Imagine a perpendicular prism with square bases and a cylinder inscribed in the prism in the way we have described. Let the prism be cut through the axis by a plane perpendicular to the plane which cuts off the section of the cylinder; this will intersect the prism containing the cylinder in the parallelogram $\alpha\beta$ [Fig.~9] and the common intersecting line of the plane which cuts off the section of the cylinder and the plane lying through the axis perpendicular to the one cutting off the section of the cylinder will be $\beta\gamma$; let the axis of the cylinder and the prism be $\gamma\delta$ which is bisected at right angles by $\epsilon\zeta$ and on $\epsilon\zeta$ let a plane be constructed perpendicular to $\gamma\delta$. This will intersect the prism in a square and the cylinder in a circle. Now let the intersection of the prism be the square $\mu\nu$ [Fig.~10], that of the cylinder, the circle $\xi o\pi\rho$ and let the circle touch the sides of the square at the points $\xi$, $o$, $\pi$ and $\rho$; let the common line of intersection of the plane cutting off the cylinder-section and that passing through $\epsilon\zeta$ perpendicular to the axis of the cylinder, be $\kappa\lambda$; this line is bisected by $\pi\theta\xi$. In the semicircle $o\pi\rho$ draw a straight line $\sigma\tau$ perpendicular to $\pi\chi$, on $\sigma\tau$ construct a plane perpendicular to $\xi\pi$ and produce it to both sides of the plane enclosing the circle $\xi o\pi\rho$; this will intersect the half-cylinder whose base is the semicircle $o\pi\rho$ and whose altitude is the axis of the prism, in a parallelogram one side of which = $\sigma\tau$ and the other = the vertical boundary of the cylinder, and it will intersect the cylinder-section likewise in a parallelogram of which one side is $\sigma\tau$ and the other $\mu\nu$ [Fig.~9]; and accordingly $\mu\nu$ will be drawn in the parallelogram $\delta\epsilon \| \beta\omega$ and will cut off $\epsilon\iota = \pi\chi$. Now because $\epsilon\gamma$ is a parallelogram and $\nu\iota \| \theta\gamma$, and $\epsilon\theta$ and $\beta\gamma$ cut the parallels, therefore $\epsilon\theta : \theta\iota = \omega\gamma : \gamma\nu = \beta\omega : \upsilon\nu$. But $\beta\omega : \upsilon\nu$ = parallelogram in the half-cylinder : parallelogram in the cylinder-section, therefore both parallelograms have the same side $\sigma\tau$; and $\epsilon\theta = \theta\pi$, $\iota\theta = \chi\theta$; and since $\pi\theta = \theta\xi$ therefore $\theta\xi : \theta\chi$ = parallelogram in half-cylinder : parallelogram in the cylinder-section. lmagine the parallelogram in the cylinder-section transferred and so brought to $\xi$ that $\xi$ is its center of gravity, and further imagine $\pi\xi$ to be a scale-beam with its center at $\theta$; then the parallelogram in the half-cylinder in its present position is in equilibrium at the point $\theta$ with the parallelogram in the