greateſt of all, and AE will therefore be leſs than AO, and AI greater. And

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.

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.

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*. ]

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.

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.

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.

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.

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