I am trying to prove Thébault's Theorem with a vector approach (I know many other approaches exist).
By Thébault's Theorem, I mean "On the sides of parallelogram ABCD erect squares -- all either on the outside or the inside of the parallelogram. Their centers then form another square."
Defining vectors a and b as emerging from a vertex along the sides of ABCD, I find that two edges of the supposed square formed (joined by a vertex) indeed have equal magnitudes, but have a dot product of $\pm$a.b, where we would expect this dot product to be zero.
In determining this, I also introduced vectors perpendicular to, but of equal length to a and b.
If anyone can show that this dot product would indeed be zero, I would appreciate the clarification.
In light of coordinates free approach, the use of complex numbers is convenient (and they are vectors over the reals). Two orthogonal vectors have the same magnitude if they are of the form $<a,b>,$ and $\pm <-b,a>$. In terms of complex numbers, they are related by $a+bi$ and $\pm i (a+bi)$ (i.e. up to multiplication by $\pm i $).
For the question, represent the vertices of the parallelogram $ABCD$ (oriented counterclockwise) by complex numbers $0,z_1,z_1+z_2,$ and $z_2$. Then the four centers of the outer squares can be found to be $$O_1=\frac 12 z_1-\frac 12z_1i,O_2=z_1+\frac 12z_2-\frac 12z_2i,O_3=z_2+\frac 12z_1+\frac 12z_1i,$$ and $$O_4=\frac 12z_2+\frac 12z_2i.$$ Abusing notations, one has $$\overrightarrow{O_1O_2}=\left(\frac 12+\frac 12i\right)z_1+\left(\frac 12-\frac 12i\right)z_2,$$ $$\overrightarrow{O_2O_3}=\left(-\frac 12+\frac 12i\right)z_1+\left(\frac 12+\frac 12i\right)z_2,~{\rm etc.}$$ Clearly $\overrightarrow{O_1O_2},\overrightarrow{O_2O_3}$ have the same magnitude and are orthogonal (because they are up to multiplication by $\pm i$), etc. QED