Question

A point OI in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D. Prove that  OB² + OD² = OC² + OA²

Answer 1

Let ABCD be the given rectangle and let O be a point within it.
Join OA, OB, OC and OD.

Through O, draw EOF || AB. Then, ABFE is a rectangle.
In right triangles ΔOEA and ΔOFC, we have
OA² = OE² + AE² and OC² = OF² + CF²
⇒ OA² + OC² = (OE² + AE²) + (OF² + CF²)
⇒ OA² + OC² = OE² + OF² + AE² + CF²   ......(i)
Now, in right triangles OFB and ODE, we have
OB² = OF² + FB² and OD² = OE² + DE²
⇒ OB² + OD² = (OF² + FB²) + (OE² + DE²)
⇒ OB² + OD2 = OE² + OF² + DE² + BF²
⇒ OB² + OD² = OE² + OF² + CF² + AE²  [∵ DE = CF and AE = BF]....(ii)
From (i) and (ii), we get
OA² + OC² = OB² + OD².