Question

prove that equal chords are equi distance from center of a circle

Answer 1

Draw two equal chords AB and CD
join AO , BO , CO and DO
To prove ang AOB = ang COD
AB = CD (given)
AO = BO(radii)
CO = DO(radii)
triangle AOB congruent to trinagle COD (by SSS)
therefore angle AOB = angle COD (Corresponding Parts of Congruent Triangles)

Answer 2

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)