In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with center 0. If ∠PRQ = 120°, then prove that OR = PR + RQ.

Given : In the figure ,

Two tangents RQ and RP are drawn from an external point R to the circle with center O and ∠PRQ = 120°

To Prove: OR = PR + RQ

Construction: Join OP and OQ

Proof :

In △△OPR and △OQR

OP = OQ [radii of same circle]

OR = OR [Common]

PR = PQ …[1]

[Tangents drawn from an external point are equal]

△OPR ≅ △OQR

[By Side - Side - Side Criterion]

∠ORP = ∠ORQ

[Corresponding parts of congruent triangles are congruent]

Also,

∠PRQ = 120°

∠ORP + ∠ORQ = 120°

∠ORP + ∠ORP = 120°

2∠ORP = 120°

∠ORP = 60°

Also, OP ⏊ PR

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

So, In right angled triangle OPR,

∴ OR = 2PR

OR = PR + PR

OR = PR + RQ [From 1]

Hence Proved.

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