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consists of as many parallelograms as are in the figure circumscribed around the portion ABC.
Since, therefore, the bases of the portion and the triangle are equal, it appears that all the parallelograms will indeed have the same width. Hence, since the parallelogram BM is to ER as BG is to EH (that is, as K is to L), and it happens that BM is less than K, then ER will also be less than L. But all the triangles of which the excess of the circumscribed figure over the triangle DEF consists are equal to the parallelogram ER; therefore, the same excess is less than the space L. But the excess by which the circumscribed figure exceeds the portion ABC is also smaller than the space K. Therefore, both excesses together will be smaller than the given space K+L. And it is clear that what was proposed can be done.
A large geometric diagram showing two sets of inscribed rectangles within curved shapes (a portion of a hyperbola/ellipse/circle and a triangle), with labels A, B, G, M, C, D, E, H, R, F, K, and L.
1 By construction.
2 14.5. Elements.
If a figure is circumscribed in an ordered manner around a portion of a hyperbola, or a portion of an ellipse or circle (not greater than a semi-ellipse or a semicircle), the center of gravity of that figure will be on the diameter of the portion.