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 analyze the situation where P is a point inside rectangle ABCD. We will derive the relationship involving the distances from point P to the vertices of the rectangle.

### Step-by-Step Solution:

1. **Draw the Rectangle and Point P**:
- Start by drawing rectangle ABCD with vertices A, B, C, and D in clockwise order.
- Place point P inside the rectangle.

2. **Label the Distances**:
- Let the distances from point P to the vertices be:
- PA = distance from P to A
- PB = distance from P to B
- PC = distance from P to C
- PD = distance from P to D

3. **Identify the Right Triangles**:
- From point P, draw lines to each vertex of the rectangle. This creates four triangles: PAB, PBC, PCD, and PDA.
- Each of these triangles contains a right angle at the vertices of the rectangle.

4. **Apply Pythagorean Theorem**:
- For triangle PAB:
\[ PA^2 = PX^2 + PY^2 \quad (1) \]
- For triangle PBC:
\[ PB^2 = PY^2 + PC^2 \quad (2) \]
- For triangle PCD:
\[ PC^2 = PY^2 + PD^2 \quad (3) \]
- For triangle PDA:
\[ PD^2 = PX^2 + PA^2 \quad (4) \]

5. **Combine the Equations**:
- Now, we can add equations (1) and (3):
\[ PA^2 + PC^2 = PX^2 + PY^2 + PY^2 + PD^2 \]
- Similarly, add equations (2) and (4):
\[ PB^2 + PD^2 = PY^2 + PC^2 + PX^2 + PA^2 \]

6. **Rearranging**:
- Rearranging gives us:
\[ PA^2 + PD^2 = PB^2 + PC^2 \]

7. **Final Result**:
- Thus, we conclude that:
\[ PA^2 + PD^2 = PB^2 + PC^2 \]

### Conclusion:
The relationship derived shows that the sum of the squares of the distances from point P to two opposite corners of the rectangle is equal to the sum of the squares of the distances to the other two corners.