1. **State the problem:** Given parallelogram ABCD with diagonals AC and BD intersecting at E, prove that $AE \cong CE$ and $BE \cong DE$. 2. **Recall properties:** In a parallelogram, opposite sides are parallel and equal in length. The diagonals bisect each other. 3. **Statements and reasons:** - Statement 1: ABCD is a parallelogram. - Reason 1: Given. - Statement 2: $AB \parallel CD$ and $AD \parallel BC$. - Reason 2: Definition of parallelogram. - Statement 3: $\angle ABE \cong \angle CDE$ and $\angle BAE \cong \angle DCE$. - Reason 3: Alternate interior angles theorem (since $AB \parallel CD$ and $AD \parallel BC$). - Statement 4: Triangles $\triangle ABE$ and $\triangle CDE$. - Reason 4: Consider these triangles formed by diagonals. - Statement 5: $AB = CD$. - Reason 5: Opposite sides of parallelogram are equal. - Statement 6: $BE = DE$ (to be proven). 4. **Use ASA congruence:** - In $\triangle ABE$ and $\triangle CDE$: - $AB = CD$ (opposite sides of parallelogram) - $\angle ABE = \angle CDE$ (alternate interior angles) - $\angle BAE = \angle DCE$ (alternate interior angles) Therefore, by ASA (Angle-Side-Angle) criterion, $\triangle ABE \cong \triangle CDE$. 5. **Conclude equal segments:** - Corresponding parts of congruent triangles are equal, so $AE = CE$ and $BE = DE$. 6. **Summary:** - The diagonals of a parallelogram bisect each other, so the segments $AE$ and $CE$ are congruent, and $BE$ and $DE$ are congruent. This completes the proof of the Parallelogram Diagonal Theorem.