(diagram 1) 6. Let there be conceived the proposed seen plane, [which is under contemplation], upright to the horizon and let there be conceived [then] things on the same side as D of line AB as being downwards, and on the other upwards, and let triangle BDG be right-angled, having its right angle at B and the side BG equal to half of the balance (AB being clearly equal to BG), and let the triangle be suspended from point BG, and let another area, Z, be suspended from the other part of the balance at A, and let area Z, suspended at A, incline equally to the BDG triangle holding where it now lies. I say that area Z is a third part of triangle BDG.

(diagram 2) For since the balance is supposed as inclining equally, line AG is parallel to the horizon, the lines drawn at right angles to AG on the plane upright to the horizon will be perpendicular to the horizon. Let line BG be cut at E in this way, so that GE is double EB, and let KE be drawn parallel to DB, and let it be bisected at Q; point Q is the center of weight of triangle BDG. For this was shown in the Mechanics.(*) And so if the hanging of triangle BDG at B, G is released, but it is suspended at E, the triangle will remain as it now holds. For each of the things suspended remains at the point from which it is fixed, so that the point of the suspended thing and the center of weight of the thing suspended will be along a perpendicular. For this too has been proved.(**) And so since triangle BDG will have this fixed setting to the balance, area Z similarly inclines equally. (diagram 3) Since Z being suspended at A and BDG at E incline equally, it is clear that they are reciprocally proportional to the lengths, and as AB is to BE so triangle BDG to area Z. But AB is three-times BE. Therefore triangle BDG is three-times area Z. It is clear also that if triangle BDG is three times area Z, it inclines equally to it.

(*) Is this a work by Archimedes or some work by someone else which is familiar enough to be referred to in this way. On Floating bodies II 2 also refers to an Elements of Mechanics for a proposition that is part of Plane Equilibria I 8, namely that if a magnitude is taken away from another magnitude, the centers of weight of the three magnitudes (the whole, the one removed and the one left) are on a straight-line. This does not actually seem to be proved in Plane Equilibria I 8, however. See the similar reference to Conic Elements after prop. 3. The theorem does not appear in this form at Plane Equilibria I 14, but part of it appears in Heron's Mechanica II 35, where the rest of the property may be derived trivially. See the note to the proposition there. Plane Equilibria II relies on the theorem proved in Quadrature of the Parabola, and so it is generally assumed that it is a later work. Plane Equilibria I is very different in its treatment of ‘equal inclination’, in that both the horizon and the balance beam play explicit roles here. Although both concern treatments of mechanical issues, the Quadrature of the Parabola seems to be based on theorems that in fact are more general than those proved in Plane Equilibria I.

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(**) See previous note. This theorem does not appear in any extant work of Archimedes; in particular it is reasonable to read Plane Equilibria as always assuming that the center of weight is on the balance beam, or rather that the balance beam just is the line connecting centers of weight.