Question

If O is any point in the interior of a rectangle, ABCD, prove that : OA² + OC² = OB² + OD².

Answer 1

Given: ABCD is a rectangle and O is any point in its interior.

To Prove: OA² + OC² = OB² + OD²

Proof [Step-wise]:

1. Place a coordinate system so that A = (0, 0), B = (1, 0), C = (l, w), D = (0, w) for positive 1, w. 

This is always possible for a rectangle.

2. Let O = (x, y) with 0 < x < 1 and 0 < y < w.

3. Compute squared distances from O to the vertices:

OA² = x² + y²

OB² = (1 – x)² + y²

OC² = (1 – x)² + (w – y)²

OD² = x² + (w – y)²

4. Form the sums:

OA² + OC²

= [x² + y²] + [(1 – x)² + (w – y)²]

= x² + (1 – x)² + y² + (w – y)²

OB² + OD²

= [(1 – x)² + y²] + [x² + (w – y)²]

= (1 – x)² + x² + y² + (w – y)²

5. The two expressions are identical.

So, OA² + OC² = OB² + OD².