TO DRAW A SQUARE, GIVEN IN POSITION AND MAGNITUDE, IN A HORIZONTAL
PLANE.
[Illustration: Fig. 20.]
Let _ABCD_, Fig. 20., be the square.
As it is given in position and magnitude, the position and magnitude
of all its sides are given.
Fix the position of the point _A_ in _a_.
Find _V_, the vanishing-point of _AB_; and _M_, the dividing-point of
_AB_, nearest _S_.
Find _V′_, the vanishing-point of _AC_; and _N_, the dividing-point of
_AC_, nearest _S_.
Draw the measuring-line through _a_, and make _ab′_, _ac′_, each equal
to the sight-magnitude of _AB_.
(For since _ABCD_ is a square, _AC_ is equal to _AB_.)
Draw _aV′_ and _c′N_, cutting each other in _c_.
Draw _aV_, and _b′M_, cutting each other in _b_.
Then _ac_, _ab_, are the two nearest sides of the square.
Now, clearing the figure of superfluous lines, we have _ab_, _ac_,
drawn in position, as in Fig. 21.
[Illustration: Fig. 21.]
And because _ABCD_ is a square, _CD_ (Fig. 20.) is parallel to _AB_.
And all parallel lines have the same vanishing-point. (Note to
Problem III.)
Therefore, _V_ is the vanishing-point of _CD_.
Similarly, _V′_ is the vanishing-point of _BD_.
Therefore, from _b_ and _c_ (Fig. 22.) draw _bV′_, _cV_, cutting each
other in _d_.
Then _abcd_ is the square required.