In the given figure, DE ║AC and DF ║ AE.
Prove that FE/BF=EC/BE
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Answer:
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.Step-by-step explanation:
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