7334An ESSAY
the Circular Baſe of the Cone will be deter-
min’d.
min’d.
Demonstration.
To prove this, draw the Lines B C and L F, cut-
ting the Line A S in the Points N and M; and make
the Line G n equal to A N, and draw the Line
n D m. It is now manifeſt, that if the Cone be
continued out above its Vertex, (that is, if the oppo-
ſite Cone be form’d) it will cut the Horizontal Plane
in a Circle equal to B E C, whoſe Seat will be BEC:
So that the Point S, in reſpect of B E C, is in the
ſame Situation as the Eye hath, with reſpect to the
Circle form’d in the Horizontal Plane, by the Conti-
nuation of the Cone. Whence it follows, that B C
is the Seat of the viſible Portion of that Circle. For,
by Conſtruction, B and C are the Points of Contact
of the Tangents to the Circle B E C, which paſs
thro’ the Point S; becauſe the Angle ABS, which
is in a Semicircle, is a right one.
ting the Line A S in the Points N and M; and make
the Line G n equal to A N, and draw the Line
n D m. It is now manifeſt, that if the Cone be
continued out above its Vertex, (that is, if the oppo-
ſite Cone be form’d) it will cut the Horizontal Plane
in a Circle equal to B E C, whoſe Seat will be BEC:
So that the Point S, in reſpect of B E C, is in the
ſame Situation as the Eye hath, with reſpect to the
Circle form’d in the Horizontal Plane, by the Conti-
nuation of the Cone. Whence it follows, that B C
is the Seat of the viſible Portion of that Circle. For,
by Conſtruction, B and C are the Points of Contact
of the Tangents to the Circle B E C, which paſs
thro’ the Point S; becauſe the Angle ABS, which
is in a Semicircle, is a right one.
Now, if a Plane be conceiv’d, as paſſing thro’ ſome
Points in the Horizontal Plane, whoſe Seats are
B and C, and which cuts the two oppoſite Cones
thro’ their Vertex; it is evident, that this Plane
continued, will cut the Geometrical Plane in a Line
parallel to B N C; and that this Line upon the
ſaid Plane, will determine the viſible Part of the
Cone’s Baſe. So, ſince G n was made equal to
A N, we have only to prove, that P m is equal to
A M: For, it follows from thence, that L M F is
the Common Section of the Geometrical Plane, and
the Plane which we have here imagin’d.
Points in the Horizontal Plane, whoſe Seats are
B and C, and which cuts the two oppoſite Cones
thro’ their Vertex; it is evident, that this Plane
continued, will cut the Geometrical Plane in a Line
parallel to B N C; and that this Line upon the
ſaid Plane, will determine the viſible Part of the
Cone’s Baſe. So, ſince G n was made equal to
A N, we have only to prove, that P m is equal to
A M: For, it follows from thence, that L M F is
the Common Section of the Geometrical Plane, and
the Plane which we have here imagin’d.
The Triangles D Q P and G H D are ſimilar, whence
D G: D P: : G H: P Q.
D G: D P: : G H: P Q.
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