on PERSPECTIVE. too obliquely, recourſe muſt be had to Problem I. , to find the Appearance of a.

127. Method II.

85. A is the Foot of the Perpendicular: The
Triangle, E P M, is drawn as directed: And T is the accidental Point of the Perpendiculars,
to the Geometrical Plane.

128. Operation .

Thro’ the Point a, the Appearance of A,
draw a Perpendicular to the Baſe Line; which
make equal in Repreſentation to the Line M E; in conſidering this laſt Line, as being
parallel to the Vertical Line. Then, from the
Extremity I of this Perſpective, to the Point of
Sight V, draw a Line cutting the Line T a, in
the Point X; which will be the Repreſentation
of the Extremity of the propos’d Line.

129. Demonstration .

Let us ſuppoſe a Line paſſing thro’ the Point
A, equal to M E, and parallel to the Verti-
cal Line. Suppoſe, moreover, that another Line
is drawn thro’ the Extremity of this Line, and
that of the propos’d Perpendicular; then this
laſt Line, by the Conſtruction of the Figure
M E P, will be parallel to the Station Line; and conſequently, its Repreſentation will paſs thro’ the Point of Sight; and its Interſection
with T a, will be the Extremity of the Repre-
ſentation ſought. But a I is the Perſpective of the firſt Line, made equal to E M; and con-
ſequently, V I is that of the ſecond. Which was
to be demonſtrated.