$O$ is any point inside a rectangle $A B C D$ . Prove that: ${O B}^{2} + {O D}^{2} = {O C}^{2} + {O A}^{2}$

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Solution

In $△ O P D, O D^{2} = O P^{2} + P D^{2} \dots (1)$
In $△ O Q B, O B^{2} = O Q^{2} + Q B^{2} \dots (2)$
$(1) + (2) ⟹ O B^{2} + O D^{2} = O P^{2} + O Q^{2} + P D^{2} + Q B^{2} = O P^{2} + O Q^{2} + C Q^{2} + A P^{2}$
In $△ O C Q, O Q^{2} + C Q^{2} = O C^{2}$
In $△ O P A, O P^{2} + P A^{2} = O A^{2}$
$⟹ O B^{2} + O D^{2} = O C^{2} + O A^{2}$

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