Given: Two circles with centres O and P intersecting at A and B.
Prove: OP is the perpendicular bisector of AB.
Construction: Join OA, OB, PA and PB. Let OP intersect AB at M.
Proof: In ∆OAP and ∆OBP,
OA = OB | Radii of a circle
PA = PB    | Radii of a circle
OP = OP    | Common
∴ ∆OAP ≅ ∆OBP    | SSS Rule
∴ ∠AOP = ∠BOP    | C.P.C.T.
∠AOM = ∠BOM    ...(1)
In ∆AOM and ∆BOM,
OA = OB    | Radii of a circle
∠AOM = ∠BOM    | From (1)
OM = OM    | Common
∴ ∆AOM = ∆BOM    | SAS Rule
∴ AM = BM    ...(2)
| C.P.C.T.
and    ∠AMO = ∠BMO    ...(3)
| C.P.C.T.
But ∠AMO + ∠BMO = 180°
| Linear Pair Axiom
∴ ∠AMO = ∠BMO = 90°    ...(4)
∴ OM, i.e., OP is the perpendicular bisector of AB.    | From (2) and (4)