If two circles intersect in two points, prove that the line through the centres is the perpendicular bisector of the common chord.

Explanation

Let two circles O and O intersect at two points A and B so that AB is the common chord of two circles and OO' is the line segment joining the centres
Let `OO` intersect `AB` at` M`
Now Draw line segments `OA`,
`OB , O'A `and `O'B`
In `triangleOAO` and `OBO` , we have
`OA = OB` (radii of same circle)
`O'A = O'B` (radii of same circle)
`O'O = OO' (common side)`
`triangleOAO' ~=triangleOBO'`(SSS congruency)
`angleAOO' = angleBOO'`
`angleAOM = angleBOM ......(i)`
Now in `triangleAOM and triangleBOM `
we have
`OA = OB` (radii of same circle)`angleAOM = angleBOM` (from (i))
`OM = OM` (common side)
`triangleAOM ~=triangleBOM` (SAS congruncy)
`AM = BM `and` angleAMO = `angleBMO` But
`angleAMO + angleBMO = 180^0`
`2angleAMO = 180^0`
`angleAMO = 90^0`
Thus, `AM = BM` and `angleAMO = angleBMO = 90^0`
Hence `OO'` is the perpendicular bisector of `AB`.