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

The problem is
THE COLLECTION. BOOK VII 415
reduced to a problem contained in Apollonius's
Determinate Section thus.
Suppose the problem solved by the semicircle DEF, BE
being equal to AD. Join E to the centre G of the semicircle
416 PAPPUS OF ALEXANDRIA
solution *of a certain quadratic equation.) Pappus observes
that the problem is always possible (requires no Siopco-fios),
and proves that it has only one solution.
(S) Lemmas on the treatise ' On
contacts' by Apollonius.
DEF. Produce DA to If, making HA equal to AD. Let K
be the middle point of DC,
Since the triangles ABC, GEC are similar,
AG 2 :GC = BE 2 2 : EC"
= AD 2 : EC 2 , by hypothesis,
= AD 2 :GC -DG : 2 (since DG = GE)
= AG -AD 2 2 :DG 2
= HG.DG:DG 2
= EG : DG.
Therefore
HG:DG = AD 2 :GC 2 -DG 2
= AD 2 :2DC.GK.
Take a straight line Z such that AD — 2 L .
HG:DG = L: GK,
HG.GK = L. DG.
therefore
or
F
c
2 DC:
Therefore, given the two straight lines HD, DK (or the
three points H, D, K on a straight line), we have to find
a point G between D and K such that
HG.GK = L. DG,
which is the second ejritagma of the third Problem in the
Determinate Section of Apollonius, and therefore may be
taken as solved. (The problem is the equivalent of the
These lemmas are all
pretty obvious except two, which are
important, one belonging to Book I of the treatise, and the other
to Book II. The two lemmas in question have already been set
out a propos of the treatise of Apollonius (see pp. 1 82-5, above).
As, however, there are several cases of the first (Props. 105,
107, 108, 109), one case (Prop. 108, pp. 836-8), different from
that before given, may be put down here : Given a circle and
tivo 'points B, E %vithin it, to draw straight lines through D, E
to a point A on the circumference in such a way that, if they
meet the circle again in B, C, BO shall be parallel to BE.
We proceed by analysis. Suppose the problem solved and
DA,EA drawn ('inflected') to A in such a way that, if AD,
AE meet the circle again in B, O,
BO is parallel to BE.
Draw the tangent at B meeting
EB produced in F.
Then Z FBB = Z AOB = lAEB;
therefore A, E, B, F are coney clic,
and consequently
FB.BE=AB.BB.
But the rectangle AB . BB is given, since it depends only
on the position of B in relation to the circle, and the circle
is given.
Therefore the rectangle FB .
BE is given.
And BE is given ; therefore FB is given, and therefore F.
If follows that the tangent FB is given in position, and
therefore B is given. Therefore BBA is given and consequently
AE also.
To solve the problem, therefore, we merely take F on EB
BE = the given rectangle made by
produced such that FB .
the segments of any chord through B, draw the tangent FB,
join BB and produce it to A, and lastly draw AE through to
O; BO is then parallel to BE.