Given : Two intersecting circles, in which OO′ is the line of centres and A and B are two points of intersection.
To prove : ∠OAO′ = ∠OBO′
Construction : Join AO, BO, AO′ and BO′.
Proof : In ∆AOO′ and ∆BOO′, we have
AO = BO [Radii of the same circle]
AO′ = BO′ [Radii of the same circle]
OO′ = OO′ [Common]
∴ ∆AOO′ ≅ ∆BOO′ [SSS axiom]
⇒ ∠OAO′ = ∠OBO′ [CPCT]
Hence, the line of centres of two intersecting circles subtends equal angles at the two points of intersection. Proved.
