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 Proof requested (Posted on 2016-04-14)
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 No Rating

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 "Complex" Solution | Comment 1 of 3
`In the following capital letters represent pointsin the complex plane and i the sqrt(-1).Let ABCD be the parallelogram ordered CCW. Therefore,   A + C = B + D                           (1)Let P, Q, R, and S be the centers of the squares onsides AB, BC, CD, and DA respectively. Therefore,   P = [ (A + B) + (A - B)*i ]/2   \      Q = [ (B + C) + (B - C)*i ]/2     \                                                (2)   R = [ (C + D) + (C - D)*i ]/2     /   S = [ (D + A) + (D - A)*i ]/2   /The quadrilateral PQRS is a parallelogram if   P + R = Q + S                           (3)and its diagonals are perpendicular and ofequal length if   (P - R)*i = Q - S                       (4)Equations (1) and (2) imply (3) and (4). Therefore,PQRS is a square.QEDNote: If the squares are constructed internalto the parallelogram, the centers also formthe vertices of a square.`
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Edited on April 14, 2016, 3:07 pm

Edited on April 14, 2016, 11:39 pm
 Posted by Bractals on 2016-04-14 13:51:23

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