equilibrium at the point $\alpha$ with the two circles produced in the sphere and the cone when they are transferred and so arranged on the scale-beam at the point $\theta$ that $\theta$ is the center of gravity of both. Therefore if cylinder, sphere and cone are filled up with such circles then the cylinder in its present position will be in equilibrium at the point $\alpha$ with the sphere and the cone together, if they are transferred and so arranged on the scale-beam at the point $\theta$ that $\theta$ is the center of gravity of both. Now since the bodies we have mentioned are in equilibrium, the cylinder with $\kappa$ as its center of gravity, the sphere and the cone transferred as we have said so that they have $\theta$ as center of gravity, then $\theta\alpha : \alpha\kappa =$ cylinder $:$ sphere $+$ cone. But $\theta\alpha = 2\alpha\kappa$, and hence also the cylinder $= 2 \times$ (sphere $+$ cone). But it is also true that the cylinder = 3 cones [Euclid, \emph{Elem.} XII, 10], hence 3 cones = 2 cones + 2 spheres. If 2 cones be subtracted from both sides, then the cone whose axes form the triangle $\alpha\epsilon\zeta =$ 2 spheres. But the cone whose axes form the triangle $\alpha\epsilon\zeta =$ 8 cones whose axes form the triangle $\alpha\beta\delta$ because $\epsilon\zeta = 2\beta\delta$, hence the aforesaid 8 cones = 2 spheres. Consequently the sphere whose greatest circle is $\alpha\beta\gamma\delta$ is four times as large as the cone with its vertex at $\alpha$, and whose base is the circle on the diatneter $\beta\delta$ perpendicular to $\alpha\gamma$. Draw the straight lines $\phi\beta\chi$ and $\psi\delta\omega \| \alpha\gamma$ through $\beta$ and $\delta$ in the parallelogram $\lambda\zeta$ and imagine a cylinder whose bases are the circles on the diameters $\phi\psi$ and $\chi\omega$ and whose axis is $\alpha\gamma$. Now since the cylinder whose axes form the parallelogram $\phi\omega$ is twice as large as the cylinder whose axes form the parallelogram $\phi\delta$ and the latter is three times as large as the cone the triangle of whose axes is $\alpha\beta\delta$, as is shown in the Elements [Euclid, \emph{Elem.} XII, 10], the cylinder whose axes form the parallelogram $\phi\omega$ is six times as large as the cone whose axes form the triangle $\alpha\beta\delta$. But it was shown that the sphere whose largest circle is $\alpha\beta\gamma\delta$ is four times as large as the same cone, consequently the cylinder is one and one half times as large as the sphere, Q. E. D.