P is a point on the side BC off the `DeltaABC` and Q is a point such that PQ is the resultant of AP,PB and PC. Then, ABQC is a

Options

A. square
B. rectangle
C. parallelogram
D. trapezium

Explanation

To solve the problem, we need to analyze the given information about the triangle \( \Delta ABC \) and the points \( P \) and \( Q \).

### Step-by-Step Solution:

1. **Identify the Points and Vectors**:
- Let \( A, B, C \) be the vertices of the triangle.
- Let \( P \) be a point on side \( BC \).
- Let \( Q \) be a point such that \( \vec{PQ} \) is the resultant of \( \vec{AP} \), \( \vec{PB} \), and \( \vec{PC} \).

2. **Express the Resultant Vector**:
- By the definition of the resultant vector, we have:
\[ \vec{PQ} = \vec{AP} + \vec{PB} + \vec{PC} \]

3. **Rearranging the Equation**:
- Rearranging gives us:
\[ \vec{PQ} - \vec{PC} = \vec{AP} + \vec{PB} \]

4. **Using Vector Notation**:
- Let’s denote \( \vec{AP} = \vec{a} \), \( \vec{PB} = \vec{b} \), and \( \vec{PC} = \vec{c} \).
- The equation becomes:
\[ \vec{PQ} - \vec{c} = \vec{a} + \vec{b} \]

5. **Substituting and Simplifying**:
- Rearranging gives:
\[ \vec{PQ} = \vec{a} + \vec{b} + \vec{c} \]

6. **Understanding the Geometry**:
- Since \( P \) is on side \( BC \), we can express \( \vec{PQ} \) in terms of \( \vec{AB} \) and \( \vec{AC} \).
- We can say:
\[ \vec{AB} + \vec{PB} + \vec{PC} = \vec{PQ} \]

7. **Conclusion**:
- From the vector relationships, we can conclude that the quadrilateral \( ABQC \) forms a parallelogram because the opposite sides are equal in vector terms.

### Final Answer:
Thus, the quadrilateral \( ABQC \) is a **parallelogram**.