In figure 3, two tangents RQ and RP are drawn from an external point R to the circle with center O. If ∠PRQ = 120°, then prove that OR = PR + RQ.

Given : Two tangents RQ and RP from an external point R to the circle with center O and

∠PRQ = 120°

To Prove : OR = PR + QR

Construction : Join OP and OQ

Proof :

PR = RQ [ Tangents drawn from an external point to a circle are equal] …[1]

Also OP ⏊PR and OQ ⏊RQ

[ as tangent drawn at a point on a circle is perpendicular to the radius through the point of contact]

∠OPR = 90°

∠OQR = 90°

In quadrilateral PORQ

∠OPR + ∠PRQ + ∠OQR + ∠QOP = 360°

90 + 120 + 90 + ∠QOP = 360

∠QOP = 60°

In △OPR and △ORQ

OP = OQ [radii of same circle]

PR = RQ [By 1]

OR = OR [common]

△OPR ≅ △ORQ [By SAS]

∠POR = ∠ROQ [By CPCT]

∠QOP = ∠POR + ∠ROQ

∠POR + ∠POR = 60°

∠POR = 30°

As △ROP is a right angled triangle

sin ∠POR = PR/OR

sin 30° = PR/OR

1/2 = PR/OR

→ OR = 2PR

= PR + PR = PR + QR [from 1]

Hence Proved.