Removing from the preceding figure the unnecessary lines, and retaining
only _R′M_ and _aV_, as in Fig. 14., produce the line _aR′_ to the other
side of _a_, and make _aX_ equal to _aR′_.
Join _Xb_, and produce _Xb_ to cut the line of sight in _N_.
[Illustration: Fig. 14.]
Then as _XR′_ is parallel to _MN_, and _aR′_ is equal to _aX_, _VN_
must, by similar triangles, be equal to _VM_ (equal to _VT_ in
Fig. 13.).
Therefore, on whichever side of _V_ we measure the distance _VT_, so as
to obtain either the point _M_, or the point _N_, if we measure the
sight-magnitude _aR′_ or _aX_ on the opposite side of the line _aV_, the
line joining _R′M_ or _XN_ will equally cut _aV_ in _b_.
The points _M_ and _N_ are called the “DIVIDING-POINTS” of the original
line _AB_ (Fig. 12.), and we resume the results of these corollaries in
the following three problems.
[14] The student will observe, in practice, that, his paper lying flat
on the table, he has only to draw the line _TV_ on its horizontal
surface, parallel to the given horizontal line _AB_. In theory,
the paper should be vertical, but the station-line _ST_
horizontal (see its definition above, page 5); in which case
_TV_, being drawn parallel to _AB_, will be horizontal also, and
still cut the sight-line in _V_.
The construction will be seen to be founded on the second
Corollary of the preceding problem.
It is evident that if any other line, as _MN_ in Fig. 9.,
parallel to _AB_, occurs in the picture, the line _TV_, drawn
from _T_, parallel to _MN_, to find the vanishing-point of _MN_,
will coincide with the line drawn from _T_, parallel to _AB_, to
find the vanishing-point of _AB_.
Therefore _AB_ and _MN_ will have the same vanishing-point.
Therefore all parallel horizontal lines have the same
vanishing-point.
It will be shown hereafter that all parallel _inclined_ lines
also have the same vanishing-point; the student may here accept
the general conclusion—“_All parallel lines have the same
vanishing-point._”
It is also evident that if _AB_ is parallel to the plane of the
picture, _TV_ must be drawn parallel to _GH_, and will therefore
never cut _GH_. The line _AB_ has in that case no
vanishing-point: it is to be drawn by the construction given in
Fig. 7.
It is also evident that if _AB_ is at right angles with the plane
of the picture, _TV_ will coincide with _TS_, and the
vanishing-point of _AB_ will be the sight-point.
[15] I spare the student the formality of the _reductio ad absurdum_,
which would be necessary to prove this.
[16] For definition of Sight-Magnitude, see Appendix I. It ought to
have been read before the student comes to this problem; but I
refer to it in case it has not.
[17] The demonstration is in Appendix II. Article II. p. 101.