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In the following figure, two tangents RQ and RP are drawn from an external point R to the circle with center O, If $\angle PRQ=120^{o}$, then prove that $OR=PR+RQ$.
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Given: Two tangents RQ and RP drawn from an external point R to the circle with center O, If $\angle PRQ = 120^{o}$.

To do: To prove that $OR = PR + RQ$.

Solution:

Let us join OR.

As known that the line joining the center and the external point is the angle bisector between the tangents.

Here given $\angle PRQ = 120^{o}$

$\angle PRO=\angle QRO=\frac{120^{o}}{2} =60^{o}$

Also we know that lengths of tangents from an external point are equal.

Thus, $PR=RQ$.

Let us join OP and OQ.

Since OP and OQ are the radii from the center O,

$OP\perp PR$ and $OQ\perp\ RQ$.

Thus, $\vartriangle OPR$ and $\vartriangle OQR$ are right angled congruent triangles.

$\therefore \ \angle POR=90^{o} -\angle PRO=90^{o}-60^{o}=30^{o}$

Similarly, $\angle QOR=90^{o}-60^{o}=30^{o}$

$sin\left( \angle POR\right) =sin30^{o} =\frac{1}{2} =\frac{PR}{OR}$

$\Rightarrow \frac{1}{2} =\frac{PR}{OR}$

$\Rightarrow OR=2PR$

$\Rightarrow OR=PR+QR$

Hence proved that $OR=PR+QR$
 

Updated on: 10-Oct-2022

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