**Archimedes,**__Quadrature of the Parabola__Prop. 9- translated by Henry Mendell (Cal. State U., L.A.)

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Prop. 8

Prop. 10

(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.