The equal chords ab and cd of a circle with centre o when produced meet at p outside the circle . Prove that. Pb=pd and pa=pc

Given,AB = CD.

To prove:BE = DE and AE = CE

Construction: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 ΔOEP and ΔOQP, we have

OE = OQ

OP = OP [common]

and OEP = OQP = 90 [By construction]

⇒ ΔOEP

ΔOQP [RHS congruency]

⇒EP = QP [c.p.c.t]

Also, AE = EB = ½

AB and CQ = QD =CD [The line joining the center of a circle and is perpendicular to the chord always bisects the chord.]

Now AB = AC implies that AE = EB = CQ = QD ... (1)

∴EP - BE = QP - BE

⇒ EP - BE = QP - QD [using (1)]

⇒ BP = DP