I am trying to understand Descartes' (original) proof of trisecting an angle. In the first part of the proof he tries to get a cubic equation, but I cannot really follow his argument. Can somebody help me?
The explanation in the (history of mathematics) book I am reading goes as follows
I do not understand this argument. I can probably apply anachronistically some trigonometrical identities, but this is not what Descartes was doing.
For those who can read French and want to read the original, see here and here.
Triangles $ONQ$ and $NQR$ have angle $\angle NQR$ in common and $\angle NOQ=\angle QNR$, because $\angle NOQ$ is a central angle subtended by an arc which is one half the arc subtending inscribed angle $\angle QNR$. Those triangles are thus similar: from $NO:NQ=NQ:QR$ one gets $QR=NQ^2/NO=z^2$.
Triangles $NQR$ and $QRS$ have angle $\angle NRQ$ in common and $\angle QNR=\angle SQR$ (because $\angle QNR=\angle NOQ=\angle QOT=\angle SQR$). They are thus similar: from $NQ:QR=QR:RS$ one gets $RS=QR^2/NQ=z^3$.
As triangle $NQR$ is isosceles and equal to $TPU$, we also have $NR=NQ=UP=z$. But $QTUS$ is a parallelogram, so that $SU=QT=z$. We have then $NP=NR+UP+SU-SR=3z-z^3$.