Question

Two equal chords AB and CD of a circle when produced intersect at point P.Prove that PB=PD.

Answer 1

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SOLUTION:--



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=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