1. **State the problem:** Given that line segment $\overline{AB}$ is parallel to line segment $\overline{DE}$, prove that triangle $\triangle ABC$ is similar to triangle $\triangle DEC$. 2. **Recall the similarity criteria:** Two triangles are similar if their corresponding angles are equal (AA criterion). 3. **Identify corresponding angles:** Since $\overline{AB} \parallel \overline{DE}$ and $D$ lies on $\overline{AC}$ and $E$ lies on $\overline{BC}$, by the Alternate Interior Angles Theorem: - $\angle BAC = \angle EDC$ (corresponding angles 1) - $\angle ABC = \angle DEC$ (corresponding angles 2) 4. **Third angle equality:** The third angles $\angle ACB$ and $\angle DCE$ are equal because the sum of angles in a triangle is $180^\circ$. 5. **Conclusion:** Since two pairs of corresponding angles are equal, by AA similarity criterion, $$\triangle ABC \sim \triangle DEC$$ This completes the proof.