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Question
From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
∠AOP = ∠BOP
Solution
In ΔAOP and ΔBOP
AP = BP ...(Tangents from P to the circle)
OP = OP ...(Common)
OA = OB ...(Radii of the same circle)
∴ By Side – Side – Side criterion of congruence,
ΔAOP ≅ ΔBOP
The corresponding parts of the congruent triangle are congruent
`=>` ∠AOP = ∠BOP ...[By c.p.c.t]
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