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 of parallelograms:** Opposite sides are equal and parallel, so $AB \cong CD$ and $AB \parallel CD$, $AD \cong BC$ and $AD \parallel BC$. 3. **Consider triangles ABE and CDE:** We want to prove these triangles are congruent to show $AE \cong CE$ and $BE \cong DE$. 4. **Identify corresponding parts:** - $AB \cong CD$ (opposite sides of parallelogram) - $\angle ABE \cong \angle CDE$ (alternate interior angles because $AB \parallel CD$ and $BD$ is a transversal) - $BE$ is common side to both triangles. 5. **Apply Side-Angle-Side (SAS) congruence:** Triangles $ABE$ and $CDE$ have two sides and the included angle congruent, so $\triangle ABE \cong \triangle CDE$. 6. **Conclude equal segments:** By CPCTC (Corresponding Parts of Congruent Triangles are Congruent), $AE \cong CE$ and $BE \cong DE$. **Final answer:** The diagonals of a parallelogram bisect each other, so $AE = CE$ and $BE = DE$.