If O is any point inside a rectangle ABCD, then prove that OA² + OC² = OB² + OD²

From the point O construct perpendiculars OP, OQ, OR and OS to the sides of the rectangle (see figure).

   

LHS = OA² + OC²
= (AS² + OS²) + (OQ² + QC²)  (considering right triangles ASO and COQ)
But AS = BQ and OQ = BP. Therefore,
(AS² + OS²) + (OQ² + QC²)
= (BQ² + OS²) + (OQ² + DS²)
= (BQ² + OQ²) + (OS² + DS²)  
=  OB² + OD²      (considering right triangles BOQ and ODS)
= RHS
Hence Proved