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greateſt of all, and AE will therefore be leſs than AO, and AI greater.
And
the ſame will hold with regard to Ao.
the ſame will hold with regard to Ao.
Here is a Limitation, which is this;
that UN or S the conſequent of the
given ratio, ſet off from R, muſt not be given greater than the difference of
the ſum of AE and AI and of a line whoſe ſquare is equal to four times their
rectangle [i. e. to expreſs it in the modern manner, UN muſt not exceed AI +
AE - √4 AI x AE*. ] This appears by Fig. 2. to this Caſe, the circle there
touching the given indefinite line, and pointing out the Limit.
given ratio, ſet off from R, muſt not be given greater than the difference of
the ſum of AE and AI and of a line whoſe ſquare is equal to four times their
rectangle [i. e. to expreſs it in the modern manner, UN muſt not exceed AI +
AE - √4 AI x AE*. ] This appears by Fig. 2. to this Caſe, the circle there
touching the given indefinite line, and pointing out the Limit.
Case III.
Let the aſſigned points be ſtill in the ſame poſition, and let the
point ſought be now required on the contrary ſide of A.
point ſought be now required on the contrary ſide of A.
Here the conſtruction is ſtill Homotactical, and UN is ſet off the ſame way as
in the laſt Caſe; and the Limitation is, that UN muſt not be given leſs than
the ſum of AI, AE, and a line whoſe ſquare is equal to four times their rect-
angle [or expreſſing it Algebraically, UN muſt not be leſs than AI + AE +
√4 AI x AE*. ]
in the laſt Caſe; and the Limitation is, that UN muſt not be given leſs than
the ſum of AI, AE, and a line whoſe ſquare is equal to four times their rect-
angle [or expreſſing it Algebraically, UN muſt not be leſs than AI + AE +
√4 AI x AE*. ]
Epitagma II.
Case IV.
Let now A be the middle point of the given
ones, and let O the point ſought be required either between A and one of the
extremes, or beyond either of the extremes.
ones, and let O the point ſought be required either between A and one of the
extremes, or beyond either of the extremes.
Here having ſet off IU = AE toward A, you may ſet off UN either way,
and uſing the Antitactical conſtruction, the ſolution will be unlimited. The
only difference is, that if UN be in the direction UI, two ſolutions will ariſe,
whereof in one the point O will fall between A and E, and in the other be-
yond I; but if UN be in the direction IU, two ſolutions will ariſe, whereof
in one the point will fall between A and I, and in the other beyond E. In
proof of which Lemma III. is to be uſed, as Lemma II. was in Caſe I. II.
and uſing the Antitactical conſtruction, the ſolution will be unlimited. The
only difference is, that if UN be in the direction UI, two ſolutions will ariſe,
whereof in one the point O will fall between A and E, and in the other be-
yond I; but if UN be in the direction IU, two ſolutions will ariſe, whereof
in one the point will fall between A and I, and in the other beyond E. In
proof of which Lemma III. is to be uſed, as Lemma II. was in Caſe I. II.
Corollary I.
If then the given ratio be that of AT to TI, or of AE to
EP ſet off from A the other way, ſo that EP be leſs than AE, I ſay then
that O will fall between E and P, as likewiſe ο between T and I, provided o
falls beyond I.
EP ſet off from A the other way, ſo that EP be leſs than AE, I ſay then
that O will fall between E and P, as likewiſe ο between T and I, provided o
falls beyond I.
For by conſtruction IU = AE, and UN = PE.
therefore IN = AP.
But by
Lemma I. oN = AO. therefore (o falling beyond I by hypotbeſis) O will fall
beyond P; but by hypotbeſis it falls ſhort of E; therefore O falls between
P and E.
Lemma I. oN = AO. therefore (o falling beyond I by hypotbeſis) O will fall
beyond P; but by hypotbeſis it falls ſhort of E; therefore O falls between
P and E.
Next to ſhew that ο will fall between T and I, we have AT:
TI:
: AE:
EP
And by Diviſion AT:
AI:
: AE:
AP
Hence AT x AP = IAE or o AO