Nice trick, but this text only handles the case of “When our extended lines from steps 2 and 3 meet”. What if they don’t, that is, what if α = β = π/4, and the triangle is isosceles and rectangular?
I haven’t seen the original text, but this proof may be incomplete.
I mean this special case is also trivial, so it seems pretty reasonable to omit it. Feels quite uncharitable to describe this proof as a nice trick and then claim its incomplete because of such a simple special case. Mathematicians wouldn't consider this incomplete when the "missing" case can be solved almost by looking at it.
I don’t see it being trivial. Of course, ‘everybody’ knows the diagonal of the unit square has length √2, but don’t we know that because of the Pythagorean theorem?
Okay so take the triangle made by taking the diagonal of the unit square. This has side lengths 1, 1, and c and has area 1/2.
Now, take four of these and arrange them in a square with the side length being c. It would be easier to draw this... basically you stick the right angles in the center. If this isn't clear I can draw a diagram.
Anyway, you just made a square with side length c but since its made of four of those original triangles we know that the area of it is 4 * (1/2) = c^2 so c^2 = 2.
Then you go to the opposite direction, and the math still holds, but now you have inverses. And those inverses, at proportionality final formulae in the text, still gives you the Pythagorean formulae. I suggest you do see the original text.
I haven’t seen the original text, but this proof may be incomplete.