ABCD is a square, A is joined to a point P on BC and D is joined to a point Q on AB. If AP=DQ and AP intersects DQ at R the, `angle DRP` is :

Options

A. `60^@`
B. `120^@`
C. `90^@`
D. can't be determined

Answer

To solve the problem, we will follow these steps:

1. **Draw the Square and Points**:
- Draw square ABCD.
- Mark point P on side BC and point Q on side AB.

2. **Identify Given Information**:
- We know that \( AP = DQ \) and that lines \( AP \) and \( DQ \) intersect at point R.

3. **Analyze Triangles**:
- Consider triangles \( ADQ \) and \( BAP \).
- Since ABCD is a square, we have \( AD = AB \) (both are sides of the square).

4. **Use Congruence Criteria**:
- In triangles \( ADQ \) and \( BAP \):
- \( AD = AB \) (sides of the square).
- \( DQ = AP \) (given).
- \( \angle DAB = \angle PAB = 90^\circ \) (angles in a square).
- By the RHS (Right angle-Hypotenuse-Side) criterion, triangles \( ADQ \) and \( BAP \) are congruent.

5. **Conclude Angles**:
- Since the triangles are congruent, we have:
- \( \angle DQA = \angle PAB \).
- Also, since \( \angle DAB = 90^\circ \), we can express:
- \( \angle BAD = \angle DAP + \angle PAB \).
- Thus, \( \angle DAP + \angle PAB = 90^\circ \).

6. **Analyze Triangle ARD**:
- In triangle \( ARD \):
- The sum of angles in a triangle is \( 180^\circ \).
- \( \angle ARD + \angle ADR + \angle DRA = 180^\circ \).
- We know \( \angle ADR + \angle DRA = 90^\circ \) (from previous steps).
- Therefore, \( \angle ARD = 180^\circ - 90^\circ = 90^\circ \).

7. **Conclusion**:
- Since \( AP \) is perpendicular to \( DQ \), we conclude that \( \angle DRP = 90^\circ \).

**Final Answer**: \( \angle DRP = 90^\circ \).

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