Let _P_, Fig. 45., be the point at which they meet. From _P_ let fall the vertical _PV_ on the sight-line, cutting the sight-line in _V_. Then the student will find experimentally that _V_ is the vanishing-point of the line _AC_.[26] Complete the rectangle of the base _AC′_, by drawing _A′C′_ to _V_, and _CC′_ to the vanishing-point of _AA′_. Join _Y′C′_. Now if _YC_ and _Y′C′_ be produced downwards, the student will find that they meet. Let them be produced, and meet in _P′_. Produce _PV_, and it will be found to pass through the point _P′_. Therefore if _AY_ (or _CY_), Fig. 45., be any inclined line drawn in perspective by Problem XV., and _AC_ the relative horizontal (_AC_ in Figs. 39, 40.), also drawn in perspective. Through _V_, the vanishing-point of _AV_, draw the vertical _PP′_ upwards and downwards. Produce _AY_ (or _CY_), cutting _PP′_ in _P_ (or _P′_). Then _P_ is the vanishing-point of _AY_ (or _P′_ of _CY_). [Illustration: Fig. 45.] The student will observe that, in order to find the point _P_ by this method, it is necessary first to draw a portion of the given inclined line by Problem XV. Practically, it is always necessary to do so, and, therefore, I give the problem in this form. Theoretically, as will be shown in the analysis of the problem, the point _P_ should be found by drawing a line from the station-point parallel to the given inclined line: but there is no practical means of drawing such a line; so that in whatever terms the problem may be given, a portion of the inclined line (_AY_ or _CY_) must always be drawn in perspective before P can be found. [26] The demonstration is in Appendix II. Article III. PROBLEM XVII. TO FIND THE DIVIDING-POINTS OF A GIVEN INCLINED LINE. [Illustration: Fig. 46.] Let _P_, Fig. 46., be the vanishing-point of the inclined line, and _V_ the vanishing-point of the relative