Let ABCD be a square with sides of unit lenght. Points E and F are taken on sides AB and AD, respectively,so that `AE=AF`. Let P be a point inside the squre ABCD.
The value of `(PA)^2-(PB)^2+(PC)^2-(PD)^2` is equal to

Options

A. 3
B. 2
C. 1
D. 0

Explanation

To solve the problem, we will follow these steps:

1. **Define the Coordinates of the Square**:
Let the square ABCD have vertices at the following coordinates:
- A(0, 0)
- B(1, 0)
- C(1, 1)
- D(0, 1)

2. **Define the Points E and F**:
Since points E and F are on sides AB and AD respectively, and AE = AF, we can denote the length of AE and AF as \( x \). Therefore:
- E(0, x)
- F(x, 0)

3. **Define the Point P Inside the Square**:
Let the coordinates of point P be (p_x, p_y), where \( 0 < p_x < 1 \) and \( 0 < p_y < 1 \).

4. **Calculate the Distances PA, PB, PC, and PD**:
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
- \( PA = \sqrt{(p_x - 0)^2 + (p_y - 0)^2} = \sqrt{p_x^2 + p_y^2} \)
- \( PB = \sqrt{(p_x - 1)^2 + (p_y - 0)^2} = \sqrt{(p_x - 1)^2 + p_y^2} \)
- \( PC = \sqrt{(p_x - 1)^2 + (p_y - 1)^2} = \sqrt{(p_x - 1)^2 + (p_y - 1)^2} \)
- \( PD = \sqrt{(p_x - 0)^2 + (p_y - 1)^2} = \sqrt{p_x^2 + (p_y - 1)^2} \)

5. **Calculate \( (PA)^2 - (PB)^2 + (PC)^2 - (PD)^2 \)**:
Now we will compute each term:
- \( PA^2 = p_x^2 + p_y^2 \)
- \( PB^2 = (p_x - 1)^2 + p_y^2 = p_x^2 - 2p_x + 1 + p_y^2 \)
- \( PC^2 = (p_x - 1)^2 + (p_y - 1)^2 = p_x^2 - 2p_x + 1 + p_y^2 - 2p_y + 1 \)
- \( PD^2 = p_x^2 + (p_y - 1)^2 = p_x^2 + p_y^2 - 2p_y + 1 \)

Now substituting these into our expression:
\[ (PA)^2 - (PB)^2 + (PC)^2 - (PD)^2 = (p_x^2 + p_y^2) - (p_x^2 - 2p_x + 1 + p_y^2) + (p_x^2 - 2p_x + 1 + p_y^2 - 2p_y + 1) - (p_x^2 + p_y^2 - 2p_y + 1) \]

Simplifying this:
\[ = p_x^2 + p_y^2 - (p_x^2 - 2p_x + 1 + p_y^2) + (p_x^2 - 2p_x + 1 + p_y^2 - 2p_y + 1) - (p_x^2 + p_y^2 - 2p_y + 1) \]
\[ = p_x^2 + p_y^2 - p_x^2 + 2p_x - 1 - p_y^2 + p_x^2 - 2p_x + 1 + p_y^2 - 2p_y + 1 - p_x^2 - p_y^2 + 2p_y - 1 \]
\[ = 0 \]

6. **Conclusion**:
Thus, the value of \( (PA)^2 - (PB)^2 + (PC)^2 - (PD)^2 \) is equal to 0.

### Final Answer:
The value is \( 0 \).