prove that equal chord of a circle are equidistant from the centre.
let a circle with centre O has two equal chords AB and CD and OD And OE are perpendicular to these chords from centre respectively.
We have to prove that OD = OE ( means perpendicular distance are equal)
in triangles ODB and OEC, we have
DB = EC (since AB = CD And perpendicular from the centre bisects the respective chord)
Angle ODB = Angle OEC = 900 ( Since OD and OE are perpendiculars)
BO = CO ( Radii of same circle)
triangle ODB is congruent to triangle OEC (by RHS critaria)
therefore OD = OE (by c.p.c.t.)