A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

ON THE SPHERE AND CYLINDER, II 47
(iii) Solution of the original problem of II. 4 by Diocles.
Diodes proceeded in a different manner, satisfying, by
a geometrical construction, not the derivative cubic equation,
but the three simultaneous relations which hold in Archimedes's
proposition, namely
HM:H'M = m:n
HA: h = r :h'
H'A'\ I! = r:hj
with the slight generalization that he substitutes for r in
these equations another length a.
48 ARCHIMEDES
Draw QMN through M parallel
meet KG produced in F.
to AK. Produce K'M to
By similar triangles,
FA:AM = K'A':A'M, or FA:h = a:h' y
whence FA = AH (k, suppose).
Similarly # A'E — A'H' (Jc, suppose).
Again, by^similar triangles,
(FA + AM) :
(A'K' + A'M) = AM: A'M
= (AK + AM):(EA' + A'M),
or (k + h):(a + h') = (a + h):(¥ + h') }
i. e. (k + h) (k' + h') = (a + h)(a + h f ). (1)
Now, by hypothesis,
m:n = (k + h):(k' + hf)
= (k + h)(k' + h'):(k' + h') 2
= (a + h) (a + hf) :
{¥ + h!f [by (I)]. (2)
Measure AR, A f R! on AA f produced both ways equal to a.
Draw RP, R'P' at right angles to RR /
as shown in the figure.
Measure along MJS T the length MV equal to MA' or h\ and
draw PP' through V, A' to meet RP, R'P'.
Then QV=k' + h', P'V= V2 (a + h'),
PV= V2(a + h),
The problem is, given a straight line A A', sl ratio m : n, and
AK (= a), to divide A A' at a point M
another straight line
and at the same time to find two points H, ffl on A A'
produced such that the above relations (with a in place
of r) hold.
The analysis leading to the construction is very ingenious.
Place AK (= a) at right angles to AA', and draw A'K' equal
and parallel to it.
Suppose the problem solved, and the points M, H, Hf
all
found.
Join KM, produce it, and complete the rectangle KGEK f .
whence PV.P'V = 2 (a + h) {a + h!) ;
and, from (2) above,
2m:n=2(a + h) (a + h') :
(¥ + h'f
= PV.P'V: QV 2 .
(3)
Therefore Q is on an ellipse in which PP' is a diameter, and
Q V is an ordinate to it.
Again, GQNK is equal to AA'K'K, whence
GQ.QN= AA'. A'K' =(h + h!) a = 2ra, (4)
and therefore Q is on the rectangular hyperbola with KF,
KK' as asymptotes and passing through A'.