95[18]
is equal to the rectangle AE, AQ;
and hence (Eu.
VI.
16.)
OQ is to AQ
as AE is to AO; therefore, by compoſition or diviſion, AO is to AQ as
EO is to AO; but by conſtruction, AQ is to R as P is to S, and ſo, by
compound ratio, the rectangle AO, AQ is to the rectangle R, AQ as the
rectangle EO, P is to the rectangle AO, S; or (Eu. V. 15. and 16.) the ſquare
on AO is to the rectangle EO, P as the rectangle AO, R is to the rectangle
AO, S; that is, as R to S.
as AE is to AO; therefore, by compoſition or diviſion, AO is to AQ as
EO is to AO; but by conſtruction, AQ is to R as P is to S, and ſo, by
compound ratio, the rectangle AO, AQ is to the rectangle R, AQ as the
rectangle EO, P is to the rectangle AO, S; or (Eu. V. 15. and 16.) the ſquare
on AO is to the rectangle EO, P as the rectangle AO, R is to the rectangle
AO, S; that is, as R to S.
Scholium.
This Problem hath three Epitagmas alſo, which I ſtill enu-
merate as before. The firſt and ſecond are conſtructed by Fig. 4, where DH
is drawn through F, the center of the circle on AQ: and theſe have no limi-
tations. The third is conſtructed as in Fig. 5, where DH is drawn parallel
to AQ; and here the given ratio of R to S muſt not be leſs than the ratio
which four times AE bears to P: for if it be, AE will be greater than
one-fourth Part of AQ (a fourth proportional to S, R and P) in which
caſe the rectangle contained by AE and AQ will be greater than the ſquare
on half AQ, and conſequently AD (a mean proportional between AE and
AQ) greater than half AQ; but it is plain when this is the caſe, that DH
will neither cut nor touch the circle on AQ, and therefore the problem is
impoſſible.
merate as before. The firſt and ſecond are conſtructed by Fig. 4, where DH
is drawn through F, the center of the circle on AQ: and theſe have no limi-
tations. The third is conſtructed as in Fig. 5, where DH is drawn parallel
to AQ; and here the given ratio of R to S muſt not be leſs than the ratio
which four times AE bears to P: for if it be, AE will be greater than
one-fourth Part of AQ (a fourth proportional to S, R and P) in which
caſe the rectangle contained by AE and AQ will be greater than the ſquare
on half AQ, and conſequently AD (a mean proportional between AE and
AQ) greater than half AQ; but it is plain when this is the caſe, that DH
will neither cut nor touch the circle on AQ, and therefore the problem is
impoſſible.
PROBLEM IV. (Fig. 6. 7. and 8.)
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 two ſquares on the
ſegments AO, EO, may obtain the Ratio of two given ſtraight lines,
R and S.
it is required to cut it in another point O, ſo that the two ſquares on the
ſegments AO, EO, may obtain the Ratio of two given ſtraight lines,
R and S.
Analysis.
Imagine the thing to be effected, and that O is really the
point required: then will the ſquare on AO be to the ſquare on EO as
R to S; or (Eu. V. 15.) the ſquare on AO will be to the ſquare on EO
as the ſquare on R is to the rectangle contained by R and S. Let DE be
made a mean proportional between EB (R) and EC (S). Then
(Eu. VI. 17.) the ſquare on AO will be to the ſquare on EO as the
ſquare on R to the ſquare on DE; and ſo (Eu. VI. 22.) AO to EO as R to
DE; and hence both AO and EO will be given by the converſe of Prop. 38.
of Eu. Data.
point required: then will the ſquare on AO be to the ſquare on EO as
R to S; or (Eu. V. 15.) the ſquare on AO will be to the ſquare on EO
as the ſquare on R is to the rectangle contained by R and S. Let DE be
made a mean proportional between EB (R) and EC (S). Then
(Eu. VI. 17.) the ſquare on AO will be to the ſquare on EO as the
ſquare on R to the ſquare on DE; and ſo (Eu. VI. 22.) AO to EO as R to
DE; and hence both AO and EO will be given by the converſe of Prop. 38.
of Eu. Data.