Lill's method is difficult to explain shortly, but you can easily find a good youtube-video by Mathodologer about it, and it is explained in texts Geometric Solutions of Algebraic Equations, by M. Riaz (American Math. Monthly, 1962), and Solving Cubics With Creases: The Work of Beloch and Lill, by Thomas Hull.
I'll ask my question assuming you are familiar with the method.
I need to prove that the method finds the root $x=-\tan\theta$. I only need the proof for 3rd degree. These mentioned texts prove the method very shortly (Hull proves it for only the case with positive coefficients): They both solve the opposite sides of an angle $\theta$, for all the similar triangles the method creates. These lead to one equation, that ends up being the original polynomial equation, where the unknown is now $x=-\tan \theta$ and so this gives a root of this equation.
The solved sides according to Riaz, where solving the last side creates n:th degree equation with a solution (Geometric solution to an equation, by Riaz):
$xa_n$; $x(a_{n-1}+xa_n)$; $x(a_{n-2}+x(a_{n-1}+xa_n))$; ...; $x(a_1+x(a_2+...+x(a_{n-2}+x(a_{n-1}+xa_n))...))=-a_0$
I understand these proofs in given basic situations, and I can follow similar steps for every individual Lill's path. But I have to do new, slightly different calculations every time, looking at the picture, because the paths created with Lill's method can be really different depending on the equation we are drawing, and the similar triangles do not form in same way every time. The steps I take are not EXACTLY the same every time, so I don't see this proof working when the path is a little different, lets say that the sides cross each other or the second path (laser path) has to hit the extensions of sides in some weird manner. Or maybe the proof works for every situation, and I just don't see the logic.
So, I am looking for a general proof for every possible scenario with 3rd degree and less, or at least a proof that does not require too many different cases proved separately. If Riaz's proof is enough, I need to understand it.