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In the given figure, the chord AB of the larger of the two concentric circles, with center O, touches the smaller circle at C. Prove that AC = CB.
Given: Two concentric circles with common center as O
To Prove: AC = CB
Construction: Join OC, OA and OB
Proof :
OC ⏊ AB
[Tangent at any point on the circle is perpendicular to the radius through point of contact]
In △OAC and △OCB, we have
OA = OB
[∵ radii of same circle]
OC = OC
[∵ common]
∠OCA = ∠OCB
[∵ Both 90° as OC ⏊ AB]
△OAC ≅ △OCB
[By Right Angle - Hypotenuse - Side]
AC = CB
[Corresponding parts of congruent triangles are congruent]
Hence Proved.