Apollonius <Pergaeus>; Lawson, John - The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

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93[16]

PROBLEM II. (Fig. 2 and 3.)

In any indefinite ſtraight line, let there be aſſigned the points A and E;
it is required to cut it in another point O, ſo that the rectangle contained
by the ſegments AO, EO may be to the ſquare on a given line P, in the
ratio of two given ſtraight lines R and S.

Analysis. Conceive the thing done, and O the point ſought: then
would the rectangle AO, EO be to the ſquare on the given line P as R is
to S, by hypotheſis. Make AQ to P as R is to S: then the rectangle AO,
EO will be to the ſquare on P as AQ to P; or (Eu. V. 15. and 16) the
rectangle AO, EO will be to the rectangle AQ, EO as the ſquare on P is
to the rectangle EO, P; and therefore AO is to AQ as P is to EO; con-
ſequently (Eu. VI. 16.) the rectangle AO, EO is equal to the rectangle
AQ, P; and hence, as the ſum, or difference of AO and EO is alſo given,
theſe lines themſelves are given by the 85th or 86th of the Data.

Synthesis. On AE deſcribe a circle, and erect at A the indefinite
perpendicular AK; and, having taken AQ a fourth proportional to S, R
and P, take AD a mean proportional between AQ and P; from D draw
DH, parallel to AE if O be required to fall between A and E; but through
F, the center of the Circle on AE, if it be required beyond A or E, cutting
the circle in H; laſtly, draw HO perpendicular to DH, meeting the inde-
finite line in O, the point required.

For it is manifeſt from the conſtruction that AD and HO are equal;
hence, and Eu. VI. 17, the rectangle AQ, P is equal to the ſquare on HO;
conſequently equal to the rectangle AO, EO (Eu. III. 35. 36) and ſo
(Eu. VI. 16.) AO is to P as AQ is to EO; but by conſtruction AQ is to
P as R to S, therefore by compound ratio, the rectangle AO, AQ is to the
ſquare on P as the rectangle AQ, R is to the rectangle EO, S: hence
(Eu. V. 15. 16.) the rectangle AO, EO is to the ſquare on P as the rect-
angle EO, R is to the rectangle EO, S, that is, as R is to S. Q. E. D.

Scholium. This Problem has three Epitagmas. Firſt, when O is ſought
beyond A; ſecondly, when it is ſought between A and E, and laſtly, when
it is ſought beyond E. The firſt and laſt of theſe are conſtructed by
Fig. 2, and have no limitations; but in the ſecond, (Fig. 3.) the given
ratio of R to S muſt not be greater than that which the ſquare on half
AE bears to the ſquare on P: ſince if it be, a third proportional to

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