In the figure, if DE || AC and DF || AE, prove that \(BF\over FE\) = \(BE\over EC\).

Solution :

In triangle BCA,similar triangles image

Given,        DE || AC

By Basic proportionality theorem, we have

\(BE\over EC\) = \(BD\over DA\)           ……..(1)

In triangle BEA,

Given,        DF ||  AE

By Basic proportionality theorem, we have

\(BF\over FE\) = \(BD\over DA\)           ……..(2)

From (1) and (2), we obtain that

\(BF\over FE\) = \(BE\over EC\)

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