If a segment is drawn between two sides in a triangle such that it divides the sides proportionally, the segment is parallel to the third side in the triangle

For $△ABC,$ since $\frac{AD}{DC}=\frac{BE}{EC}$ then $$\overline{DE}\parallel \overline{AB}.$$ The theorem is named the Converse Triangle Proportionality Theorem because the same thing holds true but in opposite order.