We have hitherto proceeded on the supposition that the given line was small enough, and near enough, to be actually drawn on our paper of its real size; as in the example given in Appendix I. We may, however, now deduce a construction available under all circumstances, whatever may be the distance and length of the line given. [Illustration: Fig. 11.] From Fig. 8. remove, for the sake of clearness, the lines _C′D′_, _bV_, and _TV_; and, taking the figure as here in Fig. 11., draw from _a_, the line _aR_ parallel to _AB_, cutting _BT_ in _R_. Then _aR_ is to _AB_ as _aT_ is to _AT_. ---- ---- as _cT_ is to _CT_. ---- ---- as _TS_ is to _TD_. That is to say, _aR_ is the sight-magnitude of _AB_.[16] [Illustration: Fig. 12.] Therefore, when the position of the point _A_ is fixed in _a_, as in Fig. 12., and _aV_ is drawn to the vanishing-point; if we draw a line _aR_ from _a_, parallel to _AB_, and make _aR_ equal to the sight-magnitude of _AB_, and then join _RT_, the line _RT_ will cut _aV_ in _b_. So that, in order to determine the length of _ab_, we need not draw the long and distant line _AB_, but only _aR_ parallel to it, and of its sight-magnitude; which is a great gain, for the line _AB_ may be two miles long, and the line _aR_ perhaps only two inches. COROLLARY III. In Fig. 12., altering its proportions a little for the sake of clearness, and putting it as here in Fig. 13., draw a horizontal line _aR′_ and make _aR′_ equal to _aR_. Through the points _R_ and _b_ draw _R′M_, cutting the sight-line in _M_. Join _TV_. Now the reader will find experimentally that _VM_ is equal to _VT_.[17] [Illustration: Fig. 13.] Hence it follows that, if from the vanishing-point _V_ we lay off on the sight-line a distance, _VM_, equal to _VT_; then draw through _a_ a horizontal line _aR′_, make _aR′_ equal to the sight-magnitude of _AB_, and join _R′M_; the line _R′M_ will cut _aV_ in _b_. And this is in practice generally the most convenient way of obtaining the length of _ab_. COROLLARY IV. 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.