Let the parallelogram $\mu\nu$ be perpendicular to the axis [of the circle] $\xi o$ [$\pi\rho$] [Fig.~11]. Draw $\theta\mu$ and $\theta\eta$ and erect upon them two planes perpendicular to the plane in which the semicircle $o\pi\rho$ lies and extend these planes on both sides. The result is a prism whose base is a triangle similar to $\theta\mu\eta$ and whose altitude is equal to the axis of the cylinder, and this prism is $\frac{1}{4}$ of the entire prism which contains the cylinder. In the semicircle $o\pi\rho$ and in the square $\mu\nu$ draw two straight lines $\kappa\lambda$ and $\tau\upsilon$ at equal distances from $\pi\xi$; these will cut % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig11.png} \begin{center} {\small Fig.~11.} \end{center} \end{wrapfigure} % the circumference of the semicircle $o\pi\rho$ at the points $\kappa$ and $\tau$, the diameter $o\rho$ at $\sigma$ and $\zeta$ and the straight lines $\theta\eta$ and $\theta\mu$ at $\phi$ and $\chi$. Upon $\kappa\lambda$ and $\tau\upsilon$ construct two planes perpendicular to $o\rho$ and extend them towards both sides of the plane in which lies the circle $\xi o\pi\rho$; they will intersect the half-cylinder whose base is the semicircle $o\pi\rho$ and whose altitude is that of the cylinder, in a parallelogram one side of which = $\kappa\sigma$ and the other = the axis of the cylinder; and they will intersect the prism $\theta\eta\mu$ likewise in a parallelogram one side of which is equal to $\lambda\chi$ and the other equal to the axis, and in the same way the half-cylinder in a parallelogram one side of which = $\tau\zeta$ and the other = the axis of the cylinder, and the prism in a parallelogram one side of which = $\nu\phi$ and the other = the axis of the cylinder.\dotfill \section*{Proposition XIII}