Let `A B C D` be a rectangle and `P` be any point in its plane. Show that `A P^2+P C^2=P B^2+P D^2dot`

Explanation

To prove that \( AP^2 + PC^2 = PB^2 + PD^2 \) for a rectangle \( ABCD \) and a point \( P \) in its plane, we will use the distance formula.

### Step-by-Step Solution:

1. **Assign Coordinates**:
- Let the coordinates of the rectangle \( ABCD \) be:
- \( A(0, 0) \)
- \( B(a, 0) \)
- \( C(a, b) \)
- \( D(0, b) \)
- Let the coordinates of point \( P \) be \( P(h, k) \).

2. **Calculate Distances**:
- Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), we calculate the distances:
- \( AP^2 = (h - 0)^2 + (k - 0)^2 = h^2 + k^2 \)
- \( PC^2 = (h - a)^2 + (k - b)^2 = (h - a)^2 + (k - b)^2 \)
- \( PB^2 = (h - a)^2 + (k - 0)^2 = (h - a)^2 + k^2 \)
- \( PD^2 = (h - 0)^2 + (k - b)^2 = h^2 + (k - b)^2 \)

3. **Add Distances**:
- Now we add \( AP^2 \) and \( PC^2 \):
\[ AP^2 + PC^2 = (h^2 + k^2) + ((h - a)^2 + (k - b)^2) \]
Expanding \( PC^2 \):
\[ = h^2 + k^2 + (h^2 - 2ah + a^2 + k^2 - 2bk + b^2) \]
\[ = 2h^2 + 2k^2 - 2ah - 2bk + a^2 + b^2 \]

4. **Calculate \( PB^2 + PD^2 \)**:
- Now we add \( PB^2 \) and \( PD^2 \):
\[ PB^2 + PD^2 = ((h - a)^2 + k^2) + (h^2 + (k - b)^2) \]
Expanding both terms:
\[ = (h^2 - 2ah + a^2 + k^2) + (h^2 + k^2 - 2bk + b^2) \]
\[ = 2h^2 + 2k^2 - 2ah - 2bk + a^2 + b^2 \]

5. **Conclusion**:
- Since both expressions for \( AP^2 + PC^2 \) and \( PB^2 + PD^2 \) are equal:
\[ AP^2 + PC^2 = PB^2 + PD^2 \]
- Therefore, we have proved that \( AP^2 + PC^2 = PB^2 + PD^2 \).