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An ESSAY
the Circular Baſe of the Cone will be deter-

min’d.

## 69. 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.

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.

The Triangles D Q P and G H D are ſimilar, whence

D G: D P: : G H: P Q.