87[10]
In the indefinite line let the four points be A, E, I, U.
It is then required
to be cut again in O ſo that OA x OU may be to OE x OI (be the Poſition
of the four given points what they may) in the ratio of AL to LE, let the
point L fall alſo as it may.
to be cut again in O ſo that OA x OU may be to OE x OI (be the Poſition
of the four given points what they may) in the ratio of AL to LE, let the
point L fall alſo as it may.
Construction.
To the three points E, L, U, and the right line UI, let
be found by the IId Problem a fourth point O, ſo that UI x OE may be to
OU x OL as AE to AL. And let ſuch a Case be choſen of the IId Problem
that, according as UO is required greater or leſs than UI, or their ſum ſhall
conſtitute OI, ſo of the three rectangles deſcribed in the Vth Lemma made
by the four points E, O, A, L, that of OE x AL may accordingly be greater
or leſs than OL x AE, or their ſum conſtitute that of OA x EL.
be found by the IId Problem a fourth point O, ſo that UI x OE may be to
OU x OL as AE to AL. And let ſuch a Case be choſen of the IId Problem
that, according as UO is required greater or leſs than UI, or their ſum ſhall
conſtitute OI, ſo of the three rectangles deſcribed in the Vth Lemma made
by the four points E, O, A, L, that of OE x AL may accordingly be greater
or leſs than OL x AE, or their ſum conſtitute that of OA x EL.
N.
B.
U being now uſed to repreſent one of the given points, in all the
following Diagrams I have ſubſtituted V in the place where U was uſed
before.
following Diagrams I have ſubſtituted V in the place where U was uſed
before.
Demonstration.
On ſuppoſition therefore that ſuch a Case of the
IId Problem is made uſe of,
We have UI x OE: OU x OL: : AE: AL
IId Problem is made uſe of,
We have UI x OE: OU x OL: : AE: AL
This Problem has three Epitagmas.
The Iſt whereof is when of the
two aſſigned points A and U, the one of them is an extreme, and the other an
alternate mean; and this admits of three Cases. The IId is when A and U
are both of them extremes; and this has four Cases. The IIId is when of
A and U one of them is an extreme, and the other is the point next to it; and
this has three Caſes. And there remain three more Anomalous Caſes, wherein
the IId Problem is of no uſe, but which may be reduced to one, as ſhall be
ſhewn in it's proper place.
two aſſigned points A and U, the one of them is an extreme, and the other an
alternate mean; and this admits of three Cases. The IId is when A and U
are both of them extremes; and this has four Cases. The IIId is when of
A and U one of them is an extreme, and the other is the point next to it; and
this has three Caſes. And there remain three more Anomalous Caſes, wherein
the IId Problem is of no uſe, but which may be reduced to one, as ſhall be
ſhewn in it's proper place.