A history of Greek mathematics - Wilbourhall.org

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
These lemmas are all
contacts' by Apollonius.
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.