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prove that the angle between the two tangents draw from an external point to a circle is supplementary to the angle substended by the line segment joining the point of contact at the centre
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To prove: BPA+BOA=180

PROOF -: quadrilateral AOBP

PBO=90∘ (as the angle between the radius and the tangent to the circle is 90∘).

Similarly, PAO=90∘

PAO+PBO+BOA+BPA=360∘ (the sum of the angles of a quadrilateral)


Therefore 90∘+90∘+BOA+BPA=360

BOA+BPA=360∘−180∘=180

HENCE PROVED....
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