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# 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.

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.