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Prop. 19
Prop. 21

(general diagram = diagram 3)

20. If a triangle is inscribed in a segment which is enclosed by a straight line and a section of a right-angled cone and has the same base as the segment and the same height, the inscribed triangle will be more than half the segment.

(diagram 1) For let segment ABG be as described, and let there be inscribed in it a triangle, ABG, having the same base as the segment and an equal height. And so since the triangle has the same base as the segment and an equal height, it is necessary that point B be the vertex of the segment. .(diagram 2) Therefore AG is parallel to the (line) touching the section at B. Let DE be drawn though B parallel to AG and from A, G, (lines) AD, GE parallel to the diameter. They will fall, in fact, outside the segment. And so since triangle ABG is half parallelogram ADEF, it obvious that it is larger than half the segment.

(diagram 3) Corollary: With this proved, it is clear that it is possible to inscribe a polygon in the segment so that the left over segments are less than (any proposed area. For always more than half being taken away, it is obvious, on account of this, that by repeatedly diminishing the remaining segments we will make these smaller than any proposed area.