In a quadrilateral ABCD, the diagonals AC, BD, intersect at right angles. Prove that:
$A B^{2} + C D^{2} = B C^{2} + D A^{2}$

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Solution

In $△ A O B$ ,
$A B^{2} = O A^{2} + O B^{2}$ ......(1)

In $△ C O D$ ,
$C D^{2} = O C^{2} + O D^{2}$ .....(2)

In $△ O A D$ ,
$D A^{2} = O A^{2} + O D^{2}$ ......(3)

In $△ B O C$ ,
$B C^{2} = O C^{2} + O B^{2}$ ......(4)

Adding 1 & 2, we get,
$A B^{2} + C D^{2} = O A^{2} + O B^{2} + O C^{2} + O D^{2}$ ....(5)

Adding 3 & 4 , we get,
$D A^{2} + B C^{2} = O A^{2} + O B^{2} + O C^{2} + O D^{2}$ ......(6)

Form 5 & 6, we get,
$A B^{2} + C D^{2} = B C^{2} + D A^{2}$

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Similar questions

The diagonals AC and BC of a quadrilateral ABCD intersect at O, at right angles. Prove that AB²+CD²=AD²+BC².

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

{AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} + 4 {PQ}^{2}

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B D

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