Let us consider a circle with center O and PA and PB are two tangents to the circle from an external point P

To Prove : Angle between two tangents is supplementary to the angle subtended by the line segments joining the points of contact at center, i.e. ∠APB + ∠AOB = 180°

Proof :

As AP and BP are tangents to given circle,

We have,

OA ⏊ AP and OB ⏊ BP

[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]

So, ∠OAP = ∠OBP = 90°

In quadrilateral AOBP, By angle sum property of quadrilateral, we have

∠OAP + ∠OBP + ∠AOB + ∠APB = 360°

90° + 90° + ∠AOB + ∠APB = 360°

∠AOB + ∠APB = 180°

Hence Proved