And the Triangles D P m and D G n are ſimilar: Wherefore
D G: D P: G n: : P m. And
G H: P Q: : G n: P m.

The Triangles B A N and L A M are ſimilar: Therefore,
BA: AL: : AN: AM. But the three firſt Terms of the two laſt Proportions,
are equal between themſelves; whence P m is alſo
equal to A M. Which was to be demonſtrated.

70. Remarks .

54. When the Height of the Cone is greater
than the Height of the Eye, the Points, G and
H, will fall below the Point D; in which Caſe,
the Lines A B and A C muſt be produc’d, till
they cut the Circle in the Points l and f, oppo-
ſite to L and F: Then lIf will be the viſible
Part of the Baſe.

When the Cone is inclin’d, ſo that T (for Ex-
ample) is the Seat of its Vertex; AT muſt be
drawn: And then having aſſum’d P D equal to
the perpendicular Height of the Cone, and Pt
equal to A T; the Line t D x muſt be drawn; and the Part T X, taken in A T, equal to G x. Alſo, X S muſt be drawn, and A s, equal and
parallel thereto.

This being done; the ſame Method muſt be
apply’d here, that I have laid down for the up-
right Cone; with this Difference only, that the
Point s muſt be us’d inſtead of the Station Point
S. But when the Height of the Cone is greater
than the Height of the Eye, the Point X muſt
be aſſum’d in the Line T A, between the Points
T and A.