Biographical note: Nicomedes (ca. 280 BCE – ca. 210 BCE). We know very little about Nicomedes, except that he used the conchoid curve to solve both the problem of trisecting an angle and of duplicating a cube. He apparently also worked on the problem of squaring the circle. Read more about Nicomedes at MacTutor.
(Heiberg 98.2) Nicomedes, in his work On Conchoid Lines, writes about the construction of a mechanism satisfying the same requirement [of finding two mean proportionals]. From the book, he seems to have been especially proud of himself while ridiculing Eratosthenes’ solutions as awkward and lacking any geometrical skill. So as to not exclude anyone who dealt with this problem, and to compare with Eratosthenes, we add it to those already mentioned. He writes essentially^{29} the following.
It is necessary to imagine two rulers joined at right angles to each other, so that they remain in one surface, just as ΑΒ and ΓΔ are. And in the ruler ΑΒ, imagine a dovetail shaped groove, through which a peg is able to slide; and in the ruler ΓΔ, near the Δ side, and in the line through the middle of it, a small cylinder attached to the ruler and extending above it a tad. And in another ruler EZ, after some small distance from the endpoint Ζ, imagine it having a cut, as ΗΘ, through which the cylinder at Δ is able to move; and in a small distance from the endpoint Ε, a round hole, through which a small bolt is passed and attached to the peg in the dovetailed shaped groove ΑΒ. And so fitting the ruler EZ to the cut ΗΘ with the cylinder Δ, and at the hole Ε, fitting it to the hole Ε with the bolt attached to the sliding knob, if one takes hold of the the extremity Κ of this ruler and moves it, first towards Α, and then towards Β, the point Ε will always remain on the ruler ΑΒ, and the cut ΗΘ will be moved about the cylinder near Δ, with the imaginary midline of ΗΘ always passing through the axis of the cylinder near Δ, and the excess ΕΚ remaining constant.
Above: Nicomedes’ First Diagram. I have simplified this diagram slightly, removing the physical parts described, such as the groove ΗΘ.
If we imagine a pencil at Κ, touching the baseplane, it will draw a certain curve, such as ΜΛΝ, which Nicomedes calls the first conchoid line, the length of ΕΚ its interval, and Δ its pole. With respect to this curve, he shows the result that it converges to the ruler ΑΒ, and if some straight line is drawn between the curve and the ruler ΑΒ, this straight line always cuts the curve. The first of these properties is easily understood by means of a different diagram. For imagining a ruler ΑΒ, pole Γ, interval ΔΕ, and conchoid line ΖΕΗ, let the lines from Γ, such as ΓΘ and ΓΖ, be drawn to meet the conchoid line [at Θ and Ζ]: with clearly the resulting lines ΚΘ and ΛΖ being equal.
I say that the perpendicular ΖΜ is less than the perpendicular ΘΝ.^{30} For since the angle ΜΛΓ is larger than the angle ΝΚΓ, the remaining angle, ΜΛΖ, being less than two right angles, is less than the remaining angle ΝΚΘ. Because of this, the angle at Ζ will be larger than the angle at Θ, with the angles at Μ and Ν being right. And if we make the angle ΜΖΞ equal to the angle at Θ, then ΚΘ, that is ΛΖ, will have to ΘΝ the same ratio which ΞΖ has to ΖM: \begin{equation} \tag{109} \text{ΚΘ : ΘΝ :: KΘ : ΛΖ :: ΞΖ : ΖΜ, } \end{equation}so that \begin{equation} \tag{110} \text{ΖΛ : ΘΝ < ΖΛ : ΖΜ, } \end{equation}and because of this, ΘΝ is larger than ΖΜ.
This could use a bit more explanation. We know that ΖΛ is larger than ΖΞ. Hence \begin{equation} \tag{111} \text{ΖΞ : ΖΜ < ΖΛ : ΖΜ. } \end{equation}Since by construction, triangle ΘΝΚ is similar to triangle ΖΜΞ, we have that \begin{equation} \tag{112} \text{ΖΞ : ΖΜ :: ΘΚ : ΘΝ. } \end{equation}Hence \begin{equation} \tag{113} \text{ΘΚ : ΘΝ < ΖΛ : ΖΜ. } \end{equation}But by definition of the conchoid line, ΘΚ and ΖΛ are equal; hence \begin{equation} \tag{114} \text{ΖΛ : ΘΝ < ΖΛ : ΖΜ.} \end{equation} 
The second property was that any straight line drawn between the ruler ΑΒ and the conchoid line cuts the conchoid line. And this becomes familiar:^{31} for the drawn line is either parallel to ΑΒ or not.
