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be uſed, and the point O will fall between E and I, and the point o beyond
L, much more beyond I.
L, much more beyond I.
Case II.
Let the given ratio, EL to LI, be inæqualitatis minoris, i.
e.
of a
leſs to a greater, and the point O ſought be required to lie between I and the
next point to it E; or elſe to fall beyond A the other extreme. For the ſame
conſtruction ſerves for both. Here Case IV. of Problem II. is to be uſed, and
the point O will fall between E and I, and o beyond A, if we uſe one of the
conſtructions there recited: but if we uſe the other, the points will ſhift places,
as was obſerved under that Caſe, viz. O will fall beyond I the other way, and
o between L and E.
leſs to a greater, and the point O ſought be required to lie between I and the
next point to it E; or elſe to fall beyond A the other extreme. For the ſame
conſtruction ſerves for both. Here Case IV. of Problem II. is to be uſed, and
the point O will fall between E and I, and o beyond A, if we uſe one of the
conſtructions there recited: but if we uſe the other, the points will ſhift places,
as was obſerved under that Caſe, viz. O will fall beyond I the other way, and
o between L and E.
Case III.
Let now the point O be ſought between A and E.
Here ſet off
the given ratio in ſuch a manner that EI may be the ſum of the terms, and
make uſe of the IIId Case of Problem II. and the Limitation here will
be evident from the Limitation there given, viz. making EI: IL: : AI: X,
the Limitation here is that X muſt not be leſs than IE + EL + √4 IEL*.
the given ratio in ſuch a manner that EI may be the ſum of the terms, and
make uſe of the IIId Case of Problem II. and the Limitation here will
be evident from the Limitation there given, viz. making EI: IL: : AI: X,
the Limitation here is that X muſt not be leſs than IE + EL + √4 IEL*.
Epitagma II.
Case IV.
Here OI the line whoſe ſquare is concerned is
to be bounded by I the middle point of the three given ones, and O or o, its
other bound is to be ſought between I and either extreine A or E. the ſame
conſtruction ſerving for both. The given ratio muſt here be ſet off in ſuch a
manner that EI may be the ſum of the terms of it; and make uſe
of Iſt Case of the IId Problem; with this caution, that of the two ſegments
AI, IE, you choſe the leſſer IE whereon to exhibit the given ratio; for then
it will appear by the work itſelf that O falling between E and L, o will alſo
fall between A and I: otherwiſe, if AI was leſs than IE, there would want
ſome proof of this. Therefore of the two extreme given points call that E
which bounds the leſſer ſegment, and then the general Demonſtration will fit
this Caſe as well as the reſt.
to be bounded by I the middle point of the three given ones, and O or o, its
other bound is to be ſought between I and either extreine A or E. the ſame
conſtruction ſerving for both. The given ratio muſt here be ſet off in ſuch a
manner that EI may be the ſum of the terms of it; and make uſe
of Iſt Case of the IId Problem; with this caution, that of the two ſegments
AI, IE, you choſe the leſſer IE whereon to exhibit the given ratio; for then
it will appear by the work itſelf that O falling between E and L, o will alſo
fall between A and I: otherwiſe, if AI was leſs than IE, there would want
ſome proof of this. Therefore of the two extreme given points call that E
which bounds the leſſer ſegment, and then the general Demonſtration will fit
this Caſe as well as the reſt.
Case V.
Let the given ratio of EL to LI be inæqualitatis minoris;
and let
the point ſought be required to lie beyond either extreme. The ſame con-
ſtruction ſerves for both. Here we muſt uſe the IVth Case of the IId Pro-
BLEM, and O being made to fall between E and L, o will fall always beyond
A, provided we call that point E which bounds the bigger ſegment. I have
in the Figure made AI = IE on purpoſe to ſhew that in that caſe the point N
will coincide with A. But if IE be greater than AI, the point N will
always fall beyond A, and conſequently the point o more ſo.
the point ſought be required to lie beyond either extreme. The ſame con-
ſtruction ſerves for both. Here we muſt uſe the IVth Case of the IId Pro-
BLEM, and O being made to fall between E and L, o will fall always beyond
A, provided we call that point E which bounds the bigger ſegment. I have
in the Figure made AI = IE on purpoſe to ſhew that in that caſe the point N
will coincide with A. But if IE be greater than AI, the point N will
always fall beyond A, and conſequently the point o more ſo.