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

MPL, LOB, are equal to half the parallelogram OD or BF, because the bases of each are equal to the base DF, and the altitude of all of them together is the altitude of the parallelogram BF. By the same reasoning, the triangles that are on the other side of the diameter are equal to half the parallelogram BF. Therefore, all the triangles together, or the aforementioned excess, is equal to the parallelogram BF, and thus smaller than the given space. But the excess of the circumscribed figure over the portion ABC was already smaller than that same excess. Therefore, this excess is much smaller than the given space. And it appears that what was proposed can be done.

THEOREM II.

Given a portion of a hyperbola, or a portion of an ellipse or circle (not greater than a semi-ellipse or a semicircle), and a given triangle that has a base equal to the base of the portion; it is possible to circumscribe around both figures a shape made of parallelograms, all of which have the same width, such that the combined excess by which the circumscribed figures exceed the portion and the triangle is less than any given space.

Let there be a given portion ABC and a triangle DEF, with bases AC and DF being equal. Let the diameter of the portion be BG, and in the triangle, let a line EH be drawn from the vertex to the middle of the base. Let both BG and EH be either perpendicular to the bases or equally inclined. And in the same ratio that BG has to EH, let the given space be divided, and let the parts be K and L. Now let the figure be circumscribed in an ordered manner around the portion ABC, as before, which exceeds the portion by an excess smaller than space K. And around the triangle DEF, let there be circumscribed a figure which

A small geometric diagram shows a portion of a curve (A B C) and a triangle (D E F) with associated lines and segments.