Question

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

• A. $5FE=10AB$ is true
• B. $7FE=10AB$ is true
• C. $3FE=10AB$ is true
• D. $5FE=5AB$ is true

Answer 1

The correct option is B $7FE=10AB$ is true

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