Question

Prove that equal chords of a circle (or congruent circles) are equidistant from the centre (or centres).

Answer 1

Theorem: Congruent chords of circle are equidistant from the center of the circle.

Given: 'O' is the center of the circle, where Chord AB ≅ Chord MN.

To prove that: CO ≅ PO

Construction: Draw radii OB and radii ON.

Proof: OC ⊥ AB and OP ⊥ MN (Given)

Therefore,

                        Seg AB = Seg PN (Given)

                        CB = 1/2 AB; PN = 1/2 MN

                      ∴ Seg CB = Seg PN ------ (i)

Now,

In ΔOCB and ΔOPN,

Seg CB ≅ Seg PN ----- From i

∠OCB ≅ ∠OPN ----- Each 90°

Seg OB ≅ Seg ON ------- Radii of circle

            ∴ ΔOCB ≅ ΔOPN (Hypo. side test)

            ∴ Seg OP = Seg CO ------- C.S.C.T

Therefore, Chords are equidistant from the center of the circle.

"Refer to the Given attachment".

Answer 2

Answer:

Given- AB= CD

To prove- OM=ON

Construction- Join OB & OD

Proof- In ∆BOM & ∆DON

OB= OD (Radius of a circle)

Angle OMB = angle OND= 90°

MB= ND

∆ BOM is congruent to ∆DON (By RHS)

OM = ON ( By CPCT)

Hence, proved

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