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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
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