\dotfill\linebreak produce $\alpha\gamma$ [Fig.~8] making $\alpha\theta = \alpha\gamma$ and $\gamma\xi$ = the radius of the sphere; imagine $\gamma\theta$ to be a scale-beam with a center at $\alpha$, and in the plane cutting % \begin{wrapfigure}{r}{0.4\textwidth} \includegraphics[width=0.4\textwidth]{fig08.png} \begin{center} {\small Fig.~8.} \end{center} \end{wrapfigure} % off the segment inscribe a circle with its center at $\eta$ and its radius = $\alpha\eta$; on this circle construct a cone with its vertex at $\alpha$ and its lateral boundaries $\alpha\epsilon$ and $\alpha\zeta$. Then draw a straight line $\kappa\lambda \| \epsilon\zeta$; let it cut the circumference of the segment at $\kappa$ and $\lambda$, the lateral boundaries of the cone $\alpha\epsilon\zeta$ at $\rho$ and $o$ and $\alpha\gamma$ at $\pi$. Now because $\alpha\gamma : \alpha\pi = \alpha\kappa^2 : \alpha\pi^2$ and $\kappa\alpha^2 = \alpha\pi^2 + \pi\kappa^2$ and $\alpha\pi^2 = \pi o^2$ (since also $\alpha\eta^2 = \epsilon\eta^2$), then $\gamma\alpha : \alpha\pi = \kappa\pi^2 + \pi o^2 : o\pi^2$. But $\kappa\pi^2 + \pi o^2 : \pi o^2$ = the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle with the diameter $o\rho$ and $\gamma\alpha = \alpha\theta$; therefore $\theta\alpha : \alpha\pi$ = the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle with the diameter $o\rho$. Now since the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle with the diameter $o\rho$ = $\alpha\theta : \pi\alpha$, let the circle with the diameter $o\rho$ be transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity; then $\theta\alpha : \alpha\pi$ = the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ in their present positions : the circle with the diameter $o\rho$ if it is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. Therefore the circles in the segment $\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ are in equilibrium at $\alpha$ with that in the cone $\alpha\epsilon\zeta$. And in the same way all circles in the segment $\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ in their present positions are in equilibrium at the point $\alpha$ with all circles in the cone $\alpha\epsilon\zeta$ if they are transferred and so arranged on the scate-beam at $\theta$ that $\theta$ is their center of gravity; then also the spherical segment $\alpha\beta\delta$ and the cone $\alpha\epsilon\zeta$ in their present positions are in equilibrium at the point $\alpha$ 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. Let the cyIinder $\mu\nu$ equal the cone whose base is the circle with the diameter