Question

prove that equal chords of a circle are equidistant from the centre of the circle

Answer 1

Given a circle with centre O and chords AB = CD
Draw OP⊥ AB and OQ ⊥ CD
Hence AP = BP = (1/2)AB and CQ = QD = (1/2)CD
Also ∠OPA = 90° and ∠OQC = 90°
Since AB = CD
⇒ (1/2) AB = (1/2) CD
⇒ AP = CQ
In Δ’s OPA and OQC,
∠OPA = ∠OQC = 90°
AP = CQ (proved)
OA = OC (Radii)
∴ ΔOPA ≅ ΔOQC (By RHS congruence criterion)
Hence OP = OQ (CPCT)

Answer 2

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.