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Two circles with centres O and P intersect each other at A
and B. Prove that the line OP bisects the common chord
AB at right angles.
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In triangle AOP and BOP
\_AOP=\_BOP
(If 2 circles intersect at 2 points, their centres lie on the perpendicular bisector of the common chord)
OP=OP (common)
AP=BP (radii of circle)
Therefore, ∆AOP ≈ ∆BOP (by RHS congruency)
AO=BO (by CPCT)
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