Question

Prove that equal chords of circle are equidistant from the centre

Answer 1

 Given a circle with centre O and chords AB = CDDraw 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

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".