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.

Question

Accepted Answer

Step 1: Let PA and PB be tangents drawn from external point P to the circle with centre O, touching at A and B respectively.
Step 2: In quadrilateral OAPB: $\angle OAP + \angle APB + \angle OBP + \angle AOB = 360°$
Step 3: $90° + \angle APB + 90° + \angle AOB = 360°$ (Tangent $\perp$ radius)
Step 4: $\angle APB + \angle AOB = 360° - 180° = 180°$. Hence proved.

Question

Solution