Question

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

Answer 1

Given that :--:

IN CIRCLE C( O, r)

AB = PQ


To prove :--:

<AOB = <POQ


Proof :--:

In Triangle AOB and POQ

AB = PQ ( Given)

AO = PO ( Radii of circle)

BO = QO ( Radii of circle)

So,

Triangle AOB is congurent to triangle POQ

=> <AOB = <POQ ( C. P. C. T)




Hence Proved :::---:---:::

Answer 2

Proof : In△AOB and △COD

AO = CO [radii of same circle]

BO = DO [radii of same circle]

Chord AB = Chord CD [given]

⇒ △AOB ≅ △COD [by SSS congruence axiom]

⇒ ∠AOB = ∠COD. [c.p.c.t.