Mathematics
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP
(ii) OP is the ⊥ bisector of chord AB.
Circles
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Answer
The circle with centre O and tangents PA and PB drawn from point P outside the circle is shown in the below figure:
(i) In ∆AOP and ∆BOP, we have
⇒ AP = BP [Tangents from an exterior point P, are equal in length.]
⇒ OA = OB [Radius of circle]
⇒ OP = OP [Common]
∴ ΔAOP ≅ ΔBOP by SSS axiom.
∴ ∠AOP = ∠BOP [By C.P.C.T.]
Hence, proved that ∠AOP = ∠BOP.
(ii) In ∆OAM and ∆OBM, we have
OA = OB [Radii of the same circle]
∠AOM = ∠BOM [Proved ∠AOP = ∠BOP]
OM = OM [Common]
∴ ΔOAM ≅ ΔOBM [By SAS axiom]
By C.P.C.T.
⇒ AM = MB and ∠OMA = ∠OMB = x
From figure,
⇒ ∠OMA + ∠OMB = 180° [As, AB is a straight line]
⇒ x + x = 180°
⇒ 2x = 180°
⇒ x =
⇒ x = 90°.
Thus, ∠OMA = ∠OMB = 90°.
Hence proved, that OP is the ⊥ bisector of chord AB.
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