Mathematics
From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of the chord AB.
Circles
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Answer
The figure is shown below:
(i) In △AOP and △BOP,
OP = OP (Common sides)
OA = OB (Radius of the circle)
∠OAP = ∠OBP (Both are equal to 90° as tangents and radius on point of contact are perpendicular to each other.)
∴ △OAP ≅ △OBP (S.A.S. axiom of congruency)
(As Corresponding parts of congruent triangles are congruent)
∴ ∠AOP = ∠BOP and ∠APO = ∠BPO.
Hence, proved that ∠AOP = ∠BOP.
(ii) In △APM and △BPM,
PM = PM (Common side)
∠APM = ∠BPM (Proved above)
AP = BP (∵ tangents from an exterior point to a circle are equal in length)
∴ △APM ≅ △BPM (S.A.S. axiom of congruency)
(Congruent parts of congruent triangles are congruent.)
∴ AM = BM and ∠AMP = ∠BMP
But ∠AMP + ∠BMP = 180°
∴ ∠AMP = ∠BMP = 90°.
Hence, proved that OP is perpendicular bisector of AB at M.
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