### Full text

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,

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.

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.

Note, When the Height of the given Point is

greater than the Height of the Eye, E B muſt

be taken from that firſt Height, to have the

Magnitude of B I.

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66.
Prob
. VI.

52. To throm a Pyramid, or Cone, into Perſpective.

Now, to throw a Pyramid into perſpective,

the Appearance of its Baſe and Center muſt be
found : After which, Lines muſt be drawn from
the Repreſentation of the Vertex, to the Ap-

pearance of thoſe Angles of the Baſe that are

viſible; and then the Perſpective ſought will be

had.

### 66.1.

And to throw a Cone into perſpective, the

Repreſentation of its Baſe and Vertex muſt be