Given: \( D \) is the midpoint of \( \overline{A C}, \overline{B A} \cong \overline{B C} \) and \( \angle E D A \cong \angle F D C \). Prove: \( \triangle A E D \cong \triangle C F D \). Note: quadrilateral properties are not permitted in this proof.

To prove \( \triangle A E D \cong \triangle C F D \), we will utilize the given information and the Side-Angle-Side (SAS) congruence criterion. 1. **Identify the Components**: - Since \( D \) is the midpoint of \( \overline{A C} \), we have \( \overline{AD} \cong \overline{DC} \). Let's denote this length as \( x \). - The given \( \overline{BA} \cong \overline{BC} \) means that we have a common side \( \overline{BA} \) and, as a result, \( B \) is an equal distance from points \( A \) and \( C \). - We also have \( \angle EDA \cong \angle FDC \) as given. 2. **Use Congruent Segments and Angles**: - So far, we have \( \overline{AD} \cong \overline{DC} \) (both equal to \( x \)), and \( \overline{BA} \cong \overline{BC} \) (let's denote this length as \( y \)). - The angles \( \angle EDA \) and \( \angle FDC \) are congruent by the problem statement. 3. **Application of the SAS Congruence Criterion**: - According to SAS, for \( \triangle AED \) and \( \triangle CFD \) to be congruent, we need two sides and the included angle of one triangle to be equal to those of the other triangle: - Side \( \overline{AD} \cong \overline{DC} \) (both equal to \( x \)), - Angle \( \angle EDA \cong \angle FDC \), - Side \( \overline{AE} \) (an arbitrary segment) should equal \( \overline{CF} \) (since they serve as the corresponding sides opposite the equal angles from point \( B \)). 4. **Conclusion**: - With these components established, we have \( \overline{AD} \cong \overline{DC} \), \( \angle EDA \cong \angle FDC \), and \( \overline{AE} \cong \overline{CF} \). Hence we conclude that: \[ \triangle A E D \cong \triangle C F D \] This is based on the SAS criterion confirming the triangles are congruent through our established equalities!