Question

In a $△ABC,D$ and $E$ are points on the sides $AB$ and $AC$ respectively such that $DE||BC$.
If $\frac{AD}{BD}=\frac{3}{4}$ and $AC=15cm$, find $AE$ .

Answer 1

$\frac{AD}{BD}=\frac{3}{4}$

$\frac{BD}{AD}=\frac{4}{3}$

$\frac{BD}{AD}+1=\frac{4}{3}+1$

$\frac{BD+AD}{AD}=\frac{3+4}{4}$

$\frac{AB}{AD}=\frac{7}{3}$

$\frac{AD}{AB}=\frac{3}{7}-\left(1\right)$

$In\Delta ADEand\Delta ABC$

$\angle DAE=\angle BAC\left(common\right)$

$\angle ADE=\angle ABC\left(correspondingangle\right)$

$so,\Delta ADE\sim \Delta ABC$

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

$\frac{3}{7}=\frac{AE}{15}$

$AE=\frac{3}{7}\times 15$

$AE=\frac{45}{7}$

$AE=6\frac{3}{7}cm$