This squaring of the circle is correct, and it is carried out by means of the equalization of twin segments and circumference arcs, by means of the segment matching method that has already been exposed and published.

Keywords: Squaring, circle, segments, arcs, circumference, equality.

I have been working and reviewing for a while the squaring of the circle, (as well as the doubling of the cube and the trisection of an angle).
I have been constant because I always intuited that all of them would have an easy solution, since I have always thought that mathematics should have no limits.
Well, here I present the squaring of the circle that I think is achieved, unless later I see that I am wrong and have to continue working on it.
I understand that the squaring the circle is very easy, and that was to be expected by means of interrelating semicircles and properly placed, since logically the number Pi must be derived from a good interrelation between the curvatures of said circles.

As we see in the drawing, the solution is extremely simple:
We construct a circle, and both on the side of the horizontal diameter (point A) and on the vertex above the vertical diameter we draw two arcs of radius = 1.
Well, when these two arcs are cut by a ruler in the appropriate position (AB), they give us two segments (a) and (a') which, when they are equal, produce the squaring of the circle.

That is, the work in simple:
Once the two arcs r =1 have been drawn, we position the ruler (fixed on point A) and move it from point (v), measuring and checking the same time with the compass if segments (a) and (a' ) are equal.
When we get it, we draw the line AB which will be equal to Pi / 2,
Then, with the compass, we pass this measure (Pi/2) to the horizontal diameter and from this we project them perpendicularly on the circumference that will cut it at point (C).
We now draw the segment AC, and this will be the side of the square that we are looking for, that is, Sqrt.Pi.

Here we have more quadratures with the component segments.

Friends, in case this work is ever accepted, I will give you the first drawing of how the squaring of the circle was made by this method.

Mathematical revision of the squaring the circle

As it is exposed in this work, to find the segment AB (Pi/2), on the given circle we draw two arcs of circumference of r = 1, from the points or vertices of the horizontal diameter A, and vertical V, in order to contain and enclose in them the segment AB that we are looking for.
And we have established that only when the segments (a) and (a'), (interior to these arcs) are equal, the segment AB, of length Pi/2, is produced.
Well, to verify that this is correct, we are going to proceed to see some segments and arcs that make up this construction are correct and complementary, and add up to the final structure of the construction.
Of course we are only going to expose the main segments and arcs, since this composition could be subdivided into a huge number of component segments, that of course, meet all them the correct quadrature and can be adjusted with the data of the fig. 1.
Since almost all component figures can be decomposed into right triangles, we are going to use both the Pythagorean theorem and trigonometry using the adequate angle.
The first thing to do is to adjust the angle formed by the inscribed Pi triangle (ABG)
We already know that this triangle must have Pi/2 = 1.570796326 ..... as hypotenuse and therefore the projection segment (AG) on the diameter will be:

AG = [ (Pi/2)^2 ]/2 = 1.23370055......

And the segment BG by Pythagoras = 0.9723086......

With these data we obtain an angle of 38.2425 .. degrees that we will use to make the adjustments for trigonometry, always with an approximation of only 5 digits that are sufficient for this explanation.

For example, to see that the angle is correct, let's look at the segment BG and AG

BG = Sen 38.2425 x Pi / 2 = 0.6189911... x 1.5707963... = 0.972309...

By Pythagoras, let's see how much the segment (a') measures.
We will not put the degrees of the angle, although we already know that it is 38.2425

(a')^2 = s^2 + j^2 = (AG - Cos.)^2 + (0.972309 - Sen.)^2 = (0.448302..)^2 + (0.353318.. )^2

a'= 0.570796....

We already see that if the quadrature is correct, that is, if we have really drawn the segment AB = Pi / 2 correctly, then the segment (a') has a value of (Pi/2) - 1 = 0.570796326 ....
We can also verify that when measuring with the compass (a = a'), the segment (a) also measures 0.57079632...
However, if we want to adjust (a) independently we can use the data from drawing 1.

a^2 = (sin - d)^2 + (cos - c)^2 = (0.353317 .. )^2 + (0.448302... )^2 = 0.325807...
a = 0.570796...

But if we want to do more checks, we can project all these component segments onto the diameter with arcs, and we can check that they all repeat themselves and measure exactly the same as each other, as seen in the drawing.
Therefore with this quadrature we can verify that all the segments are a derivation (or sub-components) of Pi, (Pi/2) - 1 = 0.570796326 ...)
Naturally, no other segment AB that was not the correct one would give us this derivation of Pi, and therefore the quadrature must be correct.

Thanks all you.