Prove that: Congruent chords of a circle are equidistant from the cent

1.	Prove that: Congruent chords of a circle are equidistant from the centre of the circle.
Answer» 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.

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

Answer»

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.

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