# 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.

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.
• 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. – Blue Oct 31 '13 at 12:21

• 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$-$$\angleDAB=\angleABC; So similarly \angleZAW=90+\angleABC. • So \bigtriangleup ZAW\cong \bigtriangleup WBX. So ZW=WX and \angle ZWA=\angle BWX. • Similarly WX=XY, XY =ZY, ZW=ZY. • Now \angle AWB=90^\circ • \angle AWB=\angle AWX+ \angle BWX= \angle AWX+ \angle ZWA=\angle ZWX=90^\circ • Now it is clear that WXYZ is a square • Why does 180-\angle DAB=\angle ABC? – David Grinberg Oct 24 '13 at 18:49 • What Im really asking I guess is why does LAQ=WBX? – David Grinberg Oct 24 '13 at 18:58 • I figured it out. Also you forgot to label Y – David Grinberg Oct 24 '13 at 19:16 • 180-DAB=ABC as DA||BC. – krishan acton Oct 25 '13 at 5:16 • LAQ=WBX.THAT IS TRUE NOT NECESSARILY.I DID NOT WRITE IT. – krishan acton Oct 25 '13 at 5:21 from here, AB mean \overrightarrow{AB}. f is counterclockwise-90^{\circ}-rotation transformation, which is linear. f(MP)=f(MQ+QB+BT+TP)=f(MQ)+f(QB)+f(BT)+f(TP) =QA+MQ+PT+TC=QA+MQ+RN+AR =MQ+QA+AR+RN=MN thus, ~\square MPON is square. Let the center of the parallelogram P:=[ABCD] be the origin of {\mathbb C}, and consider the vertices of P as complex numbers. Then C=-A, \>D=-B. The center c_W of W is then$$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.