To prove that triangles $\triangle ABD$ and $\triangle CDB$ are congruent, we can use the ASA (Angle-Side-Angle) congruence criterion. Here's the detailed proof:

---

Given:

1. Segment $AB \parallel CD$.
2. Segment $AD \parallel BC$.

---

Proof:

Step 1: Identify corresponding angles

• Since $AB \parallel CD$ and $AD$ acts as a transversal, the alternate interior angles are equal: $\angle DAB = \angle BCD$
• Similarly, since $AD \parallel BC$ and $AB$ acts as a transversal, the alternate interior angles are equal: $\angle ADB = \angle CBD$

Step 2: Identify a common side

• The segment $BD$ is a common side of $\triangle ABD$ and $\triangle CDB$.

Step 3: Use ASA congruence criterion

• In $\triangle ABD$ and $\triangle CDB$:
  1. $\angle DAB = \angle BCD$ (alternate interior angles due to $AB \parallel CD$),
  2. $\angle ADB = \angle CBD$ (alternate interior angles due to $AD \parallel BC$),
  3. $BD = BD$ (common side).

By the ASA criterion, $\triangle ABD \cong \triangle CDB$.

---

Conclusion:

The triangles $\triangle ABD$ and $\triangle CDB$ are congruent by the ASA (Angle-Side-Angle) congruence postulate.

---

Would you like me to explain any step further or provide a diagram for better understanding?
Here are 5 questions to deepen understanding:

1. How do parallel lines contribute to congruence proofs?
2. Can other congruence criteria (e.g., SAS or SSS) be used to prove $\triangle ABD \cong \triangle CDB$?
3. What is the role of transversal lines in identifying congruent angles?
4. How would the proof change if $AB$ and $CD$ were not parallel?
5. What are practical applications of proving triangle congruence?

Tip: Always start with identifying corresponding angles or sides when using congruence criteria. It simplifies the proof structure