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

$DE||BC$

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

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

$△ABC\sim △ADE$

$\Rightarrow$ $\frac{AB}{AD}=\frac{BC}{DE}$.......(i)

It is given that

$\frac{AB}{AD}=\frac{2}{3}$

$\Rightarrow$ $\frac{DB}{AD}=\frac{3}{2}$

$\Rightarrow$ $\frac{DB}{AD}+1=\frac{3}{2}+1$

$\Rightarrow$ $\frac{DB+AD}{AD}=\frac{5}{2}$

$\Rightarrow$ $\frac{AB}{AD}=\frac{5}{2}$ .............(ii)

From (i) and (ii), we get

$\frac{BC}{DE}=\frac{5}{2}$