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.


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