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Statement : There is one and only one circle passing through three given noncollinear points.
Given : AB and CD are two equal chords of the circle.
OM and ON are perpendiculars from the centre at the chords AB and CD.
To prove : OM = ON.
Construction : Join OA and OC.
In ΔAOM and ΔCON,
OA = OC . (radii of the same circle)
MA = CN . (since OM and ON are perpendicular to the chords and it bisects the chord and AM = MB, CN = ND)
∠OMA = ∠ONC = 90°
ΔAOM ≅ ΔCON (R. H. S)
OM = ON (c. p. c. t.)
Equal chords of a circle are equidistant from the centre.
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