en
triangle _A_, or five points to obtain the five-sided figure at _B_.
So can we deal with any number of figures placed at any angle.

[Illustration: Fig. 114.]

Both the above figures are placed in the same diagram, showing how any
number can be drawn by means of the same point of sight and the same
point of distance, which makes them belong to the same picture.

It is to be noted that the figures appear reversed in the perspective.
That is, in the geometrical triangle the base at _ab_ is uppermost,
whereas in the perspective _ab_ is lowermost, yet both are nearest to
the ground line.

LVI

HOW TO PUT A GIVEN SQUARE INTO ANGULAR PERSPECTIVE

Let _ABCD_ (Fig. 115) be the given square on the geometrical plane,
where we can place it as near or as far from the base and at any angle
that we wish. We then proceed to find its perspective on the picture by
finding the perspective of the four points _ABCD_ as already shown. Note
that the two sides of the perspective square _dc_ and _ab_ being
produced, meet at point _V_ on the horizon, which is their vanishing
point, but to find the point on the horizon where sides _bc_ and _ad_
meet, we should have to go a long way to the left of the figure, which
by this method is not necessary.

[Illustration: Fig. 115.]

LVII

OF MEASURING POINTS

We now have to find certain points by which to measure those vanishing
or retreating lines which are no longer at right angles to the picture
plane, as in parallel perspective, and have to be measured in a
different way, and here geometry comes to our assistance.

[Illustration: Fig. 116.]

Note that the perspective square _P_ equals the geometrical square _K_,
so that side _AB_ of the one equals side _ab_ of the other. With centre
_A_ and radius _AB_ describe arc _Bm'_ till it cuts the base line at
_m'_. Now _AB_ = _Am'_, and if we join _bm'_ then triangle _BAm'_ is an
isosceles triangle. So likewise if we join _m'b_ in the perspective
figure will m'Ab be the same isosceles triangle in perspective. Continue
line _m'b_ till it cuts the horizon in _m_, which point will be the
measuring point for the vanishing line _AbV_. For if in an isosceles
triangle we draw lines across it, parallel to its base from one side to
the other, we divide both sides in exactly the same quantities and
proportions, so that if we measure on the base line of the picture the
spaces we require, such as 1, 2, 3, on the length _Am'_, and