Question

In $\Delta ABC,AB=9cm,AC=15cm$ and $DE||BC.$ If $\frac{AB}{AD}=3$, sort in ascending order of length.

• A. $AD$
• B. $DB$
• C. $AE$
• D. $EC$

Answer 1

Given: DE$\parallel$BC, $\frac{AB}{AD}=3$, AB = 9 cm and AC = 15 cm


$\Rightarrow \frac{9}{AD}=3$
$\Rightarrow AD=3cm$
$\Rightarrow AD+DB=AB$
$\Rightarrow 3+DB=9$
$\Rightarrow DB=6$

From basic proportionality theorem
$\frac{AD}{DB}=\frac{AE}{EC}$
$\text{Now,}\frac{AB}{AD}=\frac{AD+DB}{AD}=1+\frac{DB}{AD}\Rightarrow 1+\frac{DB}{AD}=3\Rightarrow \frac{DB}{AD}=2\Rightarrow \frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{2}\Rightarrow 2AE=EC\Rightarrow AC=AE+EC-------\left(1\right)$
On substituting value of EC in (1), we get
$15=3AE\Rightarrow AE=5cm\Rightarrow EC=10cm$

$AD<AE<DB<EC$