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Theorem 2
All equal chords of a circle are equidistant from the centre.

Let AB and CD be two equal chords of a circle with
centre O. It is to be proved that the chords AB and CD
are equidistant from the centre.

Construction: Draw from Othe perpendiculars OE
andOFto the chords AB and CDrespectively. Join
O, A and O, C.

Proof:

Steps Justification

(1) OEAAB and OFACD

Therefore, AE = BE and CF = BF.

?AE=
2

1

AB and CF =

2

1

CD

[The perpendicular from

the centre bisects the

chord]

(2) But AB = DC

?AE = CF

[ supposition]

(3) Now in the right-angled triangles 'OAE and
'OCF
hypotenuse OA = hypotenuse OC and AE = CF

? 'OAE#'OCF

?OE = OF

[radius of same circle]

[Step 2]

[ RHS theorem]

(4) But OE and OF are the distances from O to the
chordsAB and CD respectively.
Therefore, the chords AB andCD are equidistant from
the centre of the circle. (Proved)

Theorem 3

Chords equidistant from the centre of a circle are equal.

Let AB and CDbe two chords of a circle with centre

O. OEand OFare the perpendiculars from Oto the

chordsAB and CDrespectively. Then OEandOF

represent the distance from centre to the chords AB

andCDrespectively.

It is to be proved that if OE = OF, AB=CD.

Construction: Join O, A and O, C.