Question

In the given 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.

Answer 1

In the given figure, chord AB of larger circle of the two concentric circles with centre O, touches the smaller circle at C.

To Prove: AC = CB

Join OC, OA and OB.

Now,

AB is tangent to the smaller circles and OC is the radius.

this implies, OC ⊥ AB.

We have two right angled triangles: ∆OAC and ∆OBC

Here OA = OB = radius of same circle

Side OC = OC = common among both the triangles

∆OAC = ∆OBC (RHS axiom)

By c.p.c.t.

AC = CB

Hence proved.