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.

Drop a perpendicular from O to both chords AB and CD.
In △OMP and △ONP
As chords are equal, perpendicular from the center would also be equal. OM=ON
OP is common.
∠OMP = ∠ONP = 90°
△OMP ≅ △ONP (RHS Congruence)
PM=PN [By CPCT] . (1)
AM=BM (Perpendicular from center bisects the chord)
Similarly, CN=DN
As AB=CD
AB-AM = CD-DN
BM=CN (2)
From (1) and (2)
BM-PM = CN-PN
PB=PC
AM=DN (Half the length of equal chords are equal)
AM+PM=DN+PN
AP=PD
Therefore, PB=PC and AP=PD is proved.