O is any point inside a rectangle ABCD, then ${O B}^{2} + {O D}^{2} = {O A}^{2} + {O C}^{2}$

True

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False

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Solution

The correct option is A True
Through O, draw $P Q ∥ B C$ so that P lies on AB and Q lies on DC.
Now, $P Q ∥ B C$
Therefore,
$P Q ⊥ A B$ and $P Q ⊥ D C [∠ B = 90^{o}$ and $∠ C = 90^{o}$ ]
Therefore, BPQC and APQD are both rectangles.
Now, from $Δ O P B$ ,
${O B}^{2} = {B P}^{2} + {O P}^{2}$ .........(i)
Similarly, from $Δ O D Q$
${O D}^{2} = {O Q}^{2} + {D Q}^{2}$ .........(ii)
From $Δ O Q C$ , we have
${O C}^{2} = {O Q}^{2} + {C Q}^{2}$ ..........(iii)
And from $Δ O A P$ , we have
${O A}^{2} = {A P}^{2} + {O P}^{2}$ ........(iv)
Adding (i) and (ii)
${O B}^{2} + {O D}^{2} = {B P}^{2} + {O P}^{2} + {O Q}^{2} + {D Q}^{2}$
$= {C Q}^{2} + {O P}^{2} + {O Q}^{2} + {A P}^{2}$
$[A s B P = C Q a n d D Q = A P]$
$= {C Q}^{2} + {O Q}^{2} + {O P}^{2} + {A P}^{2}$
$= {O C}^{2} + {O A}^{2} [F r o m (i i i) a n d (i v)]$ .
Hence Proved.

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O is any point inside a rectangle ABCD, then OB2+OD2=OA2+OC2

O is any point inside a rectangle ABCD, then ${O B}^{2} + {O D}^{2} = {O A}^{2} + {O C}^{2}$