Let the square $\alpha\beta\gamma\delta$ [Fig.~12] be the base of a perpendicular prism with square bases and let a cylinder be inscribed in the prism whose base is the circle $\epsilon\zeta\eta\theta$ which touches the sides of the parallelogram $\alpha\beta\gamma\delta$ at $\epsilon$, $\zeta$, $\eta$, and $\theta$. Pass a plane through its center and the side in the square opposite the square $\alpha\beta\gamma\delta$ corresponding to the side $\gamma\delta$; this will cut off from the whole prism a second prism which is $\frac{1}{4}$ the size of the whole prism and which will be bounded by three parallelograms and two opposite triangles. In the semicircle $\epsilon\zeta\eta$ describe a parabola whose origin is $\eta\epsilon$ and whose axis is $\zeta\kappa$, and in the parallelogram $\delta\eta$ draw $\mu\nu \| \kappa\zeta$; this will cut the circumference of the semicircle at $\xi$, the parabola at $\lambda$, and $\mu\nu \times \nu\lambda = \nu\zeta^2$ (for this is evident [Apollonios, \emph{Con.} I, 11]). Therefore $\mu\nu : \nu\lambda = \kappa\eta^2 : \lambda\sigma^2$. Upon $\mu\nu$ construct a plane parallel to $\epsilon\eta$; this will intersect the prism cut off from the whole prism in a right-angled triangle one side of which is $\mu\nu$ and the other a straight line in the plane upon $\gamma\delta$ perpendicular to $\gamma\delta$ at $\nu$ and equal to the axis of the cylinder, but whose hypotenuse is in the intersecting plane. It will intersect the portion which is cut off from the cylinder by the % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig12.png} \begin{center} {\small Fig.~12.} \end{center} \end{wrapfigure} % plane passed through $\epsilon\eta$ and the side of the square opposite the side $\gamma\delta$ in a right-angled triangle one side of which is $\mu\xi$ and the other a straight line drawn in the surface of the cylinder perpendicular to the plane $\kappa\nu$, \linebreak and the hypotenuse\dotfill\linebreak and all the triangles in the prism : all the triangles in the cylinder-section = all the straight lines in the parallelogram $\delta\eta$ : all the straight lines between the parabola and the straight line $\epsilon\eta$. And the prism consists of the triangles in the prism, the cylinder-section of those in the cylinder-section, the parallelogram $\delta\eta$ of the straight lines in the parallelogram $\delta\eta \| \kappa\zeta$ and the segment of the parabola of the straight lines cut off by the parabola and the straight line $\epsilon\eta$; hence prism : cylinder-section = parallelogram $\eta\delta$ : segment $\epsilon\zeta\eta$ that is bounded by the parabola and the straight line $\epsilon\eta$. But the parallelogram $\delta\eta = \frac{3}{2}$ the segment bounded by the parabola and the straight line $\epsilon\eta$ as indeed has been shown in the previously published work, hence also the prism is equal to one and one half times the cylinder-section.