Let a sphere be cut by a plane through its center intersecting the surface in the circle $\alpha\beta\gamma\delta$ [Fig.6], $\alpha\gamma$ and $\beta\delta$ being two diameters of the circle perpendicular to each other. Let a plane be constructed on $\beta\delta$ perpendicular to $\alpha\gamma$. Then imagine a cone whose base is the circle with the diameter $\beta\delta$, whose vertex is at $\alpha$ and its lateral boundaries are $\beta\alpha$ and $\alpha\delta$; let $\gamma\alpha$ be produced so that $\alpha\theta = \gamma\alpha$, imagine the straight line $\theta\gamma$ to be a scale-beam with its center at $\alpha$ and in the semi-circle $\beta\alpha\delta$ draw a straight line $\xi o \| \beta\delta$; let it cut the circumference of the semicircle in $\xi$ and $o$, the lateral boundaries of the cone in $\pi$ and $\rho$, and $\alpha\gamma$ in $\epsilon$. On $\xi o$ construct a plane perpendicular to $\alpha\epsilon$; it will intersect the hemisphere in a circle with the diameter $\xi o$, and the cone in a circle with the diameter $\pi\rho$. Now because $\alpha\gamma : \alpha\epsilon = \xi\alpha^2 : \alpha\epsilon^2$ and $\xi\alpha^2 = \alpha\epsilon^2 + \epsilon\xi^2$ and $\alpha\epsilon = \epsilon\pi$, therefore $\alpha\gamma : \alpha\epsilon = \xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$. But $\xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$ = the circle with the diameter $\xi o$ + the circle with the diameter $\pi\rho$ : the circle with the diameter $\pi\rho$, and $\gamma\alpha = \alpha\theta$, hence $\theta\alpha : \alpha\epsilon$ = the circle with the diameter $\xi o$ + the circle with % \begin{wrapfigure}{r}{0.33\textwidth} \includegraphics[width=0.33\textwidth]{fig06.png} \begin{center} {\small Fig.~6.} \end{center} \end{wrapfigure} % the diameter $\pi\rho$ : circle with the diameter $\pi\rho$. Therefore the two circles whose diameters are $\xi o$ and $\pi\rho$ in their present position are in equilibrium at the point $\alpha$ with the circle whose diameter is $\pi\rho$ if it is transferred and so arranged at $\theta$ that $\theta$ is its center of gravity. Now since the center of gravity of the two circles whose diameters are $\xi o$ and $\pi\rho$ in their present position [is the point $\epsilon$, but of the circle whose diameter is $\pi\rho$ when its position is changed is the point $\theta$, then $\theta\alpha : \alpha\epsilon$ = the circles whose diameters are] $\xi o$ [,$\pi\rho$ : the circle whose diameter is $\pi\rho$. In the same way if another straight line in the] hemisphere $\beta\alpha\delta$ [is drawn $\| \beta\delta$ and a plane is constructed] perpendicular to [$\alpha\gamma$ the] two [circles produced in the cone and in the hemisphere are in their position] in equilibrium at $\alpha$ [with the circle which is produced in the cone] if it is transferred and arranged on the scale at $\theta$. [Now if] the hemisphere and the cone [are filled up with circles then all circles in the] hemisphere