Show centers of squares formed by a parallelogram form a square. 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.

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.


*

*$\angle WBZ = \angle ZAW$. 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.

*$WY \perp ZX$. This brings the rhombus to a square, thus completing the proof. I dont have a clue how to do this. 

 A: 
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.
A: 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.
A: 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 square
A: 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.
