If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. -Maths 9th

If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. -Maths 9th
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Given: Two circles, with centres O and O' intersect at two points A and B. A AB is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect AB at P. To prove: OO' is the perpendicular bisector of AB. Construction: Join OA, OB, O'A and O'B  Proof: In triangles OAO' and OBO', we have OO' = OO' (Common)  OA = OB (Radii of the same circle)  O'A = O'B (Radii of the same circle) ⇒  △ OAO' ≅  △ OBO'  (SSS congruence criterion) ⇒  ​∠AOO' = ∠BOO'  (CPCT) i.e.,     ∠AOP = ∠BOP In triangles AOP and  BOP, we have OP = OP  (Common) ∠AOP = ∠BOP   (Proved above) OA = OB  (Radio of the same circle) ∴  △AOR ≅ △BOP  (By SAS congruence criterion) ⇒ AP = CP   (CPCT) and  ∠APO = ∠BPO  (CPCT) But   ∠APO + ∠BPO = 180°  (Linear pair) ∴   ∠APO + ∠APO = 18 0°  ⇒  2∠APO =  180° ⇒  ∠APO = 9 0° Thus, AP = BP and ∠APO =  ∠BPO = 9 0° Hence, OO' is the perpendicular bisector of AB.
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