2. If one magnitude is taken away from another magnitude and the center of gravity of the whole and of the part removed is not the same point, the center of gravity of the remaining portion may be found by prolonging the straight line which connects the centers of gravity of the whole and of the part removed, and setting off upon it another straight line which bears the same ratio to the straight line between the aforesaid centers of gravity, as the weight of the magnitude which has been taken away bears to the weight of the one remaining [\emph{De plan. aequil.} I, 8]. 3. If the centers of gravity of any number of magnitudes lie upon the same straight line, then will the center of gravity of all the magnitudes combined lie also upon the same straight line [Cf. \emph{ibid.} I, 5]. 4. The center of gravity of a straight line is the center of that line [Cf. \emph{ibid.} I, 4]. 5. The center of gravity of a triangle is the point in which the straight lines drawn from the angles of a triangle to the centers of the opposite sides intersect [\emph{Ibid.} I, 14]. 6. The center of gravity of a parallelogram is the point where its diagonals meet [\emph{Ibid.} I, 10]. 7. The center of gravity [of a circle] is the center [of that circle]. 8. The center of gravity of a cylinder [is the center of its axis]. 9. The center of gravity of a prism is the center of its axis. 10. The center of gravity of a cone so divides its axis that the section at the vertex is three times as great as the remainder. 11. Moreover together with the exercise here laid down I will make use of the following proposition: If any number of magnitudes stand in the same ratio to the same number of other magnitudes which correspond pair by pair, and if either all or some of the former magnitudes stand in any ratio whatever to other magnitudes, and the latter in the same ratio to the corresponding ones, then the sum of the magnitudes of the first series will bear the same ratio to the sum of those taken from the third series as the sum of those of the second series bears to the sum of those taken from the fourth series [\emph{De Conoid.} I]. \section*{Proposition I} Let $\alpha\beta\gamma$ [Fig.~1] be the segment of a parabola bounded by the straight line $\alpha\gamma$ and the parabola $\alpha\beta\gamma$. Let $\alpha\gamma$ be bisected at $\delta$, $\delta\beta\epsilon$ being parallel to the diameter, and draw $\alpha\beta$, and $\beta\gamma$. Then the segrnent $\alpha\beta\gamma$ will be $\frac{4}{3}$ as great as the triangle $\alpha\beta\gamma$.