Let it first be parallel, as ΖΗΘ, and let it be supposed, that \begin{equation} \tag{115} \text{ΔΗ : ΗΓ :: ΔΕ : some other straight line Κ.} \end{equation}With center Γ and radius Κ, let the perimeter of a circle being drawn cut ΖΗ at Η, and let ΓΖ be joined.
Above: Nicomedes’ Second Diagram. Netz notes that some of the manuscripts have Δ and Ν coinciding. Not stated is that Θ should be closer to the “center” of the conchoid line, by which I mean the vertical axis of symmetry, than Ζ: for otherwise, the role of Θ and Ζ would reverse.
Therefore, by Elements VI.2 we have \begin{equation} \tag{116} \text{ΔΗ : ΗΓ = ΛΖ : ΖΓ. } \end{equation}But \begin{equation} \tag{117} \text{ΔΗ : ΗΓ :: ΔΕ : Κ :: ΔΕ : ΓΖ [by definition of K],} \end{equation}therefore ΔΕ is equal to ΛΖ. But this is impossible, for it is necessary that Ζ be on the conchoid line.
Above: Nicomedes’ Third Diagram. Netz notes that the manuscripts only have an arc of a circle, representing the conchoid line. The actual circle (drawn here as a blue arc) is omitted. By definition of Ζ, the arc passes through Ζ. What is not obvious, at least not yet, is that Ζ must therefore also be on the conchoid line. The point Θ is not necessarily on the conchoid line: it’s merely a third point along the line ΖΗ. However, if we define Θ as the other point where the circle cuts the line ΖΗΘ, then it plays a symmetric role to Ζ. Also for Ν: it is not necessarily on either the conchoid line or the circle, but the argument that Nicomedes (through Eutocius) gives shows that the line ΜΗΝ would, if extended, meet both.
I wonder if γάρ (“for”) has been corrupted from ἄρα. Netz notes that Heiberg disliked the proof, because nowhere did Eutocius suppose that Ζ wasn’t on the conchoid line. I suppose to some extent we should look past this, since the flow of contradiction proofs such as this is, as I said above, was familiar. The essential argument here is that:

Hence let the drawn line be not parallel, as ΜΗΝ. Let ΖΗ be drawn through Η parallel to ΑΒ. Therefore ΖΗ will intersect the curve: so clearly ΜΝ will too.
Probably an appeal to the continuity of the conchoid line and the symmetry of the figure. Since Γ is the pole of this conchoid line, the line ΓΕ is the axis of symmetry for the conchoid line. So the mirror image of Ζ also passes through both the circle and the conchoid line, say, at Θ’. Thus we have a continuous line from Ε to Θ’, and straight lines ΕΗ and ΗΘ’. Since ΗΘ extends to ΗΘ’, we have “trapped” ΗΝ inside a sort of quasitriangle. Hence ΗΝ, if extended, also must intersect the conchoid line (and, if extended further, would also intersect the circle). It would have perhaps been easier had Eutocius (or Nicomedes) put Ν on the other side of the axis: then we could directly appeal to ΖΗ and ΗΕ. 
With these results following from the mechanism, its usefulness to the given problem is shown thus. Given an angle Α and a point Γ outside, to draw ΓΗ and make ΚΗ equal to a given line. Let the perpendicular ΓΘ be drawn from the point Γ to ΑΒ, and let it be extended. Let ΔΘ be equal to the given line. With pole Γ, radius ΔΘ, and ruler ΑΒ, let the first conchoid line ΕΔΖ be drawn. Therefore it intersects ΑΗ by the previous result. Let it intersect at Η, and let ΓΗ be joined: therefore ΚΗ is equal to the given line.
Above: Nicomedes’ Fourth Diagram. The angle is adjustable by moving the point Η along the conchoid line.
Having proved this, let two segments ΓΛ and ΛΑ have been given at right angles to one another. It is necessary to find two mean proportionals in succession between them. Let the parallelogram ΑΒΓΛ be completed, and let ΑΒ and ΒΓ be bisected at the points Δ and Ε (respectively). Let ΔΛ, having been joined, be extended and intersect ΓΒ (having also been joined) at Η. Let ΕΖ, having been joined, be extended and intersect ΒΓ at right angles. Let ΓΖ be extended, being equal to ΑΔ. Let ΖΗ and its parallel ΓΘ be joined. And the angle ΚΓΘ being equal to an angle, let ΖΘΚ be drawn from the given point Ζ, making ΘΚ equal to either ΑΔ or ΓΖ: for this is possible, as was shown with the conchoid line. Let ΚΛ, having been joined, be produced and intersect ΑΒ (itself having been produced) at Μ.
