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.

Let PA and PB be two tangents drawn from an external point P to a circle with centre O. We have to prove that- \( ∠ \ AOB \ + \ ∠ \ APB \ = \ 180° \)



In right \( ∆ \ OAP \) and \( OBP \), we have

\( PA \ = \ PB \)
[Tangents drawn from an external point are equal]

\( OA \ = \ OB \)
[Radii of the same circle]

\( OP \ = \ OP \) [Common]

\(\therefore ∆ \ OAP \ \cong \ ∆ \ OBP \)
[by SSS - criterion of congruence]

\(\Rightarrow ∠ \ OPA \ = \ ∠ \ OPB \)

and, \( ∠ \ AOP \ = \ ∠ \ BOP \ \)
\( \Rightarrow \ ∠ \ APB \ = \ 2 ∠ \ OPA \)

and, \( ∠ \ AOB \ = \ 2 ∠ \ AOP \)

But, \( ∠ \ AOP \ = \ 90° \ - \ ∠ \ OPA \)
[ \(\because ∆ \ OAP \) is right triangle]

\(\therefore 2 ∠ \ AOP \ = \ 180° \ - \ 2 ∠ \ OPA \)

\( \Rightarrow ∠ \ AOB \ = \ 180° \ - \ ∠ \ APB \)

\(\Rightarrow ∠ \ AOB \ + \ ∠ \ APB \ = \ 180° \)

Hence proved 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.