Question

In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC​

Answer 1

Answer:

your proof is as follows

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Step-by-step explanation:

$to \: prove \: that : \\ \frac{BF}{FE} = \frac{BE}{EC} \\ \\ so \: considering \: now \: \: △BEA \\ DF \: || AE \\ by \: basic \: proportionality \: theorem \\ we \: have \\ \\ \frac{BF }{FE} = \frac{BD }{ DA} \: \: \: \: \: \: \: \: (1)$

$similarly \: considering \: △ABC \\ ∵ \:DE \: || AC \\ by \: basic \: proportionality \: theorem \\ we \: get \\ \\ \frac{BE }{ EC } = \frac{BD }{ DA} \: \: \: \: \: \: \: \: (2)$

$equating \: (1) \: and \: (2) \\ we \: get \\ \\ \frac{BF }{ FE} = \frac{BE}{EC} \\ \\ thus \: proved$