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Question
In Fig. 5, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB.
Solution
Given: Two concentric circles C1 and C2 with centre O, and AB is the chord of C1 touching C2 at C.
To prove: AC = CB
Construction: Join OC.
Proof: AB is the chord of C1 touching C2 at C, then AB is the tangent to C2 at C with OC as its radius.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OC ⊥ AB
Considering, AB as the chord of the circle C1. So, OC ⊥ AB.
∴ OC is the bisector of the chord AB.
Hence, AC = CB (Perpendicular from the centre to the chord bisects the chord).
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