The diagonal of quadrilateral $A B C D$ intersect at right angles at $E . A E = 10, B E = 8, C E = 6,$ and $D E = 4$ . The relation between $A B^{2} + C D^{2}$ and $A D^{2} + B C^{2}$ can be given by:

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

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

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

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Cannot be determined

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Solution

The correct option is B $(A B^{2} + C D^{2}) = (A D^{2} + B C^{2})$
Given, $A E = 10, B E = 8, C E = 6$
In $△ A B E$ , using pythagoras theorem
${A B}^{2} = {A E}^{2} + {B E}^{2}$
$= 10^{2} + 8^{2}$
$= 164$
Now, in $△ C D E$
${C D}^{2} = {C E}^{2} + {D E}^{2}$
$= 6^{2} + 4^{2}$
$= 52$
In $△ A D E$ , we have
${A D}^{2} = {A E}^{2} + {D E}^{2}$
$= 10^{2} + 4^{2}$
$= 116$
In $△ B C E$ , we have
${B C}^{2} = {B E}^{2} + {C E}^{2}$
$= 8^{2} + 6^{2}$
$= 100$
Now, ${A B}^{2} + {C D}^{2} = 164 + 52 = 216$
$\Rightarrow {A D}^{2} + {B C}^{2} = 116 + 100 = 216$
Therefore both quantities are equal.
So, correct option is $C$ .

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