In a rectangle $A B C D$ , prove that
${A C}^{2} + {B D}^{2} = {A B}^{2} + {B C}^{2} + {C D}^{2} + {D A}^{2}$

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Solution

To prove:
$A C^{2} + B D^{2} = A B^{2} + B C^{2} + C D^{2} + D A^{2}$

Given:
A rectangle $A B C D$

Consider the diagram shown above.

Applying the Pythagoras theorem in the two right angled triangles:

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

In $△ B A D$ ,
$B D^{2} = A B^{2} + D A^{2}$ -----(2)

Adding (1) and (2)
$A C^{2} + B D^{2} = A B^{2} + B C^{2} + A B^{2} + D A^{2}$

Since, the given figure is a rectangle,
$A B = C D$
$A B^{2} = C D^{2}$

Therefore,
$A C^{2} + B D^{2} = A B^{2} + B C^{2} + C D^{2} + D A^{2} [H e n c e p r o v e d]$

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