108[31]
PROBLEM VII. (Fig. 32, 33, 34, &c.)
In any indeſinite ſtraight line let there be aſſigned the points A, E, I
and U; it is required to cut it in another point, O, ſo that the rectangle
contained by the ſegments AO, UO may be to that contained by the ſeg-
ments EO, IO in the ratio of two given ſtraight lines, R and S.
and U; it is required to cut it in another point, O, ſo that the rectangle
contained by the ſegments AO, UO may be to that contained by the ſeg-
ments EO, IO in the ratio of two given ſtraight lines, R and S.
Analysis.
Imagine the thing done, and O the point ſought:
then will
the rectangle AO, UO be to the rectangle EO, IO as R is to S. Make
UC to EC as R is to S; and the rectangle AO, UO will be the rectangle
EO, IO as UC is to EC. Let now OB be taken a fourth proportional to
UO, UC and IO: then (Eu. V. 15.) the rectangle AO, UO will be to
the rectangle EO, IO as the rectangle UC, OB is to the rectangle EC, OB;
or (Eu. V. 16.) the rectangle AO, UO is to the rectangle UC, OB as the
rectangle EO, IO is to the rectangle EC, OB; wherefore ſince UO is to UC
as IO to OB, by conſtruction, AO will be to BO as EO to EC; and ſo by
compoſition or diviſion, CO is to CU as IB to BO, and AB is to BO as
CO to EC: wherefore ex æquo perturb. & permut. AB is to IB as UC to
EC, that is, in the given ratio; and hence is given BC, the ſum or dif-
ference of CO and BO, as alſo the rectangle contained by them, equal to
the rectangle CU, IB, whence thoſe lines themſelves are given by the 85th
or 86th of the Data.
the rectangle AO, UO be to the rectangle EO, IO as R is to S. Make
UC to EC as R is to S; and the rectangle AO, UO will be the rectangle
EO, IO as UC is to EC. Let now OB be taken a fourth proportional to
UO, UC and IO: then (Eu. V. 15.) the rectangle AO, UO will be to
the rectangle EO, IO as the rectangle UC, OB is to the rectangle EC, OB;
or (Eu. V. 16.) the rectangle AO, UO is to the rectangle UC, OB as the
rectangle EO, IO is to the rectangle EC, OB; wherefore ſince UO is to UC
as IO to OB, by conſtruction, AO will be to BO as EO to EC; and ſo by
compoſition or diviſion, CO is to CU as IB to BO, and AB is to BO as
CO to EC: wherefore ex æquo perturb. & permut. AB is to IB as UC to
EC, that is, in the given ratio; and hence is given BC, the ſum or dif-
ference of CO and BO, as alſo the rectangle contained by them, equal to
the rectangle CU, IB, whence thoſe lines themſelves are given by the 85th
or 86th of the Data.
Synthesis.
Make AB to IB, and UC to EC in the given ratio, and de-
ſcribe on BC a circle; erect, at B the indeſinite perpendicular BK, and take
therein BD a mean proportional between AB and EC, or between IB and
and UC: from D, draw DH, parallel to BC, if O be required any where
between B and C; but through F, the center of the circle on BC, if it be
ſought any where without them, cutting the circle on BC in H. Laſtly,
draw HO perpendicular to DH, which will cut the indeſinite line in O,
the point required.
ſcribe on BC a circle; erect, at B the indeſinite perpendicular BK, and take
therein BD a mean proportional between AB and EC, or between IB and
and UC: from D, draw DH, parallel to BC, if O be required any where
between B and C; but through F, the center of the circle on BC, if it be
ſought any where without them, cutting the circle on BC in H. Laſtly,
draw HO perpendicular to DH, which will cut the indeſinite line in O,
the point required.
For it is plain from the conſtruction that HO and BD are equal, and
(Eu. VI. 17.) the rectangle AB, EC, or the rectangle IB, UC is equal to
the ſquare on BD, and therefore equal to the ſquare on HO, which (Eu.
III. 35. 36.) is equal to the rectangle BO, OC. Hence (Eu. VI. 16.) AB
is to BO as CO is to CE, and CO is to CU as IB is to BO; whence,
by compoſition or diviſion, AO is to BO as EO is to CE, and UO is
(Eu. VI. 17.) the rectangle AB, EC, or the rectangle IB, UC is equal to
the ſquare on BD, and therefore equal to the ſquare on HO, which (Eu.
III. 35. 36.) is equal to the rectangle BO, OC. Hence (Eu. VI. 16.) AB
is to BO as CO is to CE, and CO is to CU as IB is to BO; whence,
by compoſition or diviſion, AO is to BO as EO is to CE, and UO is