Prove that the line of centers of two intersecting circles subtends equal Angles at the two points of intersection.

Given: Two intersecting circles with centres A and B. Their points of intersection are P and Q.
To Prove: ∠ APB = ∠ AQB.



 Proof: In Δ APB and Δ AQB, AP = AQ                       | Radii of a circle
BP = BQ                       | Radii of a circle AB = AB                       | Common ∴ Δ APB ≅ Δ AQB           |SSS Rule ∴ ∠ APB = ∠ AQB          | CPCT