Two tangent segments $ P A $ and $ P B $ are drawn to a circle with centre $ O $ such that $ \angle A P B=120^{\circ} . $ Prove that $ O P=2 A P $.

Given:

Two tangent segments \( P A \) and \( P B \) are drawn to a circle with centre \( O \) such that \( \angle A P B=120^{\circ} . \)

To do:

We have to prove that \( O P=2 A P \).

Solution:

Join $OP$.

Take mid point of $OP$ as $M$ and join $AM$. Join $OA$ and $OB$.

In right angled triangle $OAP$,

$\angle OPA = \frac{1}{2} \angle APB = \frac{1}{2}(120^o) = 60^o$

$\angle AOP = 90^o - 60^o = 30^o$

$M$ is the mid point of hypotenuse $OP$ of $\triangle OAP$

This implies,

$MO = MA = MP$

$\angle OAM = \angle AOM = 30^o$

$\angle PAM = 90^o – 30^o = 60^o$

$\triangle AMP$ is an equilateral triangle.

$MA = MP = AP$

Also, $M$ is the mid point of $OP$.

$OP = 2 MP = 2 AP$

Hence proved.

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Updated on: 10-Oct-2022

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