Proposition I

Let αβγ [Fig. 1] be the segment of a parabola bounded by the straight line αγ and the parabola αβγ. Let αγ be bisected at δ, δβε being parallel to the diameter, and draw αβ, and βγ. Then the segment αβγ will be 4/3 as great as the triangle αβγ.

From the points α and γ draw αζ δβε, and the tangent γζ; produce [γβ to κ, and make κθ = γκ]. Think of γθ as a scale-beam with the center at κ and let μξ be any straight line whatever δ. Now since γβα is a parabola, γζ a tangent and γδ an ordinate, then εβ = βδ; for this indeed has been proved in the Elements [i.e., of conic sections, cf. Quadr. parab. 2]. For this reason and because ζα and μξ εδ, μν = νξ, and ζκ = κα. And because γα: αξ = μξ: ξo (for this is shown in a corollary, [cf. Quadr. parab. 5]), γα: αξ = γκ: κν; and γκ = κθ, therefore θκ: κν = μξ: ξo. And because ν is the center of gravity of the straight line μξ, since μν = νξ, then if we make τ η = ξo and θ as its center of gravity so that τ θ = θη, the straight line τ θη will be in equilibrium with μξ in its present position because θν is divided in inverse proportion to the weights τ η and μξ, and θκ: κν = μξ: ητ; therefore κ is the center of gravity of the combined weight of the two. In the Fig. 1. same way all straight lines drawn in the triangle ζαγ∥δ are in their present positions in equilibrium with their parts cut off by the parabola, when these are transferred to θ, so that κ is the center of gravity of the combined weight of the two. And because the triangle γζα consists of the straight lines in the triangle γζα and the segment αβγ consists of those straight lines within the segment of the parabola corresponding to the straight line ξo, therefore the triangle ζαγ in its present position will be in equilibrium at the point κ with the parabola-segment when this is transferred to θ as its center of gravity, so that κ is the center of gravity of the combined weights of the two. Now let γκ be so divided at χ that γκ = 3κχ; then χ will be the center of gravity of the triangle αζγ, for this has been shown in the Statics [cf. De plan. aequil. I, 15, p. 186, 3 with Eutokios, S. 320, 5ff.]. Now the triangle ζαγ in its present position is in equilibrium at the point κ with the segment βαγ when this is transferred to θ as its center of gravity, and the center of gravity of the triangle ζαγ is χ; hence triangle αζγ: segm. αβγ when transferred to θ as its center of gravity = θκ: κχ. But θκ = 3κχ; hence also triangle αζγ = 3 segm. αβγ. But it is also true that triangle ζαγ = 4∆αβγ because ζκ = κα and αδ = δγ; hence segm. αβγ = 4/3 the triangle αβγ. This is of course clear.

It is true that this is not proved by what we have said here; but it indicates that the result is correct. And so, as we have just seen that it has not been proved but rather conjectured that the result is correct we have devised a geometrical demonstration which we made known some time ago and will again bring forward farther on.