Let two circles with centre P and Q intersect at two distinct points say A and B.
Join A, B to form the common chord \(\overline {AB}\).
Let ‘O’ be the midpoint of AB.
Join ‘O’ with P and Q.
Now in ΔAPO and ΔBPO
AP = BP (radii)
PO = PO (common)
AO = BO (∵ O is the midpoint)
∴ ΔAPO ≅ ΔBPO (S.S.S. congruence)
Also ∠AOP = ∠BOP (CPCT)
But these are linear pair of angles.
∴ ∠AOP = ∠BOP = 90°
Similarly in ΔAOQ and ΔBOQ
AQ = BQ (radii)
AO = BO (∵ O is the midpoint of AB)
OQ = OQ (common)
∴ AAOQ ≅ ABOQ
Also ∠AOQ = ∠BOQ (CPCT)
Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)
∴ ∠AOQ = ∠BOQ = 180°/2 = 90°
Now ∠AOP + ∠AOQ = 180°
∴ PQ is a line.
Hence the proof.