1D A hath to A Z; which is the ſame that the Rectangle K E hath to
the Rectangle A G, their heights A K and K L being equal. There
fore the proportion that the Square E D hath to the Square Z G;
that is, the Square L A hath to the Square A K, the Rectangle K E
hath likewiſe to the Rectangle K Z. And in the ſelf-ſame manner
we might prove that the other Rectangles L F, M H, N I, O B are
to one another as the Squares of the Lines M A, N A, O A, P A.
Conſider we in the next place, how the Circumſcribed Figure is
compounded of certain Spaces that are to one another as the
Squares of the Lines that exceed with Exceſſes equal to the leaſt,
and how the Rectangle C P is compounded of ſo many other Spa
ces each of them equal to the Greateſt, which are all the Rectan
gles equal to O B. Therefore, by the Lemma of Archimedes, the
Circumſcribed Figure is more than the third part of the Rectangle
C P: But it was alſo leſſe, which is impoſſible: Therefore the
Mixt-Triangle is not leſſe than one third of the Rectangle C P.
I ſay likewiſe, that it is not more: For if it be more than one
third of the Rectangle C P, ſuppoſe the Space X equal to the ex
ceſſe of the Triangle above the third part of the ſaid Rectangle
C P, and the diviſion and ſubdiviſion of the Rectangle into Rect
angolets, but alwaies equal, being made, we ſhall meet with ſuch as
that one of them is leſſer than the Space X; which let be done:
and let the Rectangle B O be leſſer than X; and, having deſcribed
the Figure as before, we ſhall have inſcribed in the Mixt-Triangle
a Figure compounded of the Rectangles V O, T N, S M, N L, Q K,
which yet ſhall not be leſs
72[Figure 72]
than the third part of the
great Rectangle C P, for
the Mixt Triangle doth
much leſſe exceed the In
ſcribed Figure than it doth
exceed the third part of
the Rectangle C P; Be
cauſe the exceſſe of the
Triangle above the third part of the Rectangle C P is equal to
the Space X which is greater than the Rectangle B O, and this al
ſo is conſiderably greater than the exceſſe of the Triangle above
the Inſcribed Figure: For to the Rectangle B O, all the Rectan
golets A G, G E, E F, F H, H I, I B are equal, of which the Ex
ceſſes of the Triangle above the Inſcribed Figure are leſſe than
half: And therefore the Triangle exceeding the third part of the
Rectangle C P, by much more (exceeding it by the Space X)
than it exceedeth its inſcribed Figure, that ſame Figure ſhall alſo
be greater than the third part of the Rectangle C P: But it is leſſer,
by the Lemma preſuppoſed: For that the Rectangle C P, as being
the Rectangle A G, their heights A K and K L being equal. There
fore the proportion that the Square E D hath to the Square Z G;
that is, the Square L A hath to the Square A K, the Rectangle K E
hath likewiſe to the Rectangle K Z. And in the ſelf-ſame manner
we might prove that the other Rectangles L F, M H, N I, O B are
to one another as the Squares of the Lines M A, N A, O A, P A.
Conſider we in the next place, how the Circumſcribed Figure is
compounded of certain Spaces that are to one another as the
Squares of the Lines that exceed with Exceſſes equal to the leaſt,
and how the Rectangle C P is compounded of ſo many other Spa
ces each of them equal to the Greateſt, which are all the Rectan
gles equal to O B. Therefore, by the Lemma of Archimedes, the
Circumſcribed Figure is more than the third part of the Rectangle
C P: But it was alſo leſſe, which is impoſſible: Therefore the
Mixt-Triangle is not leſſe than one third of the Rectangle C P.
I ſay likewiſe, that it is not more: For if it be more than one
third of the Rectangle C P, ſuppoſe the Space X equal to the ex
ceſſe of the Triangle above the third part of the ſaid Rectangle
C P, and the diviſion and ſubdiviſion of the Rectangle into Rect
angolets, but alwaies equal, being made, we ſhall meet with ſuch as
that one of them is leſſer than the Space X; which let be done:
and let the Rectangle B O be leſſer than X; and, having deſcribed
the Figure as before, we ſhall have inſcribed in the Mixt-Triangle
a Figure compounded of the Rectangles V O, T N, S M, N L, Q K,
which yet ſhall not be leſs
72[Figure 72]than the third part of the
great Rectangle C P, for
the Mixt Triangle doth
much leſſe exceed the In
ſcribed Figure than it doth
exceed the third part of
the Rectangle C P; Be
cauſe the exceſſe of the
Triangle above the third part of the Rectangle C P is equal to
the Space X which is greater than the Rectangle B O, and this al
ſo is conſiderably greater than the exceſſe of the Triangle above
the Inſcribed Figure: For to the Rectangle B O, all the Rectan
golets A G, G E, E F, F H, H I, I B are equal, of which the Ex
ceſſes of the Triangle above the Inſcribed Figure are leſſe than
half: And therefore the Triangle exceeding the third part of the
Rectangle C P, by much more (exceeding it by the Space X)
than it exceedeth its inſcribed Figure, that ſame Figure ſhall alſo
be greater than the third part of the Rectangle C P: But it is leſſer,
by the Lemma preſuppoſed: For that the Rectangle C P, as being