Prove that there is only one way to make a square using all seven tangram pieces I am pretty sure there is only one way to make a square from the seven tangram pieces:

How can I prove this is the only way respecting all symmetries?
 A: This is not a exactly a proof, mostly milestones towards a solution. This is a community wiki, feel free to edit.
First, you must consider than your tan-gram has 7 pieces (You may update your title):


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*2 Large Right Triangles (A);

*1 Medium Right Triangle (B);

*2 Small Right Triangles (C);

*1 Square (D);

*1 Parallelogram (E).


Second, if we assume the tan-gram to have an area of $a^2$ there are 4 different possible edge lengths: $a,\frac{\sqrt{2}}{2}a, \frac{1}{2}a, \frac{\sqrt{2}}{4}a$. Tan-gram square (horizontal and vertical directions) edges have a length equal to $a$. All horizontal and vertical subdivisions are ratio of this fundamental length. In the same way, Tan-gram diagonal length is equal to $\sqrt{2}$, all diagonal edges are ratio of this length. 
Therefore parts with edge length that are rational product of $\sqrt{2}$  must be diagonals, and others are straight.
Third, it follows that parts A must form a bigger triangle in a corner of the tan-gram. Thus the problem resume to make an equivalent triangle with the 5 parts left (B,C,C,D,E).
Fourth, following the same logic, square D is internal part and then triangle B must be at the corner.
Fifth, triangles C must have their hypotenuses straight, so they have to share edges with square B.
Finally, there is only one place for parallelogram E.
I hope this helps!
