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

Solution :

In triangle BCA,

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

$$BF\over FE$$ = $$BE\over EC$$