Given two setS of points {${A_1,A_2,A_3,A_4}$} and {${B_1,B_2,B_3,B_4}$} in $\mathbb R^2$, denote $d_{i,j}$ as the Euclidean distance between Given two sets of points {${A_1,A_2,A_3,A_4}$} and {${B_1,B_2,B_3,B_4}$} in $\mathbb R^2$, denote $d_{i,j}$ as the Euclidean distance between $A_i$ and $A_j$, and denote $d'_{i,j}$ as the Euclidean distance between $B_i$ and $B_j$ ,$1\le i\lt j\le 4$.
Assume that $d_{i,j}-d_{k,l}=d'_{i,j}-d'_{k,l}$ for $1\le i\lt j\le 4, 1\le k\lt l\le 4.$ Do we have $d_{i,j}=d'_{i,j}$, $1\le i\lt j\le 4$?
Obviously I can set up coordinates and write down the equation and solve for it. Whereas, I believe that there have been previous studies on the related issue (for example, we can  generalize the problem to a general distance space and higher dimensions).
Is there any related material? Thanks in advance!
 A: Let me use $A(i,j)$ to mean the Euclidean Distance between Points in Set A  & $B(i,j)$ to mean the Euclidean Distance between Points in Set B.
Given $A(i,j)-A(k,l)=B(i,j)-B(k,l)$ , we can re-arrange to get :
$A(i,j)-B(i,j)=A(k,l)-B(k,l)$ when LHS has only $i,j$ & RHS has only $k,l$ which means each Side is a Constant.
$A(i,j)-B(i,j)=C=A(k,l)-B(k,l)$

Here we have 3 Point Set $A$ versus 3 Point Set $B$.
Difference between each Corresponding Pair is Constant $C$ because Equilateral triangles are Similar & (or even Congruent $C=0$).

Let us include one more Point.
The Blue lines must also have Constant Difference between the Pairs. If Corresponding triangles (made of 2 Blue lines & 1 Black line) are Similar but not Congruent, the Differences will change & not be a Constant. It will be a Constant only when the new triangles are Congruent too , that is $C=0$.
This Constant is $C=0$ in Euclidean Points , hence :
$A(i,j)=B(i,j)$
$B(k,l)=A(k,l)$
Ratios of Sides of Quadrilaterals (or Polygons) can be same to get Similar Quadrilaterals (or Similar Polygons) , but Non-Zero Differences can not be same in general.
If we had only 3 Points in each set, then Differences may not be Zero.
Consider Equilateral triangle with sides 1 in Set A & Equilateral triangle with sides 4 in Set B. These are Similar , where $C=3$.
With 4 or more Points, this will not occur. Both Sets will be Congruent Quadrilaterals or Congruent Polygons , where Differences will be $C=0$.
