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, ∠A = ∠A

Therefore, by SAS-criterion of similarity, we have

∆ABC ~ ∆ADE

`\Rightarrow \frac{AD}{AD}=\frac{BC}{DE} ….(i)`

It is given that

`\frac{AD}{DB}=\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}{DE}=\frac{5}{2} ….(ii)`

From (i) and (ii), we get

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