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.
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
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Posted by Bractals
on 2016-04-14 13:51:23 |
"Complex" Solution