Above: Nicomedes’ Fifth Diagram. Not shown in the manuscripts is the conchoid line: I have added it here in red, and the ruler for it in blue. The given angle ΚΓΘ is first drawn, with Θ otherwise undetermined. A perpendicular from Ζ to ΓΘ is drawn, serving as the length from the pole Ζ to the ruler ΓΘ. The given length is here ΑΔ (or ΓΖ, which is equal to it). In the particular configuration shown here, ΓΖ is itself that perpendicular, but this is because of the specific ratio of ΛΓ to ΛΑ (here, 2:1). Since GeoGebra is not capable of finding the intersection of the conchoid line and the like ΓΚ, I was forced to make this diagram static. The trick I used for other diagrams of using a conic to interpolate the locus is rather unsuitable here, since the conchoid line is both nearly straight and appears to have an inflection point near Κ.
I say that \begin{equation} \tag{118} \text{ΓΛ : ΚΓ :: ΚΓ : ΜΑ :: ΜΑ : ΑΛ.} \end{equation}Since ΒΓ has been bisected at Ε, and ΚΓ has been added to it, therefore [by Elements II.6] \begin{equation} \tag{119} \text{rect.(ΒΚΓ) + sq.(ΓΕ) = sq.(ΕΚ).} \end{equation}Let the square on ΕΖ be added in common: therefore [by several applications of Elements I.47] \begin{align} \tag{120} \text{rect.(ΒΚΓ) + sq.(ΓΕ) + sq.(EΖ)} & = \text{rect.(ΒΚΓ) + sq.(ΓΖ)} \\\\ & = \tag{121} \text{sq.(ΚΕ) + sq.(ΕΖ)} \\\\ & = \tag{122} \text{sq.(KZ).} \end{align}And since [by two uses of Elements VI.2] \begin{equation} \tag{123} \text{ΜΑ : ΑΒ :: ΜΛ : ΛΚ,} \end{equation}and \begin{equation} \tag{124} \text{ΜΛ : ΛΚ :: ΒΓ : ΓΚ,} \end{equation}therefore \begin{equation} \tag{125} \text{ΜΑ : ΑΒ :: Β Γ: ΓΚ. } \end{equation}And since ΑΔ is half of ΑΒ, and ΓΗ is double ΒΓ (since ΛΓ is also double ΔΒ), therefore \begin{equation} \tag{126} \text{ΜΑ : ΑΔ :: ΗΓ : ΚΓ.} \end{equation}But [using Elements VI.2 yet again] \begin{equation} \tag{127} \text{ΗΓ : ΓΚ :: ΖΘ : ΘΚ} \end{equation}since ΗΖ and ΓΘ are parallel. And therefore componendo [Elements V def. 14] \begin{equation} \tag{128} \text{ΜΔ : ΔΑ :: ΖΚ : ΚΘ.} \end{equation}But ΑΔ is established as equal to ΘΚ, since ΑΔ is also equal to ΓΖ: therefore ΜΔ is equal to ΖΚ. Therefore \begin{equation} \tag{129} \text{sq.(ΜΔ) = sq.(ΖΚ). } \end{equation}And since [by Elements II.6] \begin{equation} \tag{130} \text{sq.(ΜΔ) = rect.(ΒΜΑ) + sq.(ΔΑ),} \end{equation}and it was shown that \begin{equation} \tag{131} \text{rect.(ΒΚΓ) + sq.(ΓΖ) = sq.(ΖΚ) [by (122)],} \end{equation}and \begin{equation} \tag{132} \text{sq.(ΑΔ) = sq.(ΓΖ),} \end{equation}since ΑΔ was established as equal to ΓΖ. Therefore by Elements VI.16 \begin{equation} \tag{134} \text{ΜΒ : ΒΚ :: ΚΓ : ΑΜ.} \end{equation}But using Elements VI.2 we have \begin{equation} \tag{135} \text{ΒΜ : ΒΚ :: ΓΛ : ΓΚ,} \end{equation}and therefore \begin{equation} \tag{136} \text{ΛΓ : ΓΚ :: ΓΚ : ΑΜ. } \end{equation}But [Elements VI.2 again] \begin{equation} \tag{137} \text{ΛΓ : ΓΚ :: ΜΑ : ΑΛ, ΒΜ : ΒΚ :: ΓΛ : ΓΚ, } \end{equation}therefore \begin{equation} \tag{138} \text{ΛΓ : ΓΚ :: ΓΚ : ΑΜ :: ΑΜ : ΑΛ.} \end{equation}
29. δυνάμει here functioning adverbially.
30. This is of course only true if the perpendiculars are on the same “side” of Γ.
31. Netz has “and this is made understood as follows”. I took γνώριμον as more “familiar”, as in the argument for the second property is a familiar one.