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In trapezium PQRS, $PQ \parallel SR$ and $SR = 2 PQ$. A line segment FE drawn parallel to PQ intersects PS at F and QR at E such that $\frac{QE}{ER} = \frac{3}{4}$. Diagonal QS intersects FE at G. Prove that $\frac{FE}{PQ} = \frac{10}{7}$.
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GE $\parallel$ PQ and PQ $\parallel$ SR $\Rightarrow$ GE $\parallel$ SR
$\therefore \triangle QGE \sim \triangle QSR$
$\frac{GE}{SR} = \frac{QE}{QR} = \frac{3}{7}$
Given, $SR = 2 PQ$
$\Rightarrow \frac{GE}{PQ} = \frac{6}{7}$ ... (i)
Now, FG $\parallel$ PQ
$\therefore \triangle SFG \sim \triangle SPQ$
$\Rightarrow \frac{FG}{PQ} = \frac{SF}{SP}$
Also, $\frac{SF}{SP} = \frac{RE}{RQ} = \frac{4}{7}$
$\Rightarrow \frac{FG}{PQ} = \frac{4}{7}$ ... (ii)
Adding (i) and (ii), $\frac{FE}{PQ} = \frac{6}{7} + \frac{4}{7} = \frac{10}{7}$
$\therefore \triangle QGE \sim \triangle QSR$
$\frac{GE}{SR} = \frac{QE}{QR} = \frac{3}{7}$
Given, $SR = 2 PQ$
$\Rightarrow \frac{GE}{PQ} = \frac{6}{7}$ ... (i)
Now, FG $\parallel$ PQ
$\therefore \triangle SFG \sim \triangle SPQ$
$\Rightarrow \frac{FG}{PQ} = \frac{SF}{SP}$
Also, $\frac{SF}{SP} = \frac{RE}{RQ} = \frac{4}{7}$
$\Rightarrow \frac{FG}{PQ} = \frac{4}{7}$ ... (ii)
Adding (i) and (ii), $\frac{FE}{PQ} = \frac{6}{7} + \frac{4}{7} = \frac{10}{7}$