This being granted: Let the Parabolick Line A B be inſcribed
in this Rectangle A C B P: we are to prove the Mixt Triangle
B A P, whoſe ſides are B P and P A, and Baſe the Parabolical Line
B A, to be the third part of the whole Rectangle C P. For if it be
not ſo, it will be either more than the third part, or leſſe. Let it be
ſuppoſed that it may be
leſſe, and to that which is
71[Figure 71]
wanting ſuppoſe the Space
X to be equal. Then di
viding the Rectangle con
tinually into equal parts
with Lines parallel to the
Sides B P and C A, we
ſhall in the end arrive at
ſuch parts, as that one of them ſhall be leſſe than the Space X.
Now let one of them be the Rectangle O B, and by the Points
where the other Parallels interſect the Parabolick Line, let the Pa
rallels to A P paſſe: and here I will ſuppoſe a Figure to be cir
cumſcribed about our Mixt-Triangle, compoſed of Rectangles,
which are B O, I N, H M, F L, E K, G A; which Figure ſhall alſo
yet be leſs than the third part of the Rectangle C P, in regard that
the exceſſe of that Figure over and above the Mixed Triangle is
much leſſe than the Rectangle B O, which yet again is leſſe than
the Space X.
in this Rectangle A C B P: we are to prove the Mixt Triangle
B A P, whoſe ſides are B P and P A, and Baſe the Parabolical Line
B A, to be the third part of the whole Rectangle C P. For if it be
not ſo, it will be either more than the third part, or leſſe. Let it be
ſuppoſed that it may be
leſſe, and to that which is
71[Figure 71]wanting ſuppoſe the Space
X to be equal. Then di
viding the Rectangle con
tinually into equal parts
with Lines parallel to the
Sides B P and C A, we
ſhall in the end arrive at
ſuch parts, as that one of them ſhall be leſſe than the Space X.
Now let one of them be the Rectangle O B, and by the Points
where the other Parallels interſect the Parabolick Line, let the Pa
rallels to A P paſſe: and here I will ſuppoſe a Figure to be cir
cumſcribed about our Mixt-Triangle, compoſed of Rectangles,
which are B O, I N, H M, F L, E K, G A; which Figure ſhall alſo
yet be leſs than the third part of the Rectangle C P, in regard that
the exceſſe of that Figure over and above the Mixed Triangle is
much leſſe than the Rectangle B O, which yet again is leſſe than
the Space X.
SAGR. Softly, I pray you, for I do not ſee how the exceſſe of
this circumſcribed Figure above the Mixt Triangle is conſiderably
leſſer than the Rectangle B O.
this circumſcribed Figure above the Mixt Triangle is conſiderably
leſſer than the Rectangle B O.
SALV. Is not the Rectangle B O equal to all theſe ſmall Rect
angles by which our Parabolical Line paſſeth; I mean theſe, B I,
I H, H F, F E, E G, and G A, of which one part only lyeth with
out the Mixt Triangle? And the Rectangle B O, is it not alſo ſup
poſed to be leſſe than the Space X? Therefore if the Triangle to
gether with X did, as the Adverſary ſuppoſeth, equalize the third
part of the Rectangle C P the circumſcribed Figure that adjoyns
to the Triangle ſo much leſſe than the Space X, will remain even
yet leſſe than the third part of the ſaid Rectangle C P. But this
cannot be, for it is more than a third part, therefore it is not true
that our Mixt Triangle is leſſe than one third of the Rectangle.
angles by which our Parabolical Line paſſeth; I mean theſe, B I,
I H, H F, F E, E G, and G A, of which one part only lyeth with
out the Mixt Triangle? And the Rectangle B O, is it not alſo ſup
poſed to be leſſe than the Space X? Therefore if the Triangle to
gether with X did, as the Adverſary ſuppoſeth, equalize the third
part of the Rectangle C P the circumſcribed Figure that adjoyns
to the Triangle ſo much leſſe than the Space X, will remain even
yet leſſe than the third part of the ſaid Rectangle C P. But this
cannot be, for it is more than a third part, therefore it is not true
that our Mixt Triangle is leſſe than one third of the Rectangle.
SAGR. I underſtand the Solution of my Doubt. But it is
requiſite now to prove unto us, that the Circumſcribed Figure is
more than a third part of the Rectangle C P; which, I believe, will
be harder to do.
requiſite now to prove unto us, that the Circumſcribed Figure is
more than a third part of the Rectangle C P; which, I believe, will
be harder to do.
SALV. Not at all. For in the Parabola the Square of the Line
D E hath the ſame proportion to the Square of Z G, that the Line
D E hath the ſame proportion to the Square of Z G, that the Line