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If two circles intersect at two points,...

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

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Given : Two circles `C (O,r)` intersecting at points `A` and `B`.
To prove : `OO'` is the perpendicular bisector of `AB`.
Construction : Draw line segment `OA, OB ,O'A` and `O'B` Let `OO'` and `AB` intersect at `M`.

Proof : In `Delta OAO'` and `Delta OBO'` , we have
` OA` `=` `OB` [each equal to `r`]
`O'A` `=` `O'B` [each equal to `s` ]
`OO' `=` OO'` [common]
`:. Delta OAO' ~= Delta OBA' `[by SSS-congurence]
`implies /_AOO' = /_ BOO'`
`implies /_AOM =/_BOM` .....(i)
`[ :' /_ AMO +/_ BMO =180^(@)` and `/_ BOO' =/_BOM ]`.
In `Delta AOM` and `Delta BOM` , we have
`OA =AB ` [each equal to r ]
`/_AOM =/_BOM `[ from (i) ]
`OM =OM` [ common]
`implies Delta AOM ~= DeltaBOM ` [ by SAS - congurence ]
`implies AM =BM` and `/_ AMO = /_ BMO`
`implies AM =BM ` and `/_ AMO = /_ BMO =90^(@)`
`[ :' /_ AMO +/_ BMO =180^(@)` and `/_ AMO =/_ BMO ]`
`implies OO'` is the perpendicualr bisector of `AB`.
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