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.

Draw a circle with center O and take an external point P. PA and PB are the tangents. As the radius of the circle is perpendicular to the tangent, OA⊥PA. Similarly, OB⊥PB. ∠OBP = 90° and ∠OAP = 90°. In quadrilateral OAPB, the sum of all interior angles = 360°. →∠OAP + ∠OBP + ∠BOA + ∠APB = 360° →90° + 90° + ∠BOA + ∠APB = 360° →∠BOA + ∠APB = 180°. It proves 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 center.