Question

In the 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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Answer 1

Step-by-step explanation:

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. 

Therefore, OC⊥AB

Considering, AB as the chord of the circle C1. So, OC⊥AB.

Therefore,OC is the bisector of the chord AB.

Hence, AC=CB (Perpendicular from the centre to the chord bisects the chord). 

Answer 2

Answer:

To prove: AC=BC Construction: Join OC Proof: Since OC is the radius of the smaller circle and AB is a tangent to this circle at point C. So, OC⟂AB We know that the perpendicular drawn from the centre of a circle to any chord of the circle, bisects the chord. So, AC=BC Hence, AB is bisected at C