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* Theorem 1.
much greater to P. d. h.
or B F, as was demonstrated above *; the ratio of the portion A B C to the said excess will be greater than the ratio of the triangle A B C to the parallelogram B F, that is, greater than the ratio of A D to D F, or of L K to K E. Let M K therefore be to K E as the portion A B C is to the excess by which it is surpassed by the orderedly circumscribed figure. Thus, since K is the center of gravity of the figure circumscribed around the portion, and E is the center of gravity of the portion itself, M will be the center of gravity of all the spaces that constitute the same excess. Which cannot be; for if a line is drawn through M parallel to the diameter B D, all the spaces we have mentioned would be on one side. It is manifest, therefore, that the center of gravity of the portion A B C is on the diameter B D of the portion.
s. lib. 1. Arch. on Equilibrium.
Now let A B C be a portion of an ellipse or circle, greater than half the figure. Let the figure be completed, and let B D be produced until it meets the section at E; therefore, E D will be the diameter of the portion A E C,
Two geometric diagrams side by side depict an ellipse and a circle. Both diagrams show a horizontal chord A C intersected by a vertical diameter B E, with point D marking the intersection of the chord and the diameter. These represent the geometric construction for locating the center of gravity in a segment larger than a semicircle.
and B D E will be the diameter of the entire figure. And since the center of gravity of the entire figure is on the diameter B D E (for this will be established from what was previously demonstrated, if the entire figure is divided into two equal parts by a diameter that is parallel to A C),