1. **State the problem:** Given that ABCD is a parallelogram, CE bisects \(\angle DEB\), and \(\angle ADE \cong \angle ABE\), prove that \(\triangle DEC \cong \triangle BEC\). 2. **Given:** - ABCD is a parallelogram. - CE bisects \(\angle DEB\). - \(\angle ADE \cong \angle ABE\). 3. **Step 2:** \(\angle ADC \cong \angle ABC\) because opposite angles of a parallelogram are congruent. 4. **Step 3:** \(\angle EDC \cong \angle EBC\) since congruent angles subtracted from congruent angles remain congruent. 5. **Step 4:** \(\angle DEC \cong \angle BEC\) because CE bisects \(\angle DEB\), dividing it into two congruent angles. 6. **Step 5 (Missing Statement):** \(DC \cong BC\) because opposite sides of a parallelogram are congruent. 7. **Step 6:** \(\triangle DEC \cong \triangle BEC\) by the Angle-Angle-Side (AAS) congruence postulate, since we have two pairs of congruent angles and the included side \(EC\) common to both triangles. This completes the proof that \(\triangle DEC \cong \triangle BEC\).