1that A Z hath to Z D; by the fourth Propoſition of Archimedes, De quadratura Para
bolæ: But A Z is Seſquialter of Z D; for it is as three to two, as we ſhallanon demon-
bolæ: But A Z is Seſquialter of Z D; for it is as three to two, as we ſhallanon demon-
ſtrate: Therefore D B is Seſquialter of B V; but D B and B K are Seſquialter:
And, therefore, the Lines (c) B V and B K are equall: Which is imposſible:
Therefore the Section of the Right-angled Cone A E I, ſhall paſs thorow the Point K; which
we would demonstrate.
(c) By 9 of the
fifth,
fifth,
In regard, therefore, that the three Portions A P O L, A E I
and A T D are contained betwixt Right Lines and the Sections
of Right-angled Cones, and are Right, alike and unequall,
touching one another, upon one and the ſame Baſe.] After theſe words,
upon one and the ſame Baſe, we may ſee that ſomething is obliterated, that is to be
deſired: and for the Demonſtration of theſe particulars, it is requiſite in this place to
premiſe ſome things: which will alſo be neceſſary unto the things that follow.
and A T D are contained betwixt Right Lines and the Sections
of Right-angled Cones, and are Right, alike and unequall,
touching one another, upon one and the ſame Baſe.] After theſe words,
upon one and the ſame Baſe, we may ſee that ſomething is obliterated, that is to be
deſired: and for the Demonſtration of theſe particulars, it is requiſite in this place to
premiſe ſome things: which will alſo be neceſſary unto the things that follow.
M
LEMMA. I.
Let there be a Right Line A B; and let it be cut by two Lines,
parallel to one another, A C and D E, ſo, that as A B is to
B D. ſo A C may be to D E. I ſay that the Line that con
joyneth the Points C and B ſhall likewiſe paſs by E.
parallel to one another, A C and D E, ſo, that as A B is to
B D. ſo A C may be to D E. I ſay that the Line that con
joyneth the Points C and B ſhall likewiſe paſs by E.
274[Figure 274]For, if poſſible, let it not paſs by E, but either
above or below it. Let it first paſs below it,
as by F. The Triangles A B C and D B F ſhall
be alike: And, therefore, as (a) A B is to B D,
ſo is A C to D F: But as A B is to B D, ſo was
A C to D E: Therefore (b) D F ſhall be equall to
D E: that is, the part to the whole: Which is
abſurd. The ſame abſurditie will follow, if the
Line C B be ſuppoſed to paſs above the Point E:
And, therefore, C B muſt of necesſity paſs thorow
E: Which was required to be demonſtrated.
above or below it. Let it first paſs below it,
as by F. The Triangles A B C and D B F ſhall
be alike: And, therefore, as (a) A B is to B D,
ſo is A C to D F: But as A B is to B D, ſo was
A C to D E: Therefore (b) D F ſhall be equall to
D E: that is, the part to the whole: Which is
abſurd. The ſame abſurditie will follow, if the
Line C B be ſuppoſed to paſs above the Point E:
And, therefore, C B muſt of necesſity paſs thorow
E: Which was required to be demonſtrated.
(a) By 4. of the
ſixth.
ſixth.
(b) By 9. of the
fifth.
fifth.
LEMMA. II.
Let there be two like Portions, contained betwixt Right Lines,
and the Sections of Right-angled Cones; A B C the great
er, whoſe Diameter let be B D; and E F C the leſser, whoſe
Diameter let be F G: and, let them be ſo applyed to one
another, that the greater include the leſser; and let their
Baſes A C and E C be in the ſame Right Line, that the ſame
Point C, may be the term or bound of them both: And,
then in the Section A B C, take any Point, as H; and draw
a Line from H to C. I ſay, that the Line H C, hath to that
part of it ſelf, that lyeth betwixt C and the Section E F C, the
ſame proportion that A C hath to C E.
and the Sections of Right-angled Cones; A B C the great
er, whoſe Diameter let be B D; and E F C the leſser, whoſe
Diameter let be F G: and, let them be ſo applyed to one
another, that the greater include the leſser; and let their
Baſes A C and E C be in the ſame Right Line, that the ſame
Point C, may be the term or bound of them both: And,
then in the Section A B C, take any Point, as H; and draw
a Line from H to C. I ſay, that the Line H C, hath to that
part of it ſelf, that lyeth betwixt C and the Section E F C, the
ſame proportion that A C hath to C E.
Draw B C, which ſhall paſs thorow F, For, in regard, that the Portions are alike, the
Diameters with the Baſes contain equall Angles: And, therefore, B D and F G are parallel
to one another: and B D is to A C, as F G it to E C: and, Permutando, B D is to F G, as
A C is to C E; that is, (a) as their halfes D C to C G; therefore, it followeth, by the
preceding Lemma, that the Line B C ſhall paſs by the Point F. Moreover, from the Point
H unto the Diameter B D, draw the Line H K, parallel to the Baſe A C: and, draw a Line
Diameters with the Baſes contain equall Angles: And, therefore, B D and F G are parallel
to one another: and B D is to A C, as F G it to E C: and, Permutando, B D is to F G, as
A C is to C E; that is, (a) as their halfes D C to C G; therefore, it followeth, by the
preceding Lemma, that the Line B C ſhall paſs by the Point F. Moreover, from the Point
H unto the Diameter B D, draw the Line H K, parallel to the Baſe A C: and, draw a Line
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