Question

o is any point inside a rectangle ABCD. prove that OB^2 + OD^2 = OA^2 + OC^2

Answer 1

we draw PQ ║AB ║CD as shown in figure,
ABCD is a rectangle , it means ABPQ and PQDC are also rectangle
For , ABPQ ,
AP = BQ [opposite sides are equal ]

For, PQDC
PD = QC [ opposite sides are equal ]

now, for ∆OPD,
OD² = OP² + PD² ------(1)

For, ∆OQB,
OB² = OQ² + BQ² --------(2)

add equations (1) and (2),
OB² + OD² = (OP² + PD²)+ (OQ² + BQ²)
= (OP² + CQ²) + (OQ² + AP²)

as you can see figure,
∆OPA and ∆OQC are also right angled triangles
For, ∆OCQ ⇒OQ² + CQ² = OC²
For,∆OPA ⇒OP² + AP² = OA² , put it above

Now ,
OB² + OD² = OC² + OA²
Hence, proved //

Answer 2

FIGURE IS IN THE ATTACHMENT

Let ABCD be the given rectangle with point O within it. Join OA, OB, OC, OD. Through O draw EOF||AB.Then ABFE is a rectangle.
In right ∆OEA and ∆OFC,
OA² = OE² + AE² and OC²= OF²+CF²
OA²+OC²=(OE²+AE²) +(OF²+CF²)
OA²+OC²=OE²+OF²+AE²+CF²………….(1)
In right ∆OFB and ∆ODE,
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…………..(2)
[DE=CF & AE=BF]
From eq i & eq ii,
OA² + OC² = OB² + OD²

HOPE THIS WILL HELP YOU.