Question
In figure $DE || AC$ and $DF || AE$. Prove that $\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$
✓
Answer
In $\triangle ABE,$ we have $DF||AE,$then
$\frac{{BD}}{{AD}} = \frac{{BF}}{{FE}}\,$ $[$By BPT$] ...... (i)$
In $\triangle ABC,$ we have $DE||AC,$ then
$\frac{{BD}}{{AD}} = \frac{{BE}}{{EC}}\,$ $[$By BPT$] ...... (2)$
From $(i)$ and $(2)$, We get
$\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$
$\frac{{BD}}{{AD}} = \frac{{BF}}{{FE}}\,$ $[$By BPT$] ...... (i)$
In $\triangle ABC,$ we have $DE||AC,$ then
$\frac{{BD}}{{AD}} = \frac{{BE}}{{EC}}\,$ $[$By BPT$] ...... (2)$
From $(i)$ and $(2)$, We get
$\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$
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