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of Ancient Mathematics
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(diagram 2 = general diagram)
(diagram 1) 8. Let ABG be a balance and B its middle, and let it be suspended at B, but let triangle GDE be right-angled, having the angle at E as right, and let it be suspended from the balance at G, E, and let area Z be suspended at A and let it incline equally to GDE holding as it is now positioned, and let triangle GDE have to area K this ratio, that which AB has to BE. I say that area Z is less than triangle GDE and larger than K.
(diagram 2) Let the center of weight of triangle DEG be taken and let it be Q, and let QH be drawn parallel to DE. (diagram 3) And so, since triangle GDE inclines equally area Z, GDE has the same ratio to Z which AB has to BH. Thus Z is less than GDE. (diagram 4) And since triangle GDE has this ratio to Z which BA has to BH, but to K which BA has to BE, it is clear that triangle GDE has a larger ratio to K than to Z. Thus Z is larger than K.