C.B,
I really don't understand why you are so agressive.
Last time i talked about your demonstration your slammed the door by saying "i will not answer the same question".
This is the question you failed to answer.
The question is : Why this triangle and not another ?
You firstly said because its a right triangle.
Probably not knowing that all triangle in a semi circle with the hypothenuse = diameter and the vertex lie on the circle is a right triangle.
Which means ALL triangles with the hypothenuse = diameter on a semi circle are right triangle. Thales's theorem.
Why this kepler triangle b^2,b,1 ?
b^2, b and 1 is a kepler triangle because it is the only one with height^2 = base * hypothenuse.
And so with hypothenuse = 1 base = height^2 so we have this b^2, b, 1 triangle.
and so b = 1√ɸ
So you failed to explain why this triangle ?
you said
b = area of the circle ? and what ?
b^2 = area of a square with b as side ? and what ?
and you put b and b^2 as sides of the triangle ?
These are reasons without logical basis.
And finally you use 4b = pi to solve 4 * 1√ɸ = pi
What an easy way !
In fact you loose all logic of the initial german demonstration.
I reproduced as you know the german demonstration
here with all explanations.
The reason why this kepler triangle and not another is because of the combination of the Euclid's cathetus theorem and the Euclid's right triangle altitude theorem.
When you combine the 2 theorems on a triangle inscribed in a semi circle with the hypothenuse = diameter you got this triangle : b^2, b, 1.
You have all the algebraic demonstration.
That's why. That's the reason
After that with multiple equalities you have 4/√ɸ which the germans say "near pi".