Question

In a triangle ABC, DE is parallel to BC if $\frac{AD}{DB}=$ $\frac{2}{3}$ and $AC=18$ cm find AE.

Answer 1

Let $AE$ be $xcm$.

Then, $EC=18-xcm$

Since, $\frac{AD}{DB}=\frac{AE}{EC}$

Therefore,

$\frac{2}{3}=\frac{x}{18-x}$

$36-2x=3x$

$5x=36$

$x=7.2cm$

Hence, this is the answer.