Question

Two equal chords AB and CD of a circle when produced, intersect at point P prove that PB is equal to PD.

Answer 1

given:AB=CD
to prove:PB=PD
const: draw OE and OQ perpendicular on AB and CD respectively
proof: 
given AB and CD  are two equal chords of the same circle.
OE=OQ(equal chords of a circle are equidistant from the center.)
now in triangle OEP and OQP,
OE=OQ
OP=OP(common)
angle OEP = OQP =90 degree,by construction
therefore triangle OEP = OQP (RHS congruency)
EP = QP (CPCT)
also AE=EB=1/2 AB and CQ=QD=1/2 CD (the line joining the center of the circle is perpendicular to the chord and bisects the chord.)
Now AB = AC implies AE = EB= CQ=QD ....(1)
therefore EP-BE =QP - BE
EP - BE = QP - QD (FROM 1)
BP = PD
hence proved 

Answer 2

given AB and CD are two equal chords of the same circle
⇒ OE = OQ [Equal chords]
 in ΔOEP and ΔOQP, we have
OE = OQ
OP = OP [common]
and OEP = OQP = 90 [construction]
⇒ ΔOEP congruent to ΔOQP [RHS rule]
∴EP = QP [c.p.c.t]
Now..... AB = AC implies that AE = EB = CQ = QD ... (1)
∴EP - BE = QP - BE
∴ EP - BE = QP - QD [using (1)]
⇒ BP = DP [proved]