Prove that equal chords of a circle are equidistant from the centre.

Given: AB = CD

Construction: Drop perpendiculars OX and OY on to AB and CD respectively and join OA and OD.

Here, OX AB (perpendicular from center to chord divides it into two equal halves)

AX = BX = – – (1)

OY CD (perpendicular from center to chords divides it into equal halves

CY = DY = – – (2)

Now, given that

AB = CD

∴ =

AX = DY (from –1 and –2 ) – – (3)

In ΔAOX and ΔDOY

∠OXA = ∠OYD (right angle)

OA = OD (radius)

AX = DY (from –3 )

∴ BY RHS congruency

ΔAOXΔDOY

OX = OY (by C.P.C.T)

Hence proved.