I present here the philon line, as proposed by Philon, and two other ways of finding the same line to solve the duplication of the cube problem, one proposed by Apollonius and one by Heron.

Let AB and AC be two straight lines placed at right angles. Complete the rectangle ABDC (D is the point inside the angle through which the line shall be drawn). Let E be the center of the diagonal of the rectangle ABDC. Then a circle centered at E and going through D shall circumscribe the rectangle ABDC (note that the diameter will be AD).

Philon's wayPlace a ruler so that it passes through D and pivot it on D until it cuts AB and AC produced and the circle ABDC in points F,G,H such that the intercepts FD and HG are equal.Apollonius's way

Draw a circle centered in E and cutting the produced AB and AC in F and G respectively, but such that F,D and G are collinear.Heron's way

Place a ruler so that its edge passes through D, and move it about D until the edge intersects the produced AB and AC in points F and G respectively so that EF and EG are equals.

Obviously the three constructions compute the same points F and
G. In Philon's construction, FD=HG.
Then, the perpendicular from E on DH, which bisects DH, must also bisect
FG, so that EF = EG. Therefore Philon's
way is equivalent to Heron's
way. We now proceed to proof that Apollonius's
ways is equivalent to the other two. We will proof that (AF)(FB)
= (AG)(GC), in other words: if the power
of the point F with respect to circle
AGDC is the same as the power
of the point G and (AF)(FB) = (AG)(GC) then GH = FD.

Since the power of G and F are the same with respect to ABDC Then

(AF)(FB) = (FD)(FH) and (GC)(GA) = (GH)(GD)

if (AF)(FB) = (AG)(GC) then

(FD)(FH) = (GH)(GD)

But FH = FD + HD and GD = GH + HD

Therefore

(FD)(FD + DH)=(GH)(GH + DH)

==>

FD^{2}-GH^{2 }= HD(GH - FD)

==>

(FD + GH)(FD - GH) = HD(GH - FD)

==>

FD + GH = -DH

But two positive value can not add up to a negative value,

therefore (GH - FD) must be zero and we could not divide by it.

Hence FD = HG.

Proof that (AF)(FB)=(AG)(GC)::

a) Between Apollonius and Heron

Since K is the middle point of AB, then

(AF)(FB) + BK^{2} = FK^{2}

since FK^{2} = (FB + BK)^{2} =FB(FB + 2BK) + FK^{2}

Add KE^{2} to both sides, then

(AF)(FB) + BE^{2} = EF^{2}

Also, by symmetry

(AG)(GC) + CE^{2} = EG^{2}

But BE = CE and E F = EG

therefore

(AF)(FB) = (AG)(GC)

b) Between Philon and ApolloniusSince GH=FD, we have

(HF)(FD)=(DG)(GH)

But, since the circle GDHC passes through A, the power theorem tell us that

(AF)(FB) = (FD)(FH) and (GC)(GA) = (GH)(GD), therefore

(AF)(FB) = (AG)(GC)

With a few algebraic manipulation, we can obtain
this interesting result, that will permit us to conclude that this construction
is not feasible with Euclidean tools.

From our last result

AF / AG = CG / AG

By the similar triangle principle

FA / AG = DC / CG = FB / BD

therefore

DC / CG = CG / FB = FB / BD

or

AB / CG = CG / FB = FB / AC

We can use this last equation to set 3 equation that can relate to a solution from Menaechmus. The last equation can be rewritten a / x = x / y = y / b. From this we get 3 equation:

These represent the equation of two parabolas
and of one hyperbola.
Menaechmus
solved the problem of the two
mean proportionals by means of the points of intersection of any two
of these conics.

But, if we add the first two equations, we have

which is a circle
passing through the points common to the two parabolas
x^{2 }= ay, y^{2 }= bx.

Therefore we can equally obtain a solution by means
of the intersections of the circle
x^{2 }+ y - bx^{2 }- ay = 0 and the rectangular hyperbola
xy = ab. This is exactly what Philon
does, for if, AF and AG are the coordinate axes, the circle
x^{2 }+ y - bx^{2 }- ay = 0 is the circle
BDHC, and xy = ab is the rectangular hyperbola
with AF and AG as asymptotes and passing through D, which hyperbola
intersects the circle again in H, a point such that FD = HG. This
can also be seen by looking at the next picture. Let CP be the diameter
of the circle going through C and P and let the parabola go through P and
have CS and CR as asymptote. Then PQ produced is the Philon
Line of P for angle
RCS. For let PQ produced cut CR and CS in A and B, respectively.
Then, by a well-known property of the hyperbola,
AP = QB, and, since the circle has CP for a diameter, angle
CQP is a right angle.
Thus the CQ is the perpendicular from C on to PQ (see characterization).
But their common intersection through P, the parabola
and the circle are independent from one another. But it is known
that in general it is impossible with euclidean tools to find another point
of intersection of a conic and a circle of which only one point of intersection
is given (see Menaechmus).