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.

Let we have a circle with centre O and there is an external point P. From P draw two tangents, PA at point A and PB at point B, respectively. Now, join AO and BO in a way that subtends ∠AOB at the centre of the circle. The diagram is as follows:

It is clear from the figure that the line segments OA and PA are perpendicular.

Thus, ∠OAP = 90°

Similarly, the line segments OB ⊥ PB and Thus, ∠OBP = 90°

Now, in the quadrilateral OAPB, the sum of all interior angles will be 360°.

So, ∠APB + ∠OAP + ∠PBO + ∠BOA = 360° 

substituting the known values, we get,

∠APB + 90° + ∠BOA + 90° = 360°

∠APB + 180° + ∠BOA = 360°

Thus, ∠APB + ∠BOA = 180°

Hence proved.