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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.