Given AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.
To prove : AP = PD and PB = CP.
Construction: Draw OM perpendicular to AB and ON perpendicular CD. Join OP.

AM = MB = 1/2AB (Perpendicular bisecting the chord)
CN = ND = 1/2CD (Perpendicular bisecting the chord)
AM = ND and MB = CN (As AB = CD)
In triangle OMP and ONP, we have,
OM = MN (Equal chords are equidistant from the centre)
<OMP = <ONP (90⁰)
OP is common. Thus triangle OMP and ONP are congruent (RHS).
MP = PN (cpct)
So, AM + MP = ND + PN
or, AP = PD (i)

As MB = CN and MP = PN,
MB - MP = CN - PN 
= PB = CP (ii)

Hope that helps !!