From the points $\alpha$ and $\gamma$ draw $\alpha\zeta \| \delta\beta\epsilon$, and the tangent $\gamma\zeta$; produce [$\gamma\beta$ to $\kappa$, and make $\kappa\theta = \gamma\kappa$]. Think of $\gamma\theta$ as a scale-beam with the center at $\kappa$ and let $\mu\xi$ be any straight line whatever $\| \epsilon\delta$. Now since $\gamma\beta\alpha$ is a parabola, $\gamma\zeta$ a tangent and $\gamma\delta$ an ordinate, then $\epsilon\beta = \beta\delta$; for this indeed has been proved in the Elements [i.e., of conic sections, cf. \emph{Quadr. parab.} 2]. For this reason and because $\zeta\alpha$ and $\mu\xi \| \epsilon\delta$, $\mu\nu = \nu\xi$, and $\zeta\kappa = \kappa\alpha$. And because $\gamma\alpha : \alpha\xi = \mu\xi : \xi o$ (for this is shown in a corollary, [cf. \emph{Quadr. parab.} 5]), $\gamma\alpha : \alpha\xi = \gamma\kappa : \kappa\nu$; and $\gamma\kappa = \kappa\theta$, therefore $\theta\kappa : \kappa\nu = \mu\xi : \xi o$. And because $\nu$ is the center of gravity of the straight line $\mu\xi$, since $\mu\nu = \nu\xi$, then if we make $\tau\eta = \xi o$ and $\theta$ as its center of gravity so that $\tau\theta = \theta\eta$, the straight line $\tau\theta\eta$ will be in equilibrium with $\mu\xi$ in its present position because $\theta\nu$ is divided in inverse proportion to the weights $\tau\eta$ and $\mu\xi$, and $\theta\kappa : \kappa\nu = \mu\xi : \eta\tau$; therefore $\kappa$ is the center of gravity of the combined weight of the two. In the % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig01.png} \begin{center} {\small Fig.~1.} \end{center} \end{wrapfigure} % same way all straight lines drawn in the triangle $\zeta\alpha\gamma \| \epsilon\delta$ are in their present positions in equilibrium with their parts cut off by the parabola, when these are transferred to $\theta$, so that $\kappa$ is the center of gravity of the combined weight of the two. And because the triangle $\gamma\zeta\alpha$ consists of the straight lines in the triangle $\gamma\zeta\alpha$ and the segment $\alpha\beta\gamma$ consists of those straight lines within the segment of the parabola corresponding to the straight line $\xi o$, therefore the triangle $\zeta\alpha\gamma$ in its present position will be in equilibrium at the point $\kappa$ with the parabola-segment when this is transferred to $\theta$ as its center of gravity, so that $\kappa$ is the center of gravity of the combined weights of the two. Now let $\gamma\kappa$ be so divided at $\chi$ that $\gamma\kappa = 3\kappa\chi$; then $\chi$ will be the center of gravity of the triangle $\alpha\zeta\gamma$, for this has been shown in the Statics [cf. \emph{De plan. aequil.} I, 15, p. 186, 3 with Eutokios, S. 320, 5ff.]. Now the triangle $\zeta\alpha\gamma$ in its present position is in equilibrium at the point $\kappa$ with the segment $\beta\alpha\gamma$ when this is transferred to $\theta$ as