In the given fig., DE || AC and DF || AE. Prove that FE /BF = EC/ BE
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BASIC PROPORTIONALITY THEOREM (BPT) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. That is also known as Thales theorem.
SOLUTION:
GIVEN:
In ∆BAC, DE || AC
BE/EC = BD/DA………..(1)
[ By Thales theorem(BPT)]
In ∆BAE, DF || AE (GIVEN)
BF/FE = BD/DA………..(2)
[ By Thales theorem(BPT)]
(From eq 1 and 2)
BF/FE = BE/EC
FE/BF = EC/BE [reciprocal the terms]
Hence proved.
HOPE THIS WILL HELP YOU.....
SOLUTION:
GIVEN:
In ∆BAC, DE || AC
BE/EC = BD/DA………..(1)
[ By Thales theorem(BPT)]
In ∆BAE, DF || AE (GIVEN)
BF/FE = BD/DA………..(2)
[ By Thales theorem(BPT)]
(From eq 1 and 2)
BF/FE = BE/EC
FE/BF = EC/BE [reciprocal the terms]
Hence proved.
HOPE THIS WILL HELP YOU.....
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