The Duplicators, Part I: Eutocius’ Collection of Cube Duplications - Heron

Biographical note: Heron (or Hero) of Alexandria (ca 65 CE–ca 125 CE). Probably most familiar to modern students because of Heron’s formula for the area of a triangle in terms of its sides and semiperimeter. Heron wrote on a wide range of topics, from plane and solid geometry to optics, surveying, and mechanics; not all of his works survive. Read more about Heron at MacTutor.

(Heiberg 58.17) Let there be two given straight lines ΑΒ and ΒΓ, between which it is necessary to find two mean proportionals. Let them be placed, so that they contain a right angle at Β, and let the parallelogram ΒΔ be completed,^[11] and let ΑΓ and ΒΔ be joined.

[It is clear that they (ΑΓ and ΒΔ) are equal and they bisect each other (at E): for the circle being drawn around one of them will also pass through the extremities of the other, since the parallelogram is right-angled.]^[12]

Let ΔΓ and ΔΑ be projected [to Ζ, Η], and let a straightedge^[13] be conceived, as ΖΒΗ, being moved around some knob that remains fixed at Β, and moving, until it cuts the two segments from E equally, that is, ΕΗ and ΕΖ.

And let it (the straightedge) be conceived, having as position ΖΒΗ, and having cut equally, as was said before, the segments ΕΗ and ΕΖ. 

Above: Heron’s Diagram. I have added the circles: by moving the point Η, the radii of the circles are adjusted. When they coincide, the lines ΗΕ and ΕΖ are equal. It should be noted that this isn’t exactly what Heron is doing. The text refers to moving the ruler ΖΒΗ around Β, and hence the position of Η depends on the position of the ruler. In the diagram, I have made the ruler’s position depend on the position of Η instead. This makes the figure a bit easier to manipulate, and also seemed easier to implement in GeoGebra.

Let ΕΘ be drawn perpendicular to ΓΔ from Ε: it is clear that it bisects ΓΔ. So since it bisects ΓΔ at Θ, and ΓΖ is added, we have, by Elements II.6,  \begin{equation} \tag{11} \text{rect.(ΔZΓ) + sq.(ΓΘ) = sq.(ΘZ),} \end{equation}

Let the square on ΕΘ be added in common: therefore \begin{equation} \tag{12} \text{rect.(ΔZΓ) + sq.(ΓΘ) + sq.(ΘE) = sq.(ZΘ) + sq.(ΘE).} \end{equation}

Therefore by the Pythagorean Theorem [Elements I.47] \begin{equation} \tag{13} \text{rect.(ΔZΓ) + sq.(ΓE)  = sq.(EZ).} \end{equation}

Similarly it will be shown that \begin{equation} \tag{14} \text{rect.(ΔHA) + sq.(AE)  = sq.(EH).  } \end{equation}

And \begin{equation} \tag{15} \text{AE = EΓ, and HE = EZ,} \end{equation}

therefore \begin{equation} \tag{16} \text{rect.(ΔZΓ) = rect.(ΔHA). } \end{equation}

[But if the rectangle contained by the extremes is equal to the rectangle contained by the means, then the four straight lines are in proportion:] therefore, by Elements VI.16 \begin{equation} \tag{17} \text{ZΔ : ΔH :: AH : ΓZ} \end{equation}

But \begin{equation} \tag{18} \text{ZΔ : ΔH :: ZΓ : ΓB :: BA : AH.} \end{equation}

[Using Elements VI.2, for ΓΒ has been drawn parallel to one side of the triangle ΖΔΗ, namely ΔΗ, and ΑΒ has been drawn parallel to ΔΖ.] Therefore \begin{equation} \tag{19} \text{BA : AH :: AH : ΓZ :: ΓZ : ΓB.} \end{equation}

Therefore AH and ΓΖ are mean proportionals between ΑΒ and ΒΓ [which was required to find]. 

[11] Really, this is a rectangle. 

[12] This bracketed section is an insertion. In the Greek, ΑΓ, ΒΔ, and Ε are not explicitly referred to (only with pronouns). For clarity, I have added them here in parentheses. 

[13] I’ve chosen to translate τὸ κανόνιον here as straightedge, rather than cross-bar like I did with Plato’s construction, since Hero is not defining a sort of mechanical instrument so much as he is putting an ordinary line into a certain position.