Question

In a $△ABC,D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE\parallel BC$. If $AD=2.4cm,AE=3.2cm,DE\parallel BC$. If $AD=2.4cm,AE=3.2cm,DE=2cm$ and $BC=5cm$, find $BD$ and $CE$.

Answer 1

If a line is drawn parallel to one side of a triangle to intersect the

other two sides in distinct points, the other two sides are divided in the same ratio.

As $DE\parallel BC$

$\angle ADE=\angle ABC$

$\angle AED=\angle ACB$

So by $AAA$ $△AED\sim △ACB$

Hence $\frac{AE}{AC}=\frac{ED}{CB}$

$\therefore \frac{3.2}{AC}=\frac{2}{5}$

$\Rightarrow AC=8cm$

$AC=8cm=AE+EC=3.2+EC$

$EC=8-3.2=4.8cm$

Also $\frac{AD}{AB}=\frac{ED}{CB}$

$\therefore \frac{2.4}{AB}=\frac{2}{5}$

$AB=6cm=AD+DB=2.4+DB$

$\Rightarrow DB=3.6cm$

So $BD=3.6cm$ and $CE=4.8cm$