(diagram 1) 9. Let AG again be a balance, and is middle B, and triangle GDK be scalene having its base, DK, and height EG, 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 triangle DGK holding as it is now positioned, and let triangle GDK have this ratio to L which AB has to BE. I say that Z is larger than L, but less than DGK.

It will be proved as in the previous case.

(diagram 2) Here is a sample proof, which is unlike the previous, but whose method could have been used for Props. 8-14. Take the center of KGD. Note that it will be inside KDG, so that we do not actually have to know where it is. (see Equilibria of Planes, Assumption 7) Now take the point M on the balance that the perpendicular from the center falls on, which must be between E and G, as KD is perpendicular to AG and meets AG at E. (diagram 3) Since Z and KGD incline equally, AB : BM = KGD : Z, while AB : BE = KGD : L. Also, AB = BG.

Hence, BG : BM : BE = KGD : Z : L. BG > BM > BE. Hence, KGD > Z > L.

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