Let a segment of a right conoid cut off by a plane perpendicular to the axis be cut by another plane through the axis, and let the intersection in its surface be the parabola $\alpha\beta\gamma$ [Fig.~5] and let the common line of intersection of the plane which cut off the segment and of the intersecting plane be $\beta\gamma$; let the axis of the segment and the diameter of the parabola $\alpha\beta\gamma$ be $\alpha\delta$; produce $\delta\alpha$ so that $\alpha\theta = \alpha\delta$ and imagine $\delta\theta$ to be a scale-beam with its center at $\alpha$; % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig05.png} \begin{center} {\small Fig.~5.} \end{center} \end{wrapfigure} % then inscribe a cone in the segment with the lateral boundaries $\beta\alpha$ and $\alpha\gamma$ and in the parabola draw a straight line $\xi o \| \beta\gamma$ and let it cut the parabola in $\xi$ and $o$ and the lateral boundaries of the cone in $\pi$ and $\rho$. Now because $\xi\sigma$ and $\beta\delta$ are drawn perpendicular to the diameter of the parabola, $\delta\alpha : \alpha\sigma = \beta\delta^2 : \xi\sigma^2$ [\emph{Quadr. parab.} 3]. But $\delta\alpha : \alpha\sigma = \beta\delta : \pi\sigma = \beta\delta^2 : \beta\delta \times \pi\sigma$, therefore also $\beta\delta^2 : \xi\sigma^2 = \beta\delta^2 : \beta\delta \times \pi\sigma$. Consequently $\xi\sigma^2 = \beta\delta \times \pi\sigma$ and $\beta\delta : \xi\sigma = \xi\sigma : \pi\sigma$, therefore $\beta\delta : \pi\sigma = \xi\sigma^2 : \sigma\pi^2$. But $\beta\delta : \pi\sigma = \delta\alpha : \alpha\sigma = \theta\alpha : \alpha\sigma$, therefore also $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$. On $\xi o$ construct a plane perpendicular to $\alpha\delta$; this will intersect the segment of the right conoid in a circle whose diameter is $\xi o$ and the cone in a circle whose diameter is $\pi\rho$. Now because $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$ and $\xi\sigma^2 : \sigma\pi^2$ = the circle with the diameter $\xi o$ : the circle with the diameter $\pi\rho$, therefore $\theta\alpha : \alpha\sigma$ = the circle whose diameter is $\xi o$ : the circle whose diameter is $\pi\rho$. Therefore the circle whose diameter is $\xi o$ will in its present position be in equilibrium at the point $\alpha$ with the circle whose diameter is $\pi\rho$ when this is so transferred to $\theta$ on the scale-beam that $\theta$ is its center of gravity. Now since $\sigma$ is the center of gravity of the circle whose diameter is $\xi o$ in its present position, and $\theta$ is the center of gravity of the circle whose diameter is $\pi\rho$ if its position is changed as we have said, and inversely $\theta\alpha : \alpha\sigma$ = the circle with the diameter $\xi o$ : the circle with the diameter $\pi\rho$, then the circles are in equilibrium at the point $\alpha$. In the same way it can be shown that if another