Question

Inside a rectangle ABCD, O is any point then Prove that OB² + OD² = OA² + OC².

Answer 1

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Using Pythagorean theorem we have from the above diagram:

OA² = AH² + OH² = AH² + AE²

OC² = CG² + OG² = EB² + HD²

OB² = EO² + BE² = AH² + BE²

OD² = HD² + OH² = HD² + AE²

Adding these equalities, we get

OA² + OC² = AH² + HD² + AE² + EB²

OB² + OD² = AH² + HD² + AE² + EB²

From which we prove that for any point within the rectangle there is the relation

OA² + OC² = OB² + OD²

Hence Proved.