In this work we proceed to framing the circle and its squaring by in very approximate way, but with the ultimate purpose of showing many of the multiple segments, portions and operation that can be developed inside the circumference.
That is, framing the circle is to show some of the multiples operation y development of segments that the circumference has inside it.
by step, we see other important elements as can be the Pi triangle, etc.

My thoughts on squaring the circle:

I understand that the difficulty in squaring the circle does not come from that mathematical postulate that considers that "because Pi is a transcendental number, its square root and therefore its exact quadrature is not possible."
No, I think that this circumstance would occur at infinity, but for a digit before infinity, that root could occur and therefore that squaring.
The possibility of squaring the circle I think exists, but there is also a big problem, and it is that the direct operational relationship between Pi and the radius of the circumference only occurs at a relationship of powers and roots of Pi^17, (Pi^17 = 2Sqrt2 x 10^8) and therefore almost impossible for me to perform these operations with segments.
Nevertheless, here a very approximate method is exposed, that can be considered correct from a point of view of practical measurements on the drawing paper.

For framing and squaring the circle in a practical way, we will use the known requirement: only a ruler and a compass (preferably and for greater cleanliness and speed, we can use a double needle compass, together to the ruler).

To begin we proceed to define or obtain a circle (Fig. 2) to which we must find its quadrature, and at the same time and above all, we are going to build the triangle of Pi which will be the best product resulting from the quadrature.

Well, once the circle has been built for its quadrature, and the horizontal and vertical diameters have been marked, we think about the theory that we are going to apply and the triangle (AHB= Pi triangle) that we are going to rely on.

1.- The first thing will be (fig. 2) to mark two perpendicular points (B, B') with the compass from the horizontal diameter, (points that we intuit will be approximate to the ones that let us build the Pi triangle and inscribed rectangle described above).
These points B and B' are key for the development of the quadrature, since point B' helps us to place and align the ruler in (O-B'), and point B helps us to measure and equalize the segments (b) and (b') with the compass.

Fig. 1

2.- Once these two points have been marked (fig. 3), we align and fix the ruler with the center of the circle (O) and the point (B ').
Now with the compass we take from points (B to C) the measure of segment (b) and draw an arc towards the ruler in order to check if this segment (b) is the same as segment (b').
As it does not usually coincide in the first test, we gently turn the ruler in the proper direction (see fig. 3) and take the measurement again: Steps 1 and 2.

3.- Once the measurement matches us and we verify that (b = b') (fig. 4) we mark finally the point (B) on the circumference, and we will have managed to square the circle, since by drawing the line (AB) we verify that is equal to the side of the square sought, that is, square root of Pi (Sqrt.Pi).

For a quick and clean development, we can use a compass with two needles and once we have fixed point B on the circumference we can mark it and proceed by drawing all the elements of the quadrature.
As a practical rule, we can hold and fix the ruler on points O (center circumference) and point B' with one hand.
With the other hand we move the compass following the points indicated above, (1,2,3,4) and when we equalize the segments (b = b') we can mark point B and complete the entire quadrature.

Below is the drawing of the Pi triangle, bevel tool, which due to its construction angles, the hypotenuse represents the square root of Pi (which will be the side of the square sought for squaring the circle) and the largest leg will be half of Pi.
As is logical in this triangle the angles are invariable as detailed in the drawing, but the sides can vary depending on whether we want to make a larger or smaller triangle.
Well, by positioning the larger leg (pi/2) on the diameter of the circle to be squared from one end (A), the hypotenuse will cut the circumference at point (B), resulting a segment (AB) that will be the side of the searched square for the quadrature.

Here I will develop an explanation of the procedure to frame the circle.
However, it is so easy that for the moment I expose it with simple drawings.
Thanks.

Postscript:

As shown in the general drawing (figure 1), this procedure is not a totally exact method, but it delineates, draws and situate in its correct place to the Pi triangle, (true product and main reason for the procedure).
However and surprisingly, if we use the appropriate thickness of the needles (or marking pencil), an extraordinary accuracy is obtained in the results.
That is, if for example we are drawing, checking and squaring a 10-20 cm circle of diametre on paper, with the normal thickness of the needles or marking pencil (1-2 mm), the result of the quadrature can be considered as exact, and within the margins of error that any practical drawing has.