1. The problem states that quadrilateral $ABCD$ has diagonals $AC$ and $BD$ that bisect each other at point $E$. 2. We want to prove that $\overline{AB} \parallel \overline{CD}$ by establishing the congruence of a single pair of triangles. 3. Since $E$ is the midpoint of both diagonals, we have $\overline{AE} \cong \overline{CE}$ and $\overline{BE} \cong \overline{DE}$. 4. Consider triangles $ABE$ and $CDE$. 5. They share the angle at $E$ formed by the intersecting diagonals, so $\angle AEB \cong \angle CED$ (vertical angles are congruent). 6. We have two pairs of sides congruent: $\overline{AE} \cong \overline{CE}$ and $\overline{BE} \cong \overline{DE}$. 7. Using the Side-Angle-Side (SAS) criterion, triangles $ABE$ and $CDE$ are congruent. 8. From the congruence of these triangles, corresponding parts are congruent, so $\overline{AB} \cong \overline{CD}$ and the lines $\overline{AB}$ and $\overline{CD}$ are parallel. 9. Therefore, Craig is referring to triangles $ABE$ and $CDE$ and should use the Side-Angle-Side (SAS) criterion. Final answer: Option D: $\triangle ABE$ and $\triangle CDE$ by side-angle-side.