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E D B is equal to these three rectangles: S D P, D P B, and the rectangle under E D and P B; but the last two of these equal the rectangle E P B; therefore, the rectangle E D B is equal to two: namely, the rectangle S D P and E P B, from which it appears that the excess of the rectangle E D B over the rectangle E P B is equal to the rectangle S D P.
Given a portion of a hyperbola, or a portion of an ellipse or circle, not greater than half the figure; if a triangle of this kind is constructed at the diameter, which has its vertex at the center of the figure, and a base equal and parallel to the base of the portion; and such that the segment which extends from the vertex to the middle of the base is of such magnitude that it can equal the rectangle comprised by the lines which are placed between the base of the portion and the endpoints of the diameter of the figure. The center of gravity of the magnitude composed of the portion and the prescribed triangle will be the same point as the vertex of the triangle, namely, the center of the figure.
Let there be given a portion of a hyperbola, or a portion of an ellipse or circle not greater than half the figure, A B C. Let its diameter be B D, and the diameter of the figure be B E, in the middle of which is the center of the figure F. And let F G be taken such that it equals the rectangle B D E, and having drawn K G H equal and parallel to the base A C, which is bisected at G, let K F and F H be joined. It must be demonstrated that the center of gravity of the magnitude composed of the portion A B C and the triangle K F H is the point F.