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In the figure, MNOP is a trapezium with, MN $||$ PO and PO = $2$ MN.
A line segment FE drawn parallel to MN intersects MP at F and NO at E such that $\frac{NE}{EO} = \frac{3}{4}$. Diagonal PN intersects FE at X. Prove that $7$ FE = $10$ MN.
A line segment FE drawn parallel to MN intersects MP at F and NO at E such that $\frac{NE}{EO} = \frac{3}{4}$. Diagonal PN intersects FE at X. Prove that $7$ FE = $10$ MN.
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$\frac{NE}{EO} = \frac{3}{4} \Rightarrow \frac{NE}{NO} = \frac{3}{7}$
XE $||$ PO
Therefore, $\frac{NX}{NP} = \frac{NE}{NO} = \frac{XE}{PO} = \frac{3}{7}$
$\because$ PO = $2$ MN
$\frac{XE}{MN} = \frac{6}{7}$ --- (1)
Also, $\frac{NX}{NP} = \frac{XP}{NP} = \frac{3}{7}$
Now, FX $||$ MN
$\frac{XP}{NP} = \frac{FX}{MN} = \frac{4}{7}$ --- (2)
Using (1) and (2),
$\frac{XE}{MN} + \frac{XF}{MN} = \frac{6}{7} + \frac{4}{7}$
$\frac{EF}{MN} = \frac{10}{7}$
or $7$ FE = $10$ MN
XE $||$ PO
Therefore, $\frac{NX}{NP} = \frac{NE}{NO} = \frac{XE}{PO} = \frac{3}{7}$
$\because$ PO = $2$ MN
$\frac{XE}{MN} = \frac{6}{7}$ --- (1)
Also, $\frac{NX}{NP} = \frac{XP}{NP} = \frac{3}{7}$
Now, FX $||$ MN
$\frac{XP}{NP} = \frac{FX}{MN} = \frac{4}{7}$ --- (2)
Using (1) and (2),
$\frac{XE}{MN} + \frac{XF}{MN} = \frac{6}{7} + \frac{4}{7}$
$\frac{EF}{MN} = \frac{10}{7}$
or $7$ FE = $10$ MN