ABDC is a parallelogram and P is the point of intersection of its diagonals. If O is any point, then `overline(OA)+overline(OB)+overline(OC)+overline(OD)=`

Options

A. `overline(OP)`
B. `4overline(OP)`
C. `overline(O)`
D. `2overline(OP)`

Answer

To solve the problem, we need to find the expression \( \overline{OA} + \overline{OB} + \overline{OC} + \overline{OD} \) where \( ABCD \) is a parallelogram and \( P \) is the point of intersection of its diagonals.

### Step-by-Step Solution:

1. **Understanding the Parallelogram**:
- In a parallelogram, the diagonals bisect each other. Let \( A, B, C, D \) be the vertices of the parallelogram such that \( A \) and \( C \) are opposite vertices, and \( B \) and \( D \) are the other pair of opposite vertices.

2. **Positioning the Points**:
- Let \( O \) be any arbitrary point in the plane. We will express the vectors from point \( O \) to the vertices \( A, B, C, D \).

3. **Using the Properties of Vectors**:
- Since \( P \) is the midpoint of both diagonals \( AC \) and \( BD \), we can express the vectors as:
\[ \overline{OA} + \overline{OC} = 2\overline{OP} \quad \text{(1)} \]
\[ \overline{OB} + \overline{OD} = 2\overline{OP} \quad \text{(2)} \]

4. **Adding the Vectors**:
- Now, we can add equations (1) and (2):
\[ (\overline{OA} + \overline{OC}) + (\overline{OB} + \overline{OD}) = 2\overline{OP} + 2\overline{OP} \]
\[ \overline{OA} + \overline{OB} + \overline{OC} + \overline{OD} = 4\overline{OP} \]

5. **Conclusion**:
- Therefore, we conclude that:
\[ \overline{OA} + \overline{OB} + \overline{OC} + \overline{OD} = 4\overline{OP} \]

### Final Answer:
\[ \overline{OA} + \overline{OB} + \overline{OC} + \overline{OD} = 4\overline{OP} \]