$\epsilon\zeta$ and whose vertex is at $\alpha$ and let $\alpha\eta$ be so divided at $\phi$ that $\alpha\eta = 4\phi\eta$; then $\phi$ is the center of gravity of the cone $\epsilon\alpha\zeta$ as has been previously proved. Moreover let the cylinder $\mu\nu$ be so cut by a perpendicularly intersecting plane that the cylinder $\mu$ is in equilibrium with the cone $\epsilon\alpha\zeta$. Now since the segment $\alpha\beta\delta$ + the cone $\epsilon\alpha\zeta$ in their present positions are in equilibrium at $\alpha$ with the cone $\epsilon\alpha\zeta$ if it is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity, and cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$ and the two cylinders $\mu + \nu$ are moved to $\theta$ and $\mu\nu$ is in equilibrium with both bodies, then will also the cylinder $\nu$ be in equilibrium with the segment of the sphere at the point $\alpha$. And since the spherical segment $\beta\alpha\delta$ : the cone whose base is the circle with the diameter $\beta\delta$, and whose vertex is at $\alpha = \xi\eta : \eta\gamma$ (for this has previously been proved [\emph{De sph. et cyl.} II, 2 Coroll.]) and cone $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = the circle with the diameter $\beta\delta$ : the circle with the diameter $\epsilon\zeta = \beta\eta^2 : \eta\epsilon^2$, and $\beta\eta^2 = \gamma\eta \times \eta\alpha$, $\eta\epsilon^2 = \eta\alpha^2$, and $\gamma\eta \times \eta\alpha : \eta\alpha^2 = \gamma\eta : \eta\alpha$, therefore cone $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta = \gamma\eta : \eta\alpha$. But we have shown that cone $\beta\alpha\delta$ : segment $\beta\alpha\delta$ = $\gamma\eta : \eta\xi$, hence {\selectlanguage{greek}di' \~isou} segment $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\xi\eta : \eta\alpha$. And because $\alpha\chi : \chi\eta = \eta\alpha + 4\eta\gamma : \alpha\eta + 2\eta\gamma$ so inversely $\eta\chi : \chi\alpha = 2\gamma\eta + \eta\alpha : 4\gamma\eta + \eta\alpha$ and by addition $\eta\alpha : \alpha\chi = 6\gamma\eta + 2\eta\alpha : \eta\alpha + 4\eta\gamma$. But $\eta\xi = \frac{1}{4} (6\eta\gamma + 2\eta\alpha)$ and $\gamma\phi = \frac{1}{4} (4\eta\gamma + \eta\alpha)$; for that is evident. Hence $\eta\alpha : \alpha\chi = \xi\eta : \gamma\phi$, consequently also $\xi\eta : \eta\alpha = \gamma\phi : \chi\alpha$. But it was also demonstrated that $\xi\eta : \eta\alpha$ = the segment whose vertex is at $\alpha$ and whose base is the circle with the diameter $\beta\delta$ : the cone whose vertex is at $\alpha$ and whose base is the circle with the diameter $\epsilon\zeta$; hence segment $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\gamma\phi : \chi\alpha$. And since the cylinder $\mu$ is in equilibrium with the cone $\epsilon\alpha\zeta$ at $\alpha$, and $\theta$ is the center of gravity of the cylinder while $\phi$ is that of the cone $\epsilon\alpha\zeta$, then cone $\epsilon\alpha\zeta$ : cylinder