1be ſuppoſed to paſſe by the Point E, which ſhall make in the Cone
a Circular Section, whoſe Diameter is G E H. And becauſe upon
the Diameter I K of the Circle I B K, B D is a Perpendicular, the
Square of B D ſhall be equal to the Rectangle made by the parts
I D and D K: And likewiſe in the upper Circle which is underſtood
to paſſe by the points G F H, the Square of the Line F E is equal
to the Rectangle of the parts G E H: Therefore the Square of B D
hath the ſame proportion to the Square of F E, that the Rectangle
I D K hath to the Rectangle G E H. And becauſe the Line E D is
Parallel to H K, E H ſhall be equal to D K, which alſo are Parallels:
And therefore the Rectangle I D K ſhall have the ſame proportion
to the Rectangle G E H, as I D hath to G E; that is, that D A hath
to A E: Therefore the Rectangle I D K to the Rectangle G E H,
that is, the Square B D to the Square F E, hath the ſame proportion
that the Axis D A hath to the part A E: Which was to be de
monſtrated.
a Circular Section, whoſe Diameter is G E H. And becauſe upon
the Diameter I K of the Circle I B K, B D is a Perpendicular, the
Square of B D ſhall be equal to the Rectangle made by the parts
I D and D K: And likewiſe in the upper Circle which is underſtood
to paſſe by the points G F H, the Square of the Line F E is equal
to the Rectangle of the parts G E H: Therefore the Square of B D
hath the ſame proportion to the Square of F E, that the Rectangle
I D K hath to the Rectangle G E H. And becauſe the Line E D is
Parallel to H K, E H ſhall be equal to D K, which alſo are Parallels:
And therefore the Rectangle I D K ſhall have the ſame proportion
to the Rectangle G E H, as I D hath to G E; that is, that D A hath
to A E: Therefore the Rectangle I D K to the Rectangle G E H,
that is, the Square B D to the Square F E, hath the ſame proportion
that the Axis D A hath to the part A E: Which was to be de
monſtrated.
The other Propoſition, likewiſe neceſſary to the preſent Tract,
we will thus make out. Let us deſcribe the Parabola, of which let the
Axis C A be prolonged out unto D; and taking any point B, let the
Line B C be ſuppoſed to be continued out by the ſame Parallel un
145[Figure 145]
to the Baſe of the ſaid Parabola;
and let D A be ſuppoſed equal
to the part of the Axis C A. I ſay,
that the Right-Line drawn by
the points D and B, falleth not
within the Parabola, but without,
ſo as that it only toucheth the
ſame in the ſaid point B: For, if
it be poſſible for it to fall within,
it cutteth it above, or being pro
longed, it cutteth it below. And
in that Line let any point G be
taken, by which paſſeth the Right
Line F G E. And becauſe the
Square F E is greater than the
Square G E, the ſaid Square F E
ſhall have greater proportion to
the Square B C, than the ſaid Square G E hath to the ſaid B C. And
becauſe, by the precedent, the Square F E is to the Square B C as
E A is to A C; therefore E A hath greater proportion to A C, than
the Square G E hath to the Square B C; that is, than the Square
E D hath to the Square D C: (becauſe in the Triangle D G E as
G E is to the Parallel B C, ſo is E D to D C:) But the Line E A to
A C, that is, to A D hath the ſame proportion that four Rectangles
E A D hath to four Squares of A D, that is, to the Square C D,
we will thus make out. Let us deſcribe the Parabola, of which let the
Axis C A be prolonged out unto D; and taking any point B, let the
Line B C be ſuppoſed to be continued out by the ſame Parallel un
145[Figure 145]to the Baſe of the ſaid Parabola;
and let D A be ſuppoſed equal
to the part of the Axis C A. I ſay,
that the Right-Line drawn by
the points D and B, falleth not
within the Parabola, but without,
ſo as that it only toucheth the
ſame in the ſaid point B: For, if
it be poſſible for it to fall within,
it cutteth it above, or being pro
longed, it cutteth it below. And
in that Line let any point G be
taken, by which paſſeth the Right
Line F G E. And becauſe the
Square F E is greater than the
Square G E, the ſaid Square F E
ſhall have greater proportion to
the Square B C, than the ſaid Square G E hath to the ſaid B C. And
becauſe, by the precedent, the Square F E is to the Square B C as
E A is to A C; therefore E A hath greater proportion to A C, than
the Square G E hath to the Square B C; that is, than the Square
E D hath to the Square D C: (becauſe in the Triangle D G E as
G E is to the Parallel B C, ſo is E D to D C:) But the Line E A to
A C, that is, to A D hath the ſame proportion that four Rectangles
E A D hath to four Squares of A D, that is, to the Square C D,