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**O is any point inside a rectangle ABCD.**

Prove that: OB^{2} + OD^{2} = OC^{2} + OA^{2}.

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#### Solution

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Using Pythagorean theorem we have from the above diagram:

OA^{2 }= AH^{2 }+ OH^{2 }= AH^{2 }+ AE^{2}OC^{2 }= CG^{2 }+ OG^{2 }= EB^{2 }+ HD^{2}OB^{2 }= EO^{2 }+ BE^{2 }= AH^{2 }+ BE^{2}OD^{2 }= HD^{2 }+ OH^{2 }= HD^{2 }+ AE^{2}

Adding these equalities we get:

OA^{2 }+ OC^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2}OB^{2 }+ OD^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2}

From which we prove that for any point within the rectangle there is the relation

OA^{2 }+ OC^{2 }= OB^{2 }+ OD^{2}Hence Proved.

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