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.
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.
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.
48[Figure 48]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.
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