22. If a segment is enclosed by a straight line and a section of a right-angled cone, and areas are positioned successively, however many, in a ratio of four-times, and the largest of the areas is equal to the triangle having having the same base as the segment and height the same, then the areas altogether will be smaller than the segment.

(diagram 1) For let there be a segment, ADBEG enclosed by a straight line and a section of a right-angle cone, and let there be areas, however many, successively positioned, Z, H, Q, I, and let each leading area be four-times the following, and let Z be the largest, and let Z be equal to the triangle having its base the same as the segments and an equal height. I say that the segment is larger than areas Z, H, Q, I.

(diagram 2) Let there be a vertex of the whole segment, B, and of the left over segments, D, E. And so since triangle ABG is eight-times each of triangles ADB, BEG, it is clear that it is four-times them together. And since triangle ABG is equal to area Z, according to the same reasoning, in fact, triangles ADB, BEG are also equal to area H. It will be similarly proved that the triangles inscribed in the segments left over are also triangles having the same base as the segments and the same height (and) are equal to Q and that the triangles inscribed in the segments that come about later are equal to area I. (diagram 3) Therefore the proposed areas altogether will be equal to some inscribed polygon inscribed in the segment. And so, it is obvious that they are smaller than the segment.

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