When the circumference of a circle (c) and the perimeter of a square (8h) are equal, we call it a squared circle. Radius (r) is equal to one. (Fig. 1)
Next we simplify to the quarter squared circle which gives us our first triangle A. (Fig. 2)
Because a triangle has three sides it can be normalised in three ways, with exceptions. A normalised triangle is simply a triangle with one or more side-lengths equal to one. All sets of normalised triangles have the same angles. See the example below, normalising the 345 triangle. (Fig. 3)
Notice above that the height of triangle 1 is the reciprocal of the hypotenuse of triangle 2, the base of triangle 1 is the reciprocal of the hypotenuse of triangle 3 and the base of triangle 2 is the reciprocal of the height of triangle 3. This is the case for all sets of normalised triangles.
We can use this rule to normalise triangle A. (Fig. 4)
When the perimeter of a square (8h) and the circumference of a circle (c) are equal, the ratio of the radius (r) to an eighth of the perimeter (h) will always be 4/Pi. (Fig. 5)
We can use this rule to create three nested versions of the quarter squared circle. (Fig. 6)
This gives us a second set of triangles D, E and F. (Fig. 7)
Notice that triangles D and A and triangles E and B are identical but triangles F and C are different. Next we interlock two quarter squared circles as shown below. (Fig. 8)
We then add a quarter circle with a radius (1+x). The arrows represent the tangents. (Fig. 9)
This gives us a third set of triangles J, K and L. (Fig. 10)
The set of triangles (A, B and C) and the set of triangles (J, K and L) should be identical, because both sets are normalised. If we compare the two sets we can extrapolate the following:
As shown below, it makes sense that y equals Pi/4, because we would expect the two marked points to be at the same height. The arrows represent the tangents. (Fig. 11)
If y equals Pi/4, it means that the triangle set (D, E and F) is also normalised. By comparing the three sets we can extrapolate the following:
If we do the calculations using the current value of Pi, the equations above fail. This points to the possibility that Pi is incorrect. So the question becomes, is there a value for Pi that satisfies all of the above equations?
Yes, here it is expressed as a fraction:
If we use the above value for Pi, we find that the three sets of triangles listed above, are in fact the same normalised set. (Fig. 12)
Because the circle and the square are both perfectly symmetrical, we should expect Pi to also have perfect symmetry. This symmetry can be expressed by the following equations:
So it seems, the angle (theta1) that squares the circle, perfect symmetry, occurs when its sine is equal to the reciprocal of its tangent.
Please be kind, this has taken over a year of my life so far. Is this a valid proof?
The triangle labeled K is not an alternative "normalization" of triangle J. Triangle J is not similar to triangle K.
You constructed the hypotenuse of triangle K by connecting the center of one quarter-circle to the vertex of triangle J that lies on the other quarter-circle. There is nothing in that construction that guarantees that the hypotenuse of K will be tangent to the first quarter-circle.
Indeed the hypotenuse is not tangent to the quarter-circle. But because $4\sqrt{\sqrt{5/4}-\sqrt{1/4}}$ is between $\pi$ and $1.001\pi,$ the proportions of triangle K are "off" by less than $\frac1{10}\%,$ and it will be very difficult to see merely by looking at the figure that the line and circle are not tangent.
It is not clear how you constructed triangle L. If you did it by constructing a tangent to the quarter-circle at the exact end of the quarter-circle (perpendicular to the horizontal leg of triangle J) and then extended the hypotenuse of triangle J until it met the tangent line, triangle L will be a right triangle whose upper vertex lies nearly (but not exactly) on the larger quarter-circle. If you constructed L by extending the hypotenuse of J until it met the larger quarter-circle, took that intersection point as one vertex, and took the lower end of the quarter-circle as another vertex, then L is not even a right triangle.
As with triangle K, however, it is very unlikely that you will be able to see the difference between the actual triangle L and the figure that your proof requires it to be, even with a very accurate drawing, because the difference is so small.
I hope you did not really spend an entire year's worth of your time working on this and doing nothing else.