The problem gives us that $\angle B \cong \angle D$ and $\overline{AD} \parallel \overline{BC}$. We want to prove that $\triangle ABC \cong \triangle CDA$. We cannot use quadrilateral properties in this proof.

1. Problem Description

The problem gives us that $\angle B \cong \angle D$ and $\overline{AD} \parallel \overline{BC}$. We want to prove that $\triangle ABC \cong \triangle CDA$. We cannot use quadrilateral properties in this proof.

2. Solution Steps

Step 1: $\angle B \cong \angle D$ and $\overline{AD} \parallel \overline{BC}$. Reason: Given.

Step 2: $\angle DAC \cong \angle BCA$. Reason: If two parallel lines are cut by a transversal, then alternate interior angles are congruent. Here, $\overline{AD} \parallel \overline{BC}$ and $\overline{AC}$ is the transversal.

Step 3: $\overline{AC} \cong \overline{AC}$. Reason: Reflexive Property.

Step 4: $\triangle ABC \cong \triangle CDA$. Reason: Angle-Angle-Side (AAS) Congruence Theorem. We have $\angle B \cong \angle D$, $\angle BCA \cong \angle DAC$ and $\overline{AC} \cong \overline{AC}$.

3. Final Answer

$\triangle ABC \cong \triangle CDA$