Given triangle ABC and segment DE, if AD =2, DB =3,
AE =4 and EC =4, is DE parallel to BC?
This is converse of Basic Proportionality Theorem
See diagram.
The "Triangle Proportionality Theroem" states: If a line is parallel to one side of a triangle and intersects the other two sides in two distinct points, then it separates these sides into segments of proportional length.
So, if $\displaystyle \overline{DE} \parallel \overline{BC}$, then
$\displaystyle \frac{AD}{DB}=\frac{AE}{EC}$
$\displaystyle \frac{2}{3} \neq \frac{4}{4}$
Therefore, $\displaystyle \overline{DE} \nparallel \overline{BC}$