1.

O is any point inside a rectangle ABCD. Prove that `O B^2+O D^2=O A^2+O C^2`.

Answer» If we draw a line `EF` parallel to `AB` in the given figure,
it will divide rectangle `ABCD` into two rectangles `ABFE` and `CDEF`.
Then, `AE = BF` and `CF = DE->(1)`
Now, in `Delta OEA`,
`OA^2 = OE^2+AE^2->(2)`
In `Delta OCF`,
`OC^2 = OF^2+CF^2->(3)`
In `Delta OBF`,
`OB^2 = OF^2+BF^2->(4)`
In `Delta OED`,
`OD^2 = OE^2+DE^2->(5)`
Now, adding (2) and (3),
`OA^2+OC^2 = OE^2+AE^2+OF^2+CF^2`
From(1), `AE = BF` and `CF = DE`
So,
`OA^2+OC^2 = (OE^2+DE^2)+(OF^2+BF^2)`
`OA^2+OC^2 = OB^2+OD^2`


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