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Note:  the diagrams are conceived as being seen from above looking down on the balance (an observation of Knorr).

p. 1
(figure 1) Let there be a segment ABG enclosed by straight line AG and the sector of a right angled cone.

Note:  "the sector of a right angled cone" is Archimedes' term for a parabola.  'Parabola' occurs twice in the Archimedes' works, both times in the Method, but keep in mind that his works as we have them are the result of much editing.  Following Dijsterhuis, Archimedes (Princeton, 1987), I shall used the term 'orthotome' or right sector for 'parabola'

Here the line from AC to the parabola is the axis of the parabola.

(figure 2) Let AG be bisected at D,  (figure 3) and let there be drawn DBE parallel to the diameter.

Note:  BD is the diameter of the segment ABG, i.e. it bisects AG and is parallel to the main diameter (drawn but unlettered).

(figure 4) Join AB, BG.

p. 2

(figure 5) I say that the segment ABG is 1 1/3 the triangle ABG.

(figure 6) Let there be let from points AG, AZ parallel to DBE, GZ tangent to the orthotome

(figure 7) and extend GB to K.  and let QK = GK.  Conceive of GQ as a balance and its middle as K

(figure 8) And draw an arbitrary line  MX parallel to ED.

p. 3
(figure 9) And so since GBA is a parabola, and GZ is tangent, and GD is ordinate, EB = BD.  For this was shown in the Elements.

Note:  Archimedes draws on an introduction to conic sections which is not extant.  The theorem is Quadrature of Parabola, prop. 2, provided there without proof.

Note also the reference to the curve as a parabola.

The lost Elements treats the ordinate as any line forming a orthotomic (parabolic) segment.


an ordinate of a segment is any line parallel to the base of an orthotomic segment.

the diameter of a segment bisects the ordinate and is parallel to the main diameter of the orthotome (indicated but unlettered).

Theorem:  if the tangent (ZG) to the orthotome segment (ABC) at either of the endpoints of the ordinate and the diameter BD of the segment is extended so as to meet the tangent, the parabola bisects (at B) the diameter from the intersection to the ordinate (EB = BD)

(figure 10) For this reason (EB = BD), and because FA and MO are parallel to ED, MN is equal NO and FK is equal to KA (through similar triangles FCA, MCO, ECD).

p. 4
And since as GA is to AX so is MX : XO (for this was proved in a lemma),

Heiberg excises the reference to the lemma.  In any case, the lemma is Quadrature of the Parabola, prop. 5.

p. 5

but as AG is to AX, so is GK to KN

and QK is equal to KG

Therefore, as QK is to KN so is MX to XO.

p. 6

(figure 13) And since MN is equal to NX, if then we place TH equal to XO with its center of weight Q, so that TQ is equal to QH, MX remaining in its position will balance TQH.

(figure 14) For QN is cut in inverse proportion to weights TH and MX, and as QK is to KN so is MX to HT.  Thus K is the center of the weight composed from both.

p. 7
Similarly all such parallels to EK drawn in triangle ZAG will balance the segments cut off from them by the orthotome and transfered to Q, so that K is the center of weight composed from both.

p. 8

(figure 16) And since triangle GZA  is composed of the lines in triangle GZA and segment ABG is composed of the lines taken in the same way as XO, therefore triangle ZAG remaininging in its position will balance at point K the segment of the orthotome with its center of weight placed at Q, so that K is the center of weight of both.

(figure 17) Let GK be cut at C so that GK is 3-times KC.  Therefore, point C will be the center of weight of triangle AZG.  For this was shown in On the Equilibrium (I 15).

And so since triangle ZAG remaining in its position balances at K segment BAG with its center of weight placed at Q, and C is the center of weight of triangle ZAG, therefore as triangle AZG is to segment ABG positioned at center Q, so is QK to CK.

But QK is 3-times KC.  Therefore, triangle AZG is 3-times segment ABG.

(figure 18) But triangle ZAG is 4-times segment ABG, since ZK is equal to KA and AD is equal to DG.

ZK = KA => triangle AKG = triangle KZG = 1/2 triangle ZAG.
AD = DG and BD is parallel to AK => line KB = BG => triangle KAB = triangle BAG (their altitudes will be the same) = 1/2 triangle AKG = 1/4 triangle ZAG.

Hence, segment ABG is 1 1/3 triangle ABG.

Segment GAB = 1/3 of triangle AZG = 1/3 (4*triangle ABG).

But this is not proved through what we now said, but it has made an impression that the conclusion is true.  Because although we see it as not demonstrated, we suspect that the conclusion is true, we will prescribe the geometrical demonstration which we discovered and which was published earlier.