1. **State the problem:** Prove that the opposite angles of parallelogram ABCD are congruent, specifically that $\angle B \cong \angle D$. 2. **Given:** $\overline{AB} \parallel \overline{CD}$ and $\overline{AD} \parallel \overline{BC}$. 3. **Use properties of parallel lines:** When a transversal crosses parallel lines, alternate interior angles are congruent. 4. **Identify congruent angles:** - $\angle 2 \cong \angle 4$ because they are corresponding angles formed by transversal $AC$ crossing $\overline{AB} \parallel \overline{CD}$. - $\angle 1 \cong \angle 3$ because they are alternate interior angles formed by transversal $AC$ crossing $\overline{AD} \parallel \overline{BC}$. 5. **Triangle congruence:** Triangles $ABC$ and $CDA$ share side $AC$ and have two pairs of congruent angles ($\angle 2 \cong \angle 4$ and $\angle 1 \cong \angle 3$), so by Angle-Side-Angle (ASA) congruence, $\triangle ABC \cong \triangle CDA$. 6. **Conclusion:** Corresponding parts of congruent triangles are congruent (CPCTC), so $\angle B \cong \angle D$. This completes the proof that opposite angles of parallelogram ABCD are congruent. A similar proof applies for the other pair of opposite angles using diagonal $BD$ instead of $AC$.