1. **State the problem:** Prove that the opposite angles of parallelogram ABCD are congruent, specifically that $\angle B \cong \angle D$.\n\n2. **Given:** $AB \parallel CD$ and $AD \parallel BC$.\n\n3. **Use properties of parallel lines:** When a transversal crosses parallel lines, alternate interior angles and corresponding angles are congruent.\n\n4. **Identify angles:**\n- $\angle 2 \cong \angle 4$ because they are corresponding angles formed by transversal $AC$ crossing $AB \parallel CD$.\n- $\angle 1 \cong \angle 3$ because they are alternate interior angles formed by transversal $AC$ crossing $AD \parallel BC$.\n\n5. **Triangle congruence:** Triangles $\triangle ABC$ and $\triangle CDA$ share side $AC$, and have two pairs of congruent angles ($\angle 2 \cong \angle 4$ and $\angle 1 \cong \angle 3$). By the Angle-Side-Angle (ASA) postulate, $\triangle ABC \cong \triangle CDA$.\n\n6. **Corresponding parts of congruent triangles are congruent (CPCTC):** Therefore, $\angle B \cong \angle D$.\n\nThis completes the proof that opposite angles of parallelogram ABCD are congruent.