Christiani Hugenii ...Theoremata De Quadratura Hyperboles, Ellipsis Et Circuli, Ex Dato Portionum Gravitatis Centro

Let there be any such portion ABC, and its diameter BD; and let the figure be circumscribed around it in an ordered manner as above. It must be shown that the center of gravity of that figure will be on the diameter BD. Let lines HK, NR, PS be drawn, connecting the top sides of the parallelograms that are equally distant on either side from the diameter of the portion.

Since, therefore, FH and LK are parallel to the diameter BD, and DF and DL are equal, the line HK, which connects the two FH and LK, must be bisected by the diameter BD; wherefore the same HK will be parallel to the base AC, and EHKG will be a straight line.

A geometric diagram shows a trapezoidal shape inscribed with multiple vertical and horizontal lines, labeled with points A, F, D, L, C at the base, E, H, K, G at the sides, and P, B, Q, S, R, M, N at the top and interior.

Thus, EC is a parallelogram; since the diameter BD bisects its opposite sides, the center of gravity of the parallelogram will be on it. By the same reasoning, the parallelograms will be HM, NO, PQ, and the centers of gravity of each will be on the line BD. Therefore, the center of gravity of the figure composed of all said parallelograms must also be found on the same BD. However, that figure is the same one that had been orderedly circumscribed around the portion. Therefore, it is established that the center of gravity of the figure orderedly circumscribed around the portion is on the diameter BD of the portion. Which was to be demonstrated.

1 5. book 2. Conics.

2 9. book 1. Archimedes on Equilibrium.