7132An ESSAY
To demonſtrate which, draw the Line D L M
thro’ the Point D, parallel to a b s. Then, be-
cauſe the Triangles D M O and D L i are ſimi-
lar, we have,
thro’ the Point D, parallel to a b s. Then, be-
cauſe the Triangles D M O and D L i are ſimi-
lar, we have,
D M = as:
D L = ab:
: M O:
L i.
Again,
in the precedent Figure, the Triangles A S C and
A B E are ſimilar: Whence,
A S: A B: : C S: E B.
in the precedent Figure, the Triangles A S C and
A B E are ſimilar: Whence,
A S: A B: : C S: E B.
The three firſt Terms of theſe two Progreſſions
are the ſame: For CS is equal to M O, ſince
they are each the Difference of the Height of the
Eye, and that of the given Point; and conſe-
quently, E B is equal to L i: But B I was made
equal to B E, pl{us} FC the Height of the given
Point above the Geometrical Plane; and b i is
equal to Li, pl{us} b L; which being equal to aD,
is likewiſe the Height of the given Point above
the Geometrical Plane; whence the Lines B I
and b i are equal. Which was to be demon-
ſtrated.
are the ſame: For CS is equal to M O, ſince
they are each the Difference of the Height of the
Eye, and that of the given Point; and conſe-
quently, E B is equal to L i: But B I was made
equal to B E, pl{us} FC the Height of the given
Point above the Geometrical Plane; and b i is
equal to Li, pl{us} b L; which being equal to aD,
is likewiſe the Height of the given Point above
the Geometrical Plane; whence the Lines B I
and b i are equal. Which was to be demon-
ſtrated.
Prob. VI.