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