In figure, the chord $A B$ of the larger of the two concentric circles, with centre $O,$ touches the smaller circle at $C .$ Prove that $A C = C B .$

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Solution

Given: Two concentric circles $C_{1}$ and $C_{2}$ with centre O, and AB is the chord of $C_{1}$ touching $C_{2}$ at $C .$
To prove: $A C = C B$
Construction: Join $O C .$
Proof: $A B$ is the chord of $C_{1}$ touching $C_{2}$ at $C,$ then $A B$ is the tangent to $C_{2}$ at $C$ with $O C$ as its radius.
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, $O C ⊥ A B$
Considering, $A B$ as the chord of the circle $C_{1}$ . So, $O C ⊥ A B$ .
Therefore, $O C$ is the bisector of the chord $A B .$
Hence, $A C = C B$ (Perpendicular from the centre to the chord bisects the chord).

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