The Duplicators, Part I: Eutocius’ Collection of Cube Duplications - Menaechmus' Second Solution

In another way. 

Analysis. Let ΑΒ and ΒΓ be two given straight lines, at right angles to each other, and let the means ΔΒ and BE between them supposed as being found, so that \begin{equation} \tag{84} \text{ΓΒ : ΒΔ = ΒΔ : ΒΕ = ΒΕ : ΒΑ, } \end{equation}  and let ΔΖ and ΕΖ be drawn at right angles. 

Since \begin{equation} \tag{85} \text{ ΓΒ : ΒΔ = ΔΒ : ΒΕ, } \end{equation} therefore \begin{equation} \tag{86} \text{ rect.(ΓΒΕ) = rect.(a given segment, ΒΕ) = sq.(ΒΔ) = sq.(ΕΖ).} \end{equation}

Since the rectangle contained by the given line and ΒΕ is equal to the square on ΕΖ, therefore the point Ζ is on a parabola whose axis is the line ΒΕ.

Again, since \begin{equation} \tag{87} \text{ ΑΒ : ΒΕ = ΒΕ : ΒΔ,} \end{equation}therefore \begin{equation} \tag{88} \text{ rect.(ΑΒΔ) = rect.(a given segment, ΒΔ) = sq.(ΕΒ) = sq.(ΔΖ),} \end{equation}therefore Ζ is on a parabola whose axis is the line ΒΔ. But it is also on the other parabola whose axis is ΒE: therefore Ζ is given. So too then are ΖΔ and ΖΕ: therefore Δ and Ε are given. 

Above: Menaechmus’ Second Diagram. The construction of this diagram was similar to that for Menaechmus’ first solution. I show here only the completed one. Unlike the previous diagram, the two given magnitudes are placed on the main figure (rather than to the side). Moving the points Α and Γ will adjust the ratio of ΑΒ to ΒΓ, which by default is 2:1.

Synthesis. This will be synthesized thusly. For let two given straight lines ΑΒ and ΒΓ be at right angles to one another, and let these straight lines be produced indefinitely from Β. With axis ΒΕ, let a parabola be drawn, so that the ordinates to ΒΕ are equal in square to the areas applied to ΒΓ.

Again, let a parabola be drawn with axis ΔΒ, so that the ordinates are equal in square to the areas applied to ΑΒ. Therefore the parabolas will intersect one another. Let them intersect at Ζ, and from Ζ, let ΖΔ and ΖΕ be drawn perpendicularly. 

So since ΖΕ (that is, ΔΒ) has been drawn ordinatewise to a parabola, therefore, by Conics I.11, for the blue parabola,  \begin{equation} \tag{89} \text{ rect.(ΓΒΕ) = sq.(ΒΔ).  } \end{equation} Therefore \begin{equation} \tag{90} \text{ ΓΒ : ΒΔ = ΔΒ : ΒΕ.} \end{equation} Again, since ΖΔ (that is, ΕΒ) has been drawn ordinatewise to a parabola, therefore, by Conics I.11, for the red parabola, \begin{equation} \tag{91} \text{rect.(ΔΒΑ) = sq.(ΕΒ). } \end{equation} Therefore \begin{equation} \tag{92} \text{ ΔΒ : ΒΕ = ΒΕ : ΒΑ, } \end{equation} and so \begin{equation} \tag{93} \text{ ΓΒ : ΒΔ = ΒΔ : ΒΕ = ΕΒ : ΒΑ,} \end{equation} the very thing it was required to find.