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Question
From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
OP is the ⊥ bisector of chord AB.
Solution
In ΔOAM and ΔOBM
OA = OB ...(Radii of the same circle)
∠AOM = ∠BOM ...(Proved ∠AOP = ∠BOP)
OM = OM ...(Common)
∴ By Side – Angle – Side criterion of congruence,
ΔOAM ≅ ΔOBM
The corresponding parts of the congruent triangles are congruent.
`=>` AM = MB
And ∠OMA = ∠OMB
But, ∠OMA + ∠OMB = 180°
∴ ∠OMA = ∠OMB = 90°
Hence, OM or OP is the perpendicular bisector of chord AB.
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