Question

In a trapezium $ABCD,AB\parallel DC$ and $DC=2AB$. $EF$ drawn parallel to $AB$ cuts $AD$ in $F$ and $BC$ in $E$ such that $\frac{BE}{EC}=\frac{3}{4}$. Diagonal $DB$ intersects $EF$ at $G$. Prove that $7FE=10AB$.

Answer 1

In $\Delta DFGand\Delta DAB$,
$\angle 1=\angle 2[Corresponding\angle s\because AB\parallel FG]$
$\angle FDG=\angle ADB\left[Common\right]$
$\therefore \Delta DFG\sim \Delta DAB[By$AA $ruleofsimilarity]$
$\therefore \frac{DF}{DA}=\frac{FG}{AB}\dots \left(i\right)$
$\angle FDG=\angle ADB\left[Common\right]$
$\therefore \Delta DFG\sim \Delta DAB[By$AA$ruleofsimilarity]$
$\therefore \frac{DF}{DA}=\frac{FG}{AB}\dots \left(i\right)$
Again in trapezium $ABCD$
$EF\parallel AB\parallel DC$
$\therefore \frac{AF}{DF}=\frac{BE}{EC}$
$\Rightarrow \frac{AF}{DF}=\frac{3}{4}\left[\begin{array}{c}\frac{BE}{EC}=\frac{3}{4}\left(given\right)\end{array}\right]$
$\Rightarrow \frac{AF}{DF}+1=\frac{3}{4}+1$
$\Rightarrow \frac{AF+DF}{DF}=\frac{7}{4}$
$\Rightarrow \frac{AD}{DF}=\frac{7}{4}\Rightarrow \frac{DF}{AD}=\frac{4}{7}\dots \left(ii\right)$
From (i) and (ii), we get
$\frac{FG}{AB}=\frac{4}{7}$ i.e., $FG=\frac{4}{7}AB\dots \left(iii\right)$
In $\Delta BEGand\Delta BCD$, we have
$\angle BEG=\angle BCD[Correspondingangle\because EG\parallel CD]$
$\angle GBE=\angle DBC\left[Common\right]$
$\therefore \Delta BEG\sim \Delta BCD[By$AA$ruleofsimilarity]$
$\therefore \frac{BE}{BC}=\frac{EG}{CD}$
$\therefore \frac{3}{7}=\frac{EG}{CD}$
$\left[\begin{array}{c}\frac{BE}{EG}=\frac{3}{7}i.e.,\frac{EC}{BE}=\frac{4}{3}\Rightarrow \frac{EC+BE}{BE}=\frac{4+3}{3}\Rightarrow \frac{BC}{BE}=\frac{7}{3}\end{array}\right]$
$\therefore EG=\frac{3}{7}CD=\frac{3}{7}\left(2AB\right)\left[\begin{array}{c}\because CD=2AB\left(given\right)\end{array}\right]$
$\therefore EG=\frac{6}{7}AB\dots \left(iv\right)$
Adding (iii) and (iv), we get
$FG+EG=\frac{4}{7}AB+67AB=\frac{10}{7}AB$
$\Rightarrow EF=\frac{10}{7}ABi.e.,7EF=10AB$.
Hence proved.