An ESSAY
cauſe the Triangles THX and a F X are ſimilar,

TH — a F: a F : : Ta: a X.

And becauſe the Triangles T I x and ax L, are

alſo ſimilar, we have

TI + a L : : a L: Ta : ax.

Now let PM NR be the perſpective Plane,

O the Eye, A Q the Perpendicular, whoſe

Perſpective is requir’d, and O t a perpendicular

let fall from the Eye upon the perſpective Plane,

and ſo t will be the ſame, as the Point T in the

aforegoing Figure, Now if the Lines O Q be

drawn, it is manifeſt that A x, or A X, is the

Perſpective of A Q, according as this Line is

above or below the perſpective Plane in reſpect

to the Eye. Then becauſe the Triangles O t x

and Q A x are ſimilar, we have

O t — A Q: A Q : : t A: Ax.

And ſince the Triangles O t X and X A Q are

ſimilar,

O t + A Q: A Q : : t A: A X.

Now Ot is equal to TH or TI of the afore-

going Figure, and AQ to a F or a L of the

ſame Figure; as likewiſe At, Ta: Therefore

if theſe two laſt Proportions be compared with

the two precedent ones, we ſhall find A x = a X,

and A X = a x; which was to be demon-

ſtrated.

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151.
Remarks
.

96. When the two Circles interſect each other,

or fall within one another, and ſo this Way be-

comes uſeleſs; a Line muſt be drawn at Pleaſure,

through the Point T, equal to the Diſtance of

the Eye from the perſpective Plane; and then a

parallel equal to the given Perpendicular muſt be

drawn to the ſaid Line through the Point a, ei-

ther towards L or F, according as the Perpen-