----> ** Angle BOA + Angle BPA = 180 ^{0} **

**!!!!....Hence proved.....Hope this helps....!!!!**