$O$ is any point inside a square $P Q R S$ . Prove that ${O P}^{2} + {O R}^{2} = {O Q}^{2} + {O S}^{2}$ .

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Solution

The square is $P Q R S$ . Please see the image.

Construction : $M N | | P S | | Q R$

In Triangle SON,
$S O^{2} = O N^{2} + S N^{2}$ ...(1)

In triangle RON,
$R O^{2} = R N^{2} + O N^{2}$ ...(2)

In triangle POM,
$P O^{2} = P M^{2} + O M^{2}$ ...(3)

In triangle QOM,
$Q O^{2} = O M^{2} + Q M^{2}$ ...(4)

(1) - (2) and (3) - (4)

$S O^{2} - R O^{2} = S N^{2} - R N^{2}$ ...(5)
$P O^{2} - Q O^{2} = O M^{2} - Q M^{2}$ ...(6)

Now,
$P M = S N$
$M Q = N R$

After substituting this in (5) and (6) and adding them we get

$O P^{2} + O R^{2} = O Q^{2} + O S^{2}$

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O is any point inside a square PQRS. Prove that OP2+OR2=OQ2+OS2.

$O$ is any point inside a square $P Q R S$ . Prove that ${O P}^{2} + {O R}^{2} = {O Q}^{2} + {O S}^{2}$ .