prove that

Draw EF parallel to AB passing through O.

In right triangles OEA, OFC

OA² = OE² + AE² and

 OC² = OF²+ CF²

Adding

OA²+OC² = OE² +AE²+ OF²+CF²……..(A)

In right triangles OFB & ODE

OB² =OF² +FB².

OD² =OE² + DE².

Adding the above equations

OB²+OD²=OF²+FB² + OE²+DE^{2  ……….}  (B)

From the figure DE =CF; AE = BF

We can say both the LHS can be equated.

OA² + OC^{2 =} OB² + OD²