An ESSAY
to the Length of the ſaid Perpendiculars: And
therefore
155.
H I: T H:: a X: a T.
But in the Conſtruction of this Problem, be-
cauſe the Triangles T C D and T H G, are
ſimilar;
H G = H I: T H: : C D = a X: T D = a T; and conſequently H I and a X, repreſent Perpen-
diculars of the ſame Length. Which was to be
demonſtrated.
156.
Prob
. III.
98. To find the accidental Point of any Number
of parallel Lines inclined to the Geometrical Plane.
Let a b be the Perſpective of the Direction
of one of the given Lines.
157.
Operation
.
Draw the Line F T L, parallel to a b, through
the accidental Point T of the Lines perpendicu-
lar to the Geometrical Plane; and at the Point
T, raiſe the Perpendicular T G, which make e-
qual to the Diſtance of the Eye from the per-
ſpective Plane; then draw the Line G L, or
G F, ſo that the Angle T L G, or T F G, be e-
qual to the Angle of the Inclination of the given
Lines; and the Point L, will be the Accidental
Point ſought, if the given Lines incline to-
wards b; but if they incline towards a, F will
be the Accidental Point.
158.
Demonstration
.
It is manifeſt by Conſtruction, that if T G be
ſuppoſed to be raiſed perpendicularly to the Geo-
metrical Plane, G L or G F, will be parallel to