Question

Equidistant chords of a circle are equal why

Answer 1

Let AB and CD equidistant chords  OM perpendicular to AB and ON perpendicular to CD Am=1/2AB and CN =1/2CD AMO and CNO are congruent  Because OM=ON                 AngleM=angleN(90 degree)                 OA=OC(radii of same circle)  Hence AM=CM(c.p.c.t)  Therefore 1/2AB=1/2CD                 AB= CD So equidistant chords are equal

Answer 2

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".