Question

In the given figure, $AD=3cm,AE=5cm,BD=4cm,CE=4cm,CF=2cm,BF=2.5cm$, then find the pair of parallel lines and hence their lengths.

Answer 1

$\frac{EC}{EA}=\frac{CF}{FB}$ and $\frac{CF}{FB}=\frac{2}{2.5}=\frac{4}{5}$

$\Rightarrow \frac{EC}{EA}=\frac{CF}{FB}$

In $\Delta ABC,EF||AB$
[Converse of Thales' theorem]
Also, $\frac{CE}{CA}=\frac{4}{4+5}=\frac{4}{9}$ . . . (i)

$\frac{CF}{CB}=\frac{2}{2+2.5}=\frac{2}{4.5}=\frac{4}{9}$

$\frac{EC}{EA}=\frac{CF}{CB}$

$\angle ECF=\angle ACB$ [Common]

$\Delta CFE\sim \Delta CBA$ [SAS similarity]

$\Rightarrow \frac{EF}{AB}=\frac{CE}{CA}$

[In similar ${\Delta }^{\prime }s$, corresponding sides are proportional]

$\Rightarrow \frac{EF}{7}=\frac{4}{9}$
$[\because AB=3+4=7cm]$

$\therefore EF=\frac{28}{9}cm$ and $AB=7cm$