Question

$D$ and $E$ are respectively the points on the sides $AB$ and $AC$ of a $△ABC$ such that $AB=5.6cm,AD=1.4cm,AC=7.2cm$ and $AE=1.8cm$, show that $DE\parallel BC$.

Answer 1

We have $AB=5.6cm,AD=1.4cm,AC=7.2cm$ and $AE=1.8cm$. $BD-AB-AD-5.6-1.4-4.2cm$
and $EC=AC-AE=7.2-1.8=5.4cm$.
$\frac{AD}{DB}=\frac{1.4}{4.2}=\frac{1}{3}$ and $\frac{AE}{EC}=\frac{1.8}{5.4}=\frac{1}{3}$
$\frac{AD}{DB}=\frac{AE}{EC}$
Therefore, by converse of Basic Proportionality Theorem, we have DE is parallel to BC.
Hence proved.