Question

In trapezium ABCD, AB||DA 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=10 AB.

Answer 1

In $△DFG$ and $△DAB$, we have

$\angle 1=\angle 2$ [$\because AB||DC||EF\therefore \angle 1$ and $\angle 2$ are corresponding angles]

$\angle FDG=\angle ADB$ [Common]

So, by AA-criterion of similarity, we have

$\therefore$ $△DFG\sim △DAB$

$\Rightarrow$ $\frac{DF}{DA}=\frac{FG}{AB}$.......(i)

In trapezium ABCD, we have

EF||AB||DC

$\therefore$ $\frac{AF}{DF}=\frac{BE}{EC}$

$\Rightarrow$ $\frac{AF}{DF}=\frac{3}{4}$

[$\because \frac{BE}{EC}=\frac{3}{4}$(Given)]

$\Rightarrow$ $\frac{AF}{DF}+1=\frac{3}{4}+1$ [Adding 1 on both sides]

$\Rightarrow$ $\frac{AF+DF}{DF}=\frac{7}{4}$

$\Rightarrow$ $\frac{AD}{DF}=\frac{7}{4}$

$\Rightarrow$ $\frac{DF}{AD}=\frac{4}{7}$......(ii)

From (i) and (ii), we get

$\frac{FG}{AB}=\frac{4}{7}$

$\Rightarrow$ $FG=\frac{4}{7}AB$........(iii)

In $△BEG$ and $△BCD$, we have

$\angle BEG=\angle BCD$ [Corresponding angles]

$\angle B=\angle B$ [Common]

$\therefore$ $△BEG\sim △BCD$ [By AA-criterion of similarity]

$\Rightarrow$ $\frac{BE}{BC}=\frac{EG}{CD}$

$\Rightarrow$ $\frac{3}{7}=\frac{EG}{CD}$ [$\because \frac{BE}{EC}=\frac{3}{4}\Rightarrow \frac{EC}{BE}=\frac{4}{3}\Rightarrow \frac{EC}{BE}+1=\frac{4}{3}+1\Rightarrow \frac{BC}{BE}=\frac{7}{3}]$

$\Rightarrow$ $EG=\frac{3}{7}CD$

$\Rightarrow$ $EG=\frac{3}{7}\times 2AB$

$\Rightarrow$ $EG=\frac{6}{7}\times AB$

Adding (iii) and (iv), we get

$FG+EG=\frac{4}{7}AB+\frac{6}{7}AB$

$\Rightarrow$ $EF=\frac{10}{7}AB$

$\Rightarrow$ $7EF=10AB$ $\left[Henceproved\right]$