Question

prove that equal chords of circle subtend equal angles at the centre

Answer 1

consider a circle with centre O and two equal chords AB and CD. the radii to the ends of the chords are OA, OB and OC, OD respectively.

in ∆OAB and ∆OCD:
(i) OA = OC
(ii) OB = OD
(iii) AB = CD
therefore ∆OAB congruent to ∆OCD
therefore <AOB = <COD (CPCT)

thus, equal chords subtend equal angles at the centre.

Answer 2

Statement : Equal chords of a circle subtend equal angles at the centre.

Given : AB and CD are chords of a circle with centre O, such that AB = CD.

To prove : ∠AOC = ∠COD

Proof :

In ΔAOB and ΔCOD,

AO = CO [radii of same circle]
BO = DO [radii of same circle]
AB = CD [given]

ΔAOB ≅ ΔCOD [SSS]
∠AOB = ∠COD [C. P. C. T]

Hence, Equal chords of a circle subtend equal angles at the centre.