Synthesis
. Make EB equal to R, EC equal to S, and deſcribe on

BC a circle; erect at E the perpendicular ED, meeting the periphery of the

circle in D; alſo at A erect the perpendicular AF equal to R; draw AD,

which produce, if neceſſary, to cut the indefinite line, as in O, which will

be the point required.

For becauſe of the ſimilar triangles AOF, EOD, AO is to EO as AF

(R) is to DE; therefore the ſquare on AO is to the ſquare on EO as the

ſquare on R is to the ſquare on DE (
Eu
. VI. 22); but the ſquare on DE

is equal to the rectangle contained by R and S; therefore the ſquare on AO is

to the ſquare on EO as the ſquare on R is to the rectangle R, S; that is as

R is to S, by
Eu
. V. 15.

Q. E. D.

Scholium
. This Problem alſo hath three Epitagmas, which I enumerate

as in the laſt. The firſt is conſtructed by Fig. 6, wherein the perpendicu-

lars DE and AF are ſet off on the ſame ſide of the indefinite line; the ſecond

by Fig. 7, where they are ſet off on contrary ſides, and the third by Fig. 8,

in which they are again ſet off on the ſame ſide. The ſecond has no limits; but in the firſt R muſt be leſs, and in the third greater than S, for reaſons

too obvious to be inſiſted on; and hence, both theſe caſes are impoſſible

when the given ratio is that of equality.

##
75.
PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)

In any indefinite ſtraight line let there be aſſigned the points A, E and I; it is required to cut it in another point, O, ſo that the rectangle contained

by the ſegment AO and a given ſtraight line P may be to the rectangle

contained by the ſegments EO, IO 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, P be to the rectangle EO, IO as R to S. Make

IQ to P as S is to R; then will the rectangle AO, P be to the rectangle

EO, IO as P is to IQ; or (
Eu
. V. 15.) the rectangle AO, P be to the

rectangle EO, IO as the rectangle IO, P is to the rectangle IO, IQ; and

hence (
Eu
. V. 15. 16) AO is to EO as IO is to IQ; whence, by compoſition

or diviſion, AE is to EO as OQ is to IQ: therefore (
Eu
. VI. 16.) the