In the given figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.

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Solution

Construction: Join OA, OC and OB

We know that the radius and tangent are perperpendular at their point of contact
∴ ∠OCA = ∠OCB = 90^∘
Now, In △OCA and △OCB
∠OCA = ∠OCB = 90^∘
OA = OB (Radii of the larger circle)
OC = OC (Common)
By RHS congruency
△OCA ≅ △OCB
∴ CA = CB

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