96[19]
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.
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.
(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.
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.
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.
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.)
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.)