Question

In $△ABC$, DE is parallel to base BC, with D on AB and E on AC. If $\frac{AD}{DB}=\frac{2}{3}$, find $\frac{BC}{DE}$.

Answer 1

In $△ABC$, we have

$DE\parallel BC\Rightarrow \frac{AB}{AD}=\frac{AC}{AE}$
Thus, in triangles ABC and ADE, we have
$\frac{AB}{AD}=\frac{AC}{AE}and,\angle A=\angle A$
Therefore, by SAS-criterion of similarity, we have
$△ABC\sim △ADE$
$\Rightarrow \frac{AD}{AD}=\frac{BC}{DE}$ .......(i)
$\frac{AB}{AD}=\frac{BC}{DE}$
$\frac{AD+DB}{AD}=\frac{BC}{DE}$
So $\frac{BC}{DE}=1+\frac{3}{2}=\frac{5}{2}$