## 86.PROBLEM III.

In this, the point O is ſought without all the given ones, and the three
Epitagmas are as in Problem I.

Epitagma I. There are here eight Caſes, viz. four when the order of
the given points is the ſame as ſpecified in Epitagma I. of Problem I, and
O ſought beyond the given point which bounds the antecedent rectangle; and four others when O is ſought beyond that which bounds the conſe-
quent one: the Conſtructions of the four firſt are ſhewn by the ſmall
letters b and o in Fig. 32, 34, 36 and 38; and the four latter ones by
the ſame letters in Fig. 33, 35, 37 and 39; and the demonſtrations that
o will fall as required by the Problem are exactly the ſame as thoſe made
uſe of in the laſt mentioned Epitagma. It is farther obſervable, that the
four firſt Caſes are not poſſible, unleſs the given ratio be of a leſs to a
greater; nor the four latter, unleſs it be of a greater to a leſs, as is ma-
nifeſt without farther illuſtration.

Epitagma II. Here, as in the ſecond Epitagma of Problem I, the points
A and U are one an extreme, and the other an adjacent mean, and there
are eight Caſes; but it will be ſufficient to exhibit the conſtructions of
four of them, the others being not eſſentially different; and theſe are ſhewn
by the ſmall b and o in Fig. 40, 41, 42 and 43; the demonſtrations that o
will fall as required need not be pointed out here; but it may be neceſſary
to remark that the firſt and third are not poſſible unleſs the given ratio be
of a leſs to a greater, nor the ſecond and fourth unleſs it be of a greater to
a leſs, as is obvious enough.

Epitagma III. In which the points A and U are both means, or both
extremes; and there are here eight Caſes, viz. four wherem theſe points
are extremes, and four others wherein they are means: but theſe laſt being
reducible to the former by the ſame method that was uſed in the third Epi-
tagmas of the two preceding Problems, I ſhall omit them.

All the Caſes of this Epitagma are conſtructed by making B fall beyond
I, and C beyond E, with reſpect to A and U; and drawing DH parallel to
BC. That O will fall beyond A in Fig. 58 and 60, and beyond U in Fig. 59 and 61 appears hence. Draw AG perpendicular to BC, meeting the
circle on BC in G: by Eu. VI. 13. 17, the ſquare on AG is equal to the

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