• # question_answer 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.

 $\frac{EC}{EA}=\frac{CF}{FB}$  and $\frac{CF}{FB}=\frac{2}{2.5}=\frac{4}{5}$ $\Rightarrow \frac{EC}{EA}=\frac{CF}{FB}$ In $\Delta \text{ }ABC,\text{ }EF\parallel AB$ [Converse of Thales' theorem] Also,     $\frac{CE}{CA}=\frac{4}{4+5}=\frac{4}{9}$ $\frac{CE}{CB}=\frac{2}{2+2.5}=\frac{2}{4.5}=\frac{4}{9}$ $\frac{EC}{EA}=\frac{CF}{CB}$ $\angle ECF=\angle ACB$                                  [Common] $\Delta \text{ }CFE\tilde{\ }\Delta \text{ }CBA$                   [SAS similarity] $\Rightarrow \frac{EF}{AB}=\frac{CE}{CA}$ [In similar $\Delta 's$, corresponding sides are proportional] $\Rightarrow \frac{EF}{7}=\frac{4}{9}$ $[\because ~\,AB=3+4=7\text{ }cm]$ $\therefore EF=\frac{28}{9}cm$ and $AB=7\,cm$