8739on PERSPECTIVE.
the Station Point, and A the Point wherein the
Perpendicular meets the Geometrical Plane.
Perpendicular meets the Geometrical Plane.
Operation,
Without Compaſſes.
Without Compaſſes.
Having firſt found the Perſpective a of 1131.
Point A, draw the Line A S cutting the Baſe
Line in E, through which Point E draw the
Line Ea; then from the Point B draw a Line
B a to the Point a, cutting the Horizontal Line
in F. Again through F draw a Line to the
Point L, cutting E a in I; and a I is the Repre-
ſentation ſought.
Line in E, through which Point E draw the
Line Ea; then from the Point B draw a Line
B a to the Point a, cutting the Horizontal Line
in F. Again through F draw a Line to the
Point L, cutting E a in I; and a I is the Repre-
ſentation ſought.
Demonstration.
To prove this, let G N be a Perpendicular to
the Baſe Line drawn from the Point G, wherein
the ſaid Baſe Line is cut by the Line B F; alſo
let G D be equal to the Perpendicular whoſe Ap-
pearance is ſought, and a H parallel to the Baſe
Line.
the Baſe Line drawn from the Point G, wherein
the ſaid Baſe Line is cut by the Line B F; alſo
let G D be equal to the Perpendicular whoſe Ap-
pearance is ſought, and a H parallel to the Baſe
Line.
It is plain that the Perſpective of E A is
E a: But E A paſſes through the Station Point;
and conſequently its Repreſentation is 2241. dicular to the Baſe Line; therefore we are 3356. to prove, that a I is equal to a H.
E a: But E A paſſes through the Station Point;
and conſequently its Repreſentation is 2241. dicular to the Baſe Line; therefore we are 3356. to prove, that a I is equal to a H.
Now the Triangles B G C and B F M are ſimi-
lar; and ſo
lar; and ſo
B C :
B M :
: B G:
B F.
But B M by Conſtruction is the double of B C;
whence B F is alſo the double of B G, which,
conſequently, is equal to G F.
whence B F is alſo the double of B G, which,
conſequently, is equal to G F.
Becauſe the Triangles F G N and F B L are
ſimilar, therefore
ſimilar, therefore
F G :
F B :
:
G N :
B L.