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Answer :
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.