Question

Prove that: Congruent chords of a circle are equidistant from the centre of the circle.​

Answer 1

Statement : There is one and only one circle passing through three given noncollinear points.

Given : AB and CD are two equal chords of the circle.

OM and ON are perpendiculars from the centre at the chords AB and CD.

To prove : OM = ON.

Construction : Join OA and OC.

Proof :

In ΔAOM and ΔCON,

OA = OC . (radii of the same circle)

MA = CN . (since OM and ON are perpendicular to the chords and it bisects the chord and AM = MB, CN = ND)

∠OMA = ∠ONC = 90°

ΔAOM ≅ ΔCON (R. H. S)

OM = ON (c. p. c. t.)

Equal chords of a circle are equidistant from the centre.

Answer 2

Step-by-step explanation:

I have solved it in my words, I Hope this helps you..