In a quadrilateral ABCD, if the diagonals AC, BD intersect at right angles, then

A B^{2} + B C^{2} = D C^{2} + D A^{2}

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A B^{2} + C D^{2} = B C^{2} + D A^{2}

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A B^{2} + A D^{2} = C B^{2} + C D^{2}

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A B^{2} + B C^{2} = 2 (D C^{2} + D A^{2})

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Solution

The correct option is B $A B^{2} + C D^{2} = B C^{2} + D A^{2}$

Here, $A B C D$ is a quadrilateral, diagonals $A C$ and $B D$ bisect each perpendicularly.
In right angles $△ A O B$
$\Rightarrow$ $O A^{2} + O B^{2} = A B^{2}$ ----( 1 ) [ By using Pythagoras theorem ]
In right angled $△ B O C,$
$\Rightarrow$ $O B^{2} + O C^{2} = B C^{2}$ ---- ( 2 ) [ By using Pythagoras theorem ]
In right angled $△ C O D,$
$\Rightarrow$ $O C^{2} + O D^{2} = C D^{2}$ ---- ( 3 ) [ By using Pythagoras theorem ]
In right angled $△ D O A,$
$\Rightarrow$ $O D^{2} + O A^{2} = D A^{2}$ ----- ( 4 ) [ By using Pythagoras theorem ]
Adding ( 1 ), ( 2 ), ( 3 ) and ( 4 ) we get,
$\Rightarrow$ $2 (O A^{2} + O C^{2} + O B^{2} + O D^{2}) = A B^{2} + B C^{2} + C D^{2} + D A^{2}$
$\Rightarrow$ $2 B C^{2} + 2 D A^{2} = A B^{2} + B C^{2} + C D^{2} + D A^{2}$
$\Rightarrow$ $B C^{2} + D A^{2} = A B^{2} + C D^{2}$
$∴$ $A B^{2} + C D^{2} = B C^{2} + D A^{2}$

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