Question

two equal chords ab and CD of a circle when produced meet at e prove that

1. BE=DE
2. AE=CE

Answer 1

first prove congruent to Triangle ;
Δope=Δoqe
pe=qd by cpct


a b is equal to CD because equal chords lie equal distance

ab+be=cd+de
ae=ce

pe=pb+be
qe=qd+de




Pb is equal to qd equal chords lie on same distance

so b is equal to de

Answer 2

hlo ...mate...


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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

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