Given: $\overline{A B} \| \overline{C D}$ and $\overline{B D}$ bisects $\overline{A C}$. Prove: $\triangle A B E \cong \triangle C D E$.

Step 1 ：Given that line segment AB is parallel to line segment CD and that line segment BD bisects line segment AC.

Step 2 ：We deduce that the alternate interior angles ABD and CBD are equal because alternate interior angles are equal when a transversal intersects two parallel lines.

Step 3 ：We also know that angle ABE is equal to angle CDE because they are vertical angles and vertical angles are always equal.

Step 4 ：Finally, we know that line segment BE is equal to itself.

Step 5 ：By the Angle-Angle-Side (AAS) congruence theorem, we can conclude that triangle ABE is congruent to triangle CDE.

Step 6 ：\(\boxed{\triangle A B E \cong \triangle C D E}\)