Question

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

Answer 1

Sol: GE II DC

$\Delta$BGE$\sim$$\Delta$BDC (AA condition)

So,$\frac{BG}{BD}=\frac{BE}{BC}=\frac{GE}{DC}$

or,$\frac{BG}{BD}=\frac{GE}{DC}=\frac{3}{7}[\frac{BE}{EC}=\frac{3}{4}\Rightarrow \frac{BE}{BC}=\frac{3}{7}]$

$\Rightarrow \frac{GE}{DC}=\frac{3}{7},also\frac{BG}{BD}=\frac{3}{7}$

$\Rightarrow GE=\frac{3}{7}DC$

$\Rightarrow GE=\frac{6}{7}AB----\left(i\right),\therefore \frac{DG}{BD}=\frac{4}{7}$

Also, $\Delta$DFG ~ $\Delta$DAB

$\therefore \frac{FG}{AB}=\frac{DG}{BD}\Rightarrow \frac{FG}{AB}=\frac{4}{7}$

$FG=\frac{4}{7}AB----------\left(ii\right)$

Adding (i) and (ii), we get

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

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

$\Rightarrow 7FE=10AB$