Question

proof that equal chords of a cirlcle are equidistant from the centre

Answer 1

In the figure,

Given:- A circle with centre O and chords AB = CD

Construction:- 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 ΔOPA and ΔOQC,
∠OPA = ∠OQC = 90°
AP = CQ (proved)
OA = OC (Radii)

Therefore, ΔOPA ≅ ΔOQC (By RHS congruence criterion)

Hence OP = OQ (CPCT)

So, equal chords are equidistant from the centre is proved.