prove that equal chords of a circle subtend equal angles at the centre.
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Given : In a circle C(O,r), chord AB = chord CD.
To Prove : ∠AOB = ∠COD.
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.]
To Prove : ∠AOB = ∠COD.
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.]
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Hope the image helps you
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Peaches15manny:
wheres the image ??
It's in answer
kk got it tq
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