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

Consider,

Proof :

Join P and Q, PQ is the chord of the circle centered at O and it is also the chord of the circle centered at O’.
Let OO' intersect PQ at M.
In 
Δ OPO' and  Δ OQO', we have
OP = OQ (radii of same circle)
O'P = O'Q (radii of same circle)
OO' = OO' (common)
Δ OPO'  ≅  Δ  OQO'(SSS congruency)
∠ POO' = ∠  QOO'
∠ POM =  ∠ QOM -----(i)
Now, in  Δ POM and  Δ QOM we have
OP = OQ (radii of same circle)
∠ POM =  ∠ QOM
OM = OM (common)
△ POM  ≅ △ QOM (SAS congruency)
PM = QM and 
∠ PMO =  ∠ QMO
But
∠ PMO + ∠ QMO =  180 ∘  (linear pair)
2 ∠ PMO =  180 
∠ PMO =  90
Hence, OO' is perpendicular bisector.