In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm.

GIVEN: DE is parallel to BC, AD = 1cm and BD = 2cm.

TO FIND: Ratio of ΔABC to area of ΔADE

According to BASIC PROPORTIONALITY THEOREM, if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$

$\frac{1}{2+1}=\frac{\mathrm{AE}}{\mathrm{AC}}$

$\Rightarrow \frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{3}$

So

$\frac{\operatorname{ar}(\triangle \mathrm{ABC})}{\operatorname{ar}(\triangle \mathrm{ADE})}=\left(\frac{\mathrm{AC}}{\mathrm{AE}}\right)^{2}$

$=\left(\frac{3}{1}\right)^{2}$

$=\frac{9}{1}$