Return to Vignettes of Ancient Mathematics
Return to Archimedes, Quadrature of the Parabola, content
Prop. 16
Prop. 18

(general diagram = diagram 3)

17. With this proved, it is obvious that every segment enclosed by a straight-line and section of a right-angled cone is a third again the triangle having a base that is the same as the segment and an equal height.

(diagram 1) For let there be a segment contained by a straight-line and a section of a right-angled cone, and let point Q be its vertex, and let triangle BQG be inscribed in it, having the same base as the segment and equal height. (diagram 2) And so, since point Q is the vertex of the segment, the straight line drawn from Q parallel to the diameter bisects BG and BG is parallel to the line touching the section at Q.(*1*) And let EQ be drawn parallel to the diameter (diagram 3) and let BD also be drawn from B parallel to the diameter, and let GD be drawn from G and touching the section of the cone at G.

And so since KQ is parallel to the diameter, and GD is touching the section at G, and EG is parallel to the line touching the section at Q, triangle BDG is four-times triangle BQG.(*2*) But since triangle BDG is three-times segment BQG (Prop. 16), and four-times triangle BQG, it is clear that segment BQG is a third again triangle BQG.

(*1*) The logical structure of this statement is somewhat problematic. The inference is:

Q is a vertex => QE (diameter or parallel to the diameter) bisects BG & the tangent at Q is parallel to BG.

First, we do not know what a vertex is until after this proposition. So, perhaps as part of the mechanical features of the proof or something like that, we suppose what a vertex is. As Heiberg notes, the statement is the converse of Prop. 18, which will be proved there. The proof is trivial if we have the assumptions that each segment has only one vertex and at each point there is but one tangent to the curve (proved as part of Prop. 2). Both consequences are required for the theorem, or so it seems (see next note).

back

(*2*) KQ is parallel to the diameter & GD is tangent at G & EG is parallel to the tangent at Q => triangle BDG is 4*triangle BQG. The logical structure of this argument is a little obscure. It is enough to show that BQG = EGK and that EGK = 1/4*BDG. So it is needs to be broken down into several inferences. Here is but one suggestion.

  1. KQ is parallel to the diameter => KQ is parallel to BD -- this means that triangle BDG : triangle EKG = T(BD) : T(EK) = T(BG) : T(EG)
  2. KQ is parallel to the diameter & GD is tangent at G => KQ = EQ => triangle EQG = triangle GQK
  3. EG is parallel to the tangent at Q => EG = BE => triangle BEQ = triangle EQG
  4. triangle EQG = triangle GQK & triangle BEQ = triangle EQG => triangles BEQ + EQG = triangles EQG + GQK => triangle BQG = triangle EKG
  5. BE = EG => triangle EGK = 1/4* triangle BDG

back

top