e two drawings of
triangles which explain themselves. To put a triangle into perspective
we have but to find three points, such as _fEP_, Fig. 148 A, and then
transfer these points to the perspective square 148 B, as there shown,
and form the perspective triangle; but these figures explain themselves.
Any other triangle or rectilineal figure can be worked out in the same
way, which is not only the simplest method, but it carries its
mathematical proof with it.

[Illustration: Fig. 148 A.]

[Illustration: Fig. 148 B.]

[Illustration: Fig. 149 A.]

[Illustration: Fig. 149 B.]

LXXX

PERSPECTIVE OF A SQUARE PLACED AT AN ANGLE NEW METHOD

As we have drawn a triangle in a square so can we draw an oblique square
in a parallel square. In Figure 150 A we have drawn the oblique square
_GEPn_. We find the points on the base _Am_, as in the previous figures,
which enable us to construct the oblique perspective square _n'G'E'P'_
in the parallel perspective square Fig. 150 B. But it is not necessary
to construct the geometrical figure, as I will show presently. It is
here introduced to explain the method.

[Illustration: Fig. 150 A.]

[Illustration: Fig. 150 B.]

Fig. 150 B. To test the accuracy of the above, produce sides _G'E'_ and
_n'P'_ of perspective square till they touch the horizon, where they
will meet at _V_, their vanishing point, and again produce the other
sides _n'G'_ and _P'E'_ till they meet on the horizon at the other
vanishing point, which they must do if the figure is correctly drawn.

In any parallel square construct an oblique square from a given
point--given the parallel square at Fig. 150 B, and given point _n'_ on
base. Make _A'f'_ equal to _n'm'_, draw _f'S_ and _n'S_ to point of
sight. Where these lines cut the diagonal _AC_ draw horizontals to _P'_
and _G'_, and so find the four points _G'E'P'n'_ through which to draw
the square.

LXXXI

ON A GIVEN LINE PLACED AT AN ANGLE TO THE BASE DRAW A SQUARE IN ANGULAR
PERSPECTIVE, THE POINT OF SIGHT, AND DISTANCE, BEING GIVEN.

[Illustration: Fig. 151.]

Let _AB_ be the given line, _S_ the point of sight, and _D_ the distance
(Fig. 151, 1). Through _A_ draw _SC_ from point of sight to base (Fig.
151, 2 and 3). From _C_ draw _CD_ to point of distance. Draw _Ao_
parallel to base till it cuts _CD_ at _o_, through _O_ draw _SP_, from
_B_ mark off _BE_ equal to _CP_. From _E_ draw _ES_ intersecting _CD_ at
_K_, from _K_