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Answer:
Proven that, OO¹ is the perpendicular bisector of common chord AB.
Step-by-step explanation:
Between the triangles ΔAOO¹ and ΔBOO¹:
OO¹ is a common side
OA = OB [radii of same circle]
O¹A = O¹B [radii of same circle]
∴ ΔAOO¹ ≅ ΔBOO¹
∴ ∠AON = ∠BON and ∠AO¹N = ∠BO¹N
Now, Between the triangles ΔAON and ΔBON:
ON is a common side
OA = OB [radii of same circle]
And angle between them are equal [shown earlier above]
∴ ΔAON ≅ ΔBON
∴ ∠ANO = ∠BNO
Since they are supplementary angles, and hence are right angles.
Also, AN = BN.
Similarly, between the triangles ΔAO¹N and ΔBO¹N:
O¹N is a common side
O¹A = O¹B [radii of same circle]
And angle between them are equal [shown earlier above]
∴ ΔAO¹N ≅ ΔBO¹N
∴ ∠ANO¹ = ∠BNO¹
Since they are supplementary angles, and hence are right angles.
∴ OO¹ is the perpendicular bisector of common chord AB.