**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.**