Construct (with ruler and compass) a square given one point from each side. Construct (with ruler and compass) a square given one point from each side.
I see a very interesting question.Answer, but there is no resolution, I do not know the reason why such a mapping?Is not there other mapping methods?

 A: Well... Assuming it's a proof you want, here you go:
Draw a square, and let $A, B, C$ be 3 points, on 3 different sides of the square, in that order (so that they are in the same order as on your picture above).
Draw the perpendicular to $(AC)$ passing through $B$, and let $D$ denote the intersection of this perpendicular and the 4th side of the square.
Now what we want to prove is that $BD = AC$.
I'm sure there is a way to do this formally with Pythagore's theorem, but the easiest way is to deduce it from geometrical transformations : translate the $(AC)$ line until $A$ is in one corner of the square, and the $BD$ line until $B$ is in the same corner. Now $C$ is the image of $D$ by rotation of center $A\equiv B$ and angle $90°$. This shows that $BD=AC$.
To be 100% rigorous, we need to check that you indeed get a square with the construction, but the argument is identical (translate the $[AC]$ and $[BD']$ segments until they are two sides of the square).
This proves that there are actually 6 squares going through $A$, $B$, $C$ and $D$ in the wanted way, and that you can build all of them with the method given : for the first line drawn, you can choose either $(AB)$ or $(AD)$ as well as $(AC)$. And when you draw the circle of center $B$ and radius $AC$, there are 2 intersections, leading to 2 different $D'$ and 2 different squares.
A: This May Explain Why?! They Decided to Construct What They Did
It is clear which points are on opposite sides, as quadrilateral has only one pair of diagonals. In the figure, AD and BC are the pairs. 
You want perpendicular axes such that the projection of AC onto one axis is the same length as the projection of BD onto the other. If you turn both AC (to get the image A'C') and its axis through $90^\circ$, then you now wish to find a single axis onto which the projections of A'C' and BD are both the same. 
The easiest way to do this is to translate A'C' so that A' and B coincide (the new A'C' is called BD'). Then simply join D' and D and this line will be perpendicular to the axis of projection (so that the projected distances are equal). Since BD' is the axis of projection for BD, we must have that D and D' are on the same side.
