$A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ are two squares with their vertices arranged clockwise.The perpendicular bisector of segment $A_1B_1,A_2B_2,A_3B_3,A_4B_4$ and the perpendicular bisector of segment $A_2B_2,A_3B_3,A_4B_4,A_1B_1$ intersect at point $P,Q,R,S$ respectively.Show that:$PR\perp QS$.
For $i\in \{1,2,3,4\}$,let $C_i$ be the midpoint of $A_iB_i$,let line $A_iB_i,A_{i+1}B_{i+1}$ intersect at $D_i$($i \mod 4$).Denote the similarity center of square $A_1A_2A_3A_4$ and square $B_1B_2B_3B_4$ as $O$.
Claim1:$C_1C_2C_3C_4$ is a square.
Claim2:For $i\in \{1,2,3,4\}$,$O,D_i,C_i,C_{i+1}$ are concyclic($i \mod 4$).Let $O_i$ be the center of the circle.Moreover,$P\in \odot O_1,Q\in \odot O_2,R\in \odot O_3,S\in \odot O_4$,and $PD_1,QD_2,RD_3,SD_4$ are diameters respectively.
Claim3:$\triangle OO_1O_3 \sim \triangle OPR$,$\triangle OO_2O_4 \sim \triangle OQS$.
Denote $\odot O_1 \cap \odot O_3=X,\odot O_2 \cap \odot O_4=Y$.Notice that $\angle OC_4B_4=\angle OC_1D_1$,so $\angle OO_3D_3=\angle OO_1D_1$,which implies $\triangle OO_1D_1 \sim \triangle OO_3D_3$,thus $\triangle OO_1O_3 \sim \triangle OD_1D_3$,which leads to the fact that $D_1,X,D_3$ are collinear.Also we have:$\angle D_1XP=\angle D_1XR=90^{\circ}$,so $X,P,R$ are collinear.This implies that $\triangle OO_1O_3 \sim \triangle OPR$.Likewise $\triangle OO_2O_4 \sim \triangle OQS$.
Finish:Notice that $O_1O_3\perp O_2O_4$,so it's enough to show that $\angle O_3OR=\angle O_4OS$.By angle chasing:
$\angle OO_3R=2\angle OC_4R=2\angle OC_4S=\angle OO_4S.$
So we are done.$\Box$
Call M the center of the spiral similarity from $A_1$ to $B_1$,$A_2$ to $B_2$, etc...
It clearly exist since the two figures are directly similar.
Now by gliding the midpoint $G_1, G_2, G_3, G_4$ of the segments form a square and we have that the condition of perpendicolar bisector becomes(i changed the letter so look to the picture below):
$$\angle MG_4A = \angle MG_3B = \angle MG_2C = \angle MG_1D$$this implies the following ciclicities: $(G_1MG_4A),(G_4MG_3B),(G_3MG_2C),(G_2MG_1D)$
Now by angle chasing:
$$\angle G_1G_4M=\angle DAM, \angle G_4G_1M=\angle BAM$$and ciclically, in particular we can notice that $\angle AMD+\angle CMB= \pi$, by this $M$ has a isogonal conjugate in $ABCD$ then call this $M_*$, we can se by the angle condition that $\angle AM_*B=\frac{\pi}2$ and ciclical, so this must be the intersection of the diagonals (the sum of 2 consecutive angle is $\pi$), and they are orthogonal, Q.E.D.
