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Question
O is any point inside a rectangle ABCD.
Prove that: OB2 + OD2 = OC2 + OA2.
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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:
OA2 = AH2 + OH2 = AH2 + AE2
OC2 = CG2 + OG2 = EB2 + HD2
OB2 = EO2 + BE2 = AH2 + BE2
OD2 = HD2 + OH2 = HD2 + AE2
Adding these equalities we get:
OA2 + OC2 = AH2 + HD2 + AE2 + EB2
OB2 + OD2 = AH2 + HD2 + AE2 + EB2
From which we prove that for any point within the rectangle there is the relation
OA2 + OC2 = OB2 + OD2
Hence Proved.
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