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In the given figure, $DE \parallel AC$ and $DF \parallel AE$. Prove that : $\frac{BF}{FE} = \frac{BE}{EC}$
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Sol. In $\triangle BEA$, $FD \parallel EA$
$\therefore \frac{BF}{FE} = \frac{BD}{DA}$ ...(i) (I) (1 Mark)
In $\triangle BCA$, $ED \parallel CA$
$\therefore \frac{BE}{EC} = \frac{BD}{DA}$ ...(ii) (II) ($\frac{1}{2}$ Mark)
Using (i) and (ii)
$\frac{BF}{FE} = \frac{BE}{EC}$ (III) ($\frac{1}{2}$ Mark)
$\therefore \frac{BF}{FE} = \frac{BD}{DA}$ ...(i) (I) (1 Mark)
In $\triangle BCA$, $ED \parallel CA$
$\therefore \frac{BE}{EC} = \frac{BD}{DA}$ ...(ii) (II) ($\frac{1}{2}$ Mark)
Using (i) and (ii)
$\frac{BF}{FE} = \frac{BE}{EC}$ (III) ($\frac{1}{2}$ Mark)