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Proof requested (Posted on 2016-04-14) Difficulty: 3 of 5
Prove the following theorem:

Given any parallelogram, construct on its sides four squares
external to the parallelogram - the quadrilateral formed by joining the centers
of those four squares is a square.

No Solution Yet Submitted by Ady TZIDON    
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Solution Analytical solution | Comment 2 of 3 |
Lets the vertices be
(0,0), (a,0), (a+b,c), (b,c)
the centers of the squares are 
(a/2,-a/2), (a+b/2+c/2, c/2-b/2), (a/2+b, a/2+c), (b/2-c/2, b/2+c/2)

The diagonals of the quadrilateral:
have slopes (a+c)/b and -b/(a+c) so are perpendicular,
have the same midpoint: (a/2+b/2, c/2)
have the same length: sqrt((a+c)^2 + b^2)

Only a square satisfies all of these.

  Posted by Jer on 2016-04-15 13:45:37
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