If P is a point within a rectangle ABCD, then:

Options

A. `AP^2+PC^2=BP^2+PD^2`
B. `AP^2+BP^2=PC^2+PD^2`
C. AP+PC=BP+PD
D. `APxxPC=BPxxPD`

Answer

To solve the problem, we need to establish the relationship between the distances from point P to the vertices of rectangle ABCD. Let's denote the distances as follows:

- AP = a
- BP = b
- CP = c
- DP = d

We will analyze the options given in the question.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a rectangle ABCD with vertices A, B, C, and D.
- Point P is located inside the rectangle.

2. **Using the Distance Formula**:
- The distances from point P to the vertices can be represented as:
- AP = distance from P to A
- BP = distance from P to B
- CP = distance from P to C
- DP = distance from P to D

3. **Analyzing the Options**:
- **Option 1**: \( AP^2 + PC^2 = BP^2 + PD^2 \)
- **Option 2**: \( AP^2 + BP^2 = PC^2 + PD^2 \)
- **Option 3**: \( AP + PC = BP + PD \)
- **Option 4**: \( AP \cdot PC = BP \cdot PD \)

4. **Verifying Option 1**:
- According to the properties of rectangles, we can use the Pythagorean theorem to establish relationships between these distances.
- By applying the theorem, we can derive that:
\[ AP^2 + PC^2 = BP^2 + PD^2 \]
- This holds true as we can visualize that the sum of the squares of the distances from point P to opposite corners of the rectangle are equal.

5. **Conclusion**:
- Therefore, the correct relationship is given by **Option 1**:
\[ AP^2 + PC^2 = BP^2 + PD^2 \]