10. 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 center.

First, draw a circle with centre O. Choose an external point P and draw two tangents PA and PB at point A and point B respectively. Now, join A and B to make AB in a way that it subtends ∠AOB at the center of the circle. The diagram is as follows: From the above diagram, it is seen that the line segments OA and PA are perpendicular. So, ∠OAP = 90° In a similar way, the line segments OB ⊥ PB and so, ∠OBP = 90° Now, in the quadrilateral OAPB, ∴∠APB+∠OAP +∠PBO +∠BOA = 360° (since the sum of all interior angles will be 360°) By putting the values we get, ∠APB + 180° + ∠BOA = 360° So, ∠APB + ∠BOA = 180° (Hence proved).