Whence

G I: G C: : G F: G O.

Again, becauſe the Triangles G I E and G C D
are ſimilar, we have

G I: G C: : G E: G D.

And conſequently

G F: G O: : G E: G D.

And ſo the Triangles G F E, and G O D are
ſimilar; and the Line F E A is parallel to O D: Whence it follows , that the Perſpective of E A, is a Part of E a D. We demonſtrate in
the ſame Manner, that B a is the Perſpective
of B A, and ſo the Perſpective of the Point A,
the common Section of E A and B A, is a, the
Interſection of the Appearances of the ſaid two
Lines.

119. Prob . IV.

80. To find the Repreſentation of a Line, per-
pendicular to the Geometrical Plane, when the per-
ſpective Plane is above the Eye.

In the Baſe Line B E, aſſume the Line E D,
equal in Length to the propoſed Perpendicular; and draw C L, parallel to the Baſe Line, and
diſtant therefrom (for Example) {1/4} of the Height
of the Eye; make F L equal to {3/4} of D E, and
draw the Lines E L and D F. Note, if the
Diſtance from C L to B E, had been aſſumed
equal to a fifth Part of the Height of the Eye,
F L muſt have been aſſumed equal to {4/5} Parts of
E D. Now let a be the Perſpective of the Foot
of the propoſed Perpendicular; through which
draw a H parallel to the Baſe Line, and a I per-
pendicular to the ſaid Line; then make a I equal
to G H, and the propoſed Perſpective will be
had. The Demonſtration of this Operation is
manifeſt , in conſidering that D F and E L