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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 "Complex" Solution | Comment 1 of 3

In the following capital letters represent points
in 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 on
sides 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 of
equal length if

   (P - R)*i = Q - S                       (4)

Equations (1) and (2) imply (3) and (4). Therefore,
PQRS is a square.

QED

Note: If the squares are constructed internal
to the parallelogram, the centers also form
the vertices of a square.

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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