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Prop. 4
Prop. 6

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Statement (general diagram)

(diagram 1) Let there be a segment ABG enclosed by a straight-line and a section of a right-angled cone,Theorem 5 and from A and parallel to the diameter let ZA be drawn, and from G and touching the section of the right-angled cone at G let GZ be drawn. If in fact some line parallel to AZ be drawn in triangle ZAG, the line drawn will be cut in the same ratio by the section of a right-angled cone as AG by the line drawn [proportionally], but the segment of AG at A will be homologous (same parts of their ratios) as the segment of the line drawn at A.

The property Archimedes describes is: QK : LQ = AK : KG.

Proof
(diagram 2) Let some line DE parallel to AZ be drawn, and first let DE bisect AG.  And so since ABG is a section of a right-angled cone and BD is drawn parallel to the diameter, but AD and DG are equal, then the line touching the section of a right-angled cone at B will be parallel to AG (Prop. 1).  Again, since DE is parallel to the diameter, and GE was drawn from G and touching the section of a right-angled cone at G, but DG is parallel to the line touching at B, therefore EB is equal to BD (Prop. 2).(*1*) Thus, AD has the same ratio to DG which DB has to BE.  And so it has been proved if the drawn line bisects AG.(*2*)

(diagram 3) Otherwise, some other line KL will be parallel to AZ.  And so we must prove that AK has the same ratio to KG which KQ has to QL.  (diagram 4) For since BE is equal to BD, IL is also equal to KI.  Therefore LK has the same ratio to KI which AG has to DA.  (diagram 5) But KI also has to KQ the same ratio which DA has to AK.  For this was proved previously (Prop. 4).  (diagram 6) Thus KQ has the same ratio to QL which AK has to KG.(*3*)  And so the proposed has been proved.

(*1*) It is an interesting fact worth noting that the tangent is taken as given, although there is no procedure for constructing than using the property that the line from the bisection of the base to the vertex equals the extension of this line from the vertex to the tangent. We may imagine, however, that the tangent was constructed from some other segment of the parabola, or that it is just given.

(*2*) It becomes clear why this trivial case is required in the proof of the non-trivial case. Prop. 4 does not cover the situation where the line parallel to the diameter bisects the base.

Theorem 5(*3*) It seems that to prove the step, we need to break it up into cases, where QK < IK and where QK > IK.

Let QK < IK.
1. KI : QI = AD : KD (Prop. 4)
2. LK : IK = AG : AD (proved above)
3. KI : KI-IQ = AD : AD-DK (1 separando)
4. KI : KQ = AD : AK
5. LK : KQ = AG : AK (2, 4 ex aequali)
6. LQ : KQ = GK : AK (5 separando)

Let QK > IK
1. KI : QI = AD : KD (Prop. 4)
2. LK : IK = AG : AD (proved above)
3. IK : QK = AD : KD+AD = AD : AK (1 componendo)
4. LK : QK = AG : AK (ex aequali)
5. LQ : QK = GK : AK (separando)

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