If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Let PQ and CD be the 2 equal chords. PQ = CD. Let the chords intersect at point E. Join OE.

To prove PE = CE and QE = DE.

Draw perpendiculars from the center O to the chords. This Perpendicular bisects the chord PQ at M and CD at N.

Thus, PM = MQ = CN = DN ……(1)

In ∆OME and ∆ONE

∠M = ∠N = 90°

OE = OE

OM = ON (Equal chords are equidistant from the center.)

By RHS criteria, ∆OME and ∆ONE are congruent.

So by CPCT, ME = NE ….. (2)

We know that: CE = CN + NE and PE = AM + ME

From (1) and (2), it is evident CE = PE

DE = CD - CE and PE = PQ - AE

PQ and CD are equal, CE and PE are equal. So, DE and PE are also equal.