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Theorem 4(i) - Equal chords of a Circle are Equidistant from the Centre

Proof :

Given :

and are two equal chords of a circle.


and are perpendiculars from the centre to the chords and

Construction: Join and .

To Prove: .


Consider the and .

(radii of the same circle)

(since and are perpendicular to the chords and it bisets the chord and )

Equal chords of a circle are equidistant from the centre.

Theorem 4 (ii) :

Equal chords of congruent circles are equidistant from the corresponding centres.
To Prove: OE = O'F where OE AB and O'F CD
Construction: Join OA and O'C

Proof :

In right-triangles OAE and O'CF AE = AB and CF = CD
AE = CF                    [... AB = CD]
  OA = O'C = r             [Radii of the congruent circles]

∴ ΔOAE ≅ ΔO'CF          (R.H.S)
Hence, OE = O'F (CPCT)




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