# Thread: DE Parallel to BC?

1. ## DE Parallel to BC?

Given triangle ABC and segment DE, if AD =2, DB =3,
AE =4 and EC =4, is DE parallel to BC?

2. This is converse of Basic Proportionality Theorem

3. Originally Posted by magentarita
Given triangle ABC and segment DE, if AD =2, DB =3,
AE =4 and EC =4, is DE parallel to BC?
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 $\overline{DE} \parallel \overline{BC}$, then

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

$\frac{2}{3} \neq \frac{4}{4}$

Therefore, $\overline{DE} \nparallel \overline{BC}$

4. ## ok...

Originally Posted by masters
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 $\overline{DE} \parallel \overline{BC}$, then

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

$\frac{2}{3} \neq \frac{4}{4}$

Therefore, $\overline{DE} \nparallel \overline{BC}$
I thank both replies. I want to thank masters for always answering my questions quickly and professionally.