Question

In the trapezium $ABCD$
$AB\parallel DC,EF\parallel AB,DC=2AB$ and $\frac{BE}{EC}=\frac{3}{4}$. Prove that $7EF=10AB$.

Answer 1

Let us consider that

$\frac{BE}{EC}=\frac{AF}{FD}=\frac{3}{4}$

From $△BDC$,

$\frac{GE}{CD}=\frac{3}{7}$

$\Rightarrow \frac{GE}{2AB}=\frac{3}{7}$

$\therefore GE=\frac{6AB}{7}$[Since $CD=2AB$]....(i)

From $△DAB$

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

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

$\therefore GF=\frac{4AB}{7}$....(ii)

Adding (i) and (ii),

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

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

$\Rightarrow 7FE=10AB$