CAT Equal Chords and Their Distances from the Centre - (Part 2) - Practice Questions & MCQ

The converse of theorem 6.

Theorem 7 : Chords of circle (or of congruent circles) which are equidistant from the centre (or centers) of a circle are equal in length.

Let's prove this

Let AB and CD are two chords of a circle C(O,r), OL is perpendicualr to AB and OM is perpendicualr to CD such that OL = OM.

We need to prove AB = CD

Join OA and OC

We know that the perpendicular from the centre of a circle to a chord bisects the chord

Now, in the right triangle OLA and OMC, we have

              OA = OC              (radius of circle)

              OL = OM              (given)

        ∠ OLA = ∠ OMC = 90°

Therefore,   ∆ OLA ≅ ∆ OMC    ([by RHS-congruence)

Therefore,   AL = CM  or   2AL = 2CM

Hence, AB = CD