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A homework question I have been having some issue with -

Given parallelogram $ABCD$, generate 4 squares from the sides of the $ABCD$. Given the 4 centers of the squares $W, X, Y, Z$ (formed by their respective squares defined by sides $AB, BC, CD, DA$ respectively), show the centers form a new square.

A diagram of the problem

The first step would be to show two sides, say $ZW$ and $WX$, are equal. This can be done by similar triangles. $AW$ and $WB$ are equal because they are both half the length of the diagonal of the same square. $BX$ and $AZ$ are equal because of the symmetry line $DB$ (flip the parallelogram over and its equal).

Which leads me to where Im stuck at. I need to show 2 more things to finish the proof.

  1. $\angle WBZ = \angle ZAW$. This completes the side-angle-side, proving the triangles are congruent and their third side is equal => the centers form at least a rhombus. I believe this is done using interior/exterior angles, but I dont see how.
  2. $WY \perp ZX$. This brings the rhombus to a square, thus completing the proof. I dont have a clue how to do this.
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  • $\begingroup$ Fun facts: (1) This works even if the squares are constructed on the "inner" sides of each edge. (2) Note that a parallelogram is the image of a square (a regular $4$-gon) under a linear transformation. If you hit a regular $n$-gon with a linear transformation, then the centers of regular $n$-gons constructed on the sides of the result give you another regular $n$-gon. (Again, this works if the regular $n$-gons are constructed on the "inner" sides of each edge.) When $n=3$, this is Napoleon's Theorem; the general result is called the Napoleon-Barlotti Theorem. $\endgroup$ – Blue Oct 31 '13 at 12:21
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let AB=CD=a and AD=BC=b

  • Now 4 squares are drawn as shown in the figure.we join WA and WB & ZA;XB.
  • Now AW=BW=a$\sqrt[2]{2}$/2; AZ=BX=b$\sqrt[2]{2}$/2; and $\angle$ZAW=$\angle$WBX [$\angle$WBX=45+45+$\angle$ABC=90 +$\angle$ABC: now $\angle$LAQ=360$-$(90+90+$\angle$DAB)=180$-$$\angle$DAB=$\angle$ABC; So similarly $\angle$ZAW=90+$\angle$ABC.
  • So $\bigtriangleup ZAW\cong \bigtriangleup WBX$. So $ZW=WX$ and $\angle ZWA=\angle BWX$.
  • Similarly $WX=XY, XY =ZY, ZW=ZY$.
  • Now $\angle AWB=90^\circ$
  • $\angle AWB=\angle AWX+ \angle BWX= \angle AWX+ \angle ZWA=\angle ZWX=90^\circ$
  • Now it is clear that WXYZ is a squareenter image description here
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  • $\begingroup$ Why does 180-$\angle DAB$=$\angle ABC$? $\endgroup$ – David Grinberg Oct 24 '13 at 18:49
  • $\begingroup$ What Im really asking I guess is why does LAQ=WBX? $\endgroup$ – David Grinberg Oct 24 '13 at 18:58
  • $\begingroup$ I figured it out. Also you forgot to label $Y$ $\endgroup$ – David Grinberg Oct 24 '13 at 19:16
  • $\begingroup$ 180-DAB=ABC as DA||BC. $\endgroup$ – krishan acton Oct 25 '13 at 5:16
  • $\begingroup$ LAQ=WBX.THAT IS TRUE NOT NECESSARILY.I DID NOT WRITE IT. $\endgroup$ – krishan acton Oct 25 '13 at 5:21
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enter image description here

from here, $AB$ mean $\overrightarrow{AB}$.
$f$ is counterclockwise-$90^{\circ}$-rotation transformation, which is linear.
$f(MP)=f(MQ+QB+BT+TP)=f(MQ)+f(QB)+f(BT)+f(TP)$
$=QA+MQ+PT+TC=QA+MQ+RN+AR$
$=MQ+QA+AR+RN=MN$

thus, $~\square MPON$ is square.

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Let the center of the parallelogram $P:=[ABCD]$ be the origin of ${\mathbb C}$, and consider the vertices of $P$ as complex numbers. Then $C=-A$, $\>D=-B$. The center $c_W$ of $W$ is then $$c_W={A+B\over2}+i{B-A\over2}={1-i\over2}A+{1+i\over2}B\ .$$ Similarly $$c_Y={D+C\over2}-i{C-D\over2}={-1+i\over2}A+{-1-i\over2}B=-c_W\ .$$ In the same way one obtains $$c_Z=-c_X={1+i\over2}A+{-1+i\over2}B=i\> c_W\ .$$ Altogether we see that the four centers $c_k$ stand in the relation $c_{k+1}=i\> c_k$, whence they form a square.

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This can also be done with a basic high school coordinate proof (here is an outline):

Let the vertices of the parallelogram going clockwise from the origin be at (0,0), (2b,2c), (2a+2b,2c), and (2a,0).

Calculate the vertices of the constructed squares by noting that each side is the same length as and perpendicular to a side of the parallelogram. The midpoints of the diagonals of those squares will be the four points in question: (b-c,b+c), (a+2b,a+2c), (2a+b+c,c-b), and (a,-a).

It is a fairly simple matter to check that the diagonals have the same midpoint (thus it's a parallelogram), the diagonals are perpendicular (thus it's a rhombus), and any two sides are perpendicular (thus it's a rectangle). If it's all those things, it must be a square.

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