To prove that triangles \( \triangle DBC \) and \( \triangle APQ \) are congruent, we need to use the properties of parallelograms and some fundamental concepts of triangle congruence. Here's a step-by-step approach:

### 1. Understanding the Setup
- \( ABCD \) is a parallelogram.
- \( P \) and \( Q \) are points on the sides \( AD \) and \( BC \), respectively, with lines \( PQ \) drawn inside the parallelogram.

### 2. Properties of Parallelogram
- Opposite sides of a parallelogram are equal and parallel:
  - \( AB \parallel CD \) and \( AB = CD \).
  - \( AD \parallel BC \) and \( AD = BC \).
- Diagonals of a parallelogram bisect each other.

### 3. Analyzing Triangles \( \triangle DBC \) and \( \triangle APQ \)
To prove that \( \triangle DBC \cong \triangle APQ \), we can use the properties of corresponding parts of a parallelogram and parallel lines.

### 4. Using Corresponding Angles and Sides
- Since \( AB \parallel CD \) and \( PQ \) is a transversal, we have:
  - \( \angle APQ \) and \( \angle DBC \) are corresponding angles. Thus, \( \angle APQ = \angle DBC \).

- Similarly, because \( AD \parallel BC \) and considering the lines intersected by the transversal \( PQ \):
  - \( \angle PAQ \) and \( \angle BDC \) are corresponding angles. Hence, \( \angle PAQ = \angle BDC \).

### 5. Equality of Sides
- \( PQ \parallel AB \) and \( PQ \parallel CD \), indicating that:
  - The segment \( PQ \) corresponds in length to both segments \( AB \) and \( CD \).
- Since \( AD \parallel BC \) and \( AP \) and \( BD \) are within the same structure, \( AP = BD \) by the property of the parallelogram.

### 6. Proving Congruence by ASA (Angle-Side-Angle)
Using the ASA (Angle-Side-Angle) criterion for triangle congruence:

- In \( \triangle APQ \) and \( \triangle DBC \):
  - \( \angle APQ = \angle DBC \) (corresponding angles due to parallel lines).
  - \( PQ = BC \) (segment parallel and equal to each other).
  - \( \angle PAQ = \angle BDC \) (corresponding angles).

Since we have two angles and the included side of one triangle equal to the corresponding two angles and the included side of the other triangle:

\[
\triangle DBC \cong \triangle APQ \quad \text{(by ASA criterion)}
\]

### Conclusion
Triangles \( \triangle DBC \) and \( \triangle APQ \) are congruent by the Angle-Side-Angle (ASA) postulate.

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Would you like more details or have any questions?

### Related Questions:
1. What is the definition and properties of a parallelogram?
2. How does the ASA (Angle-Side-Angle) criterion work for proving triangle congruence?
3. What other methods can be used to prove triangles congruent, besides ASA?
4. How can corresponding angles be used to prove other geometric properties?
5. What are the properties of transversals when intersecting parallel lines?

### Tip:
When proving triangles congruent in geometric figures, always check if you can use parallel lines and transversals to identify corresponding angles, as these often provide a straightforward way to establish triangle congruence.

Learn how to prove triangles DBC and APQ congruent using parallelogram properties and the Angle-Side-Angle (ASA) criterion. This problem involves analyzing the relationship between triangles inside a parallelogram, suitable for high school students studying geometry.

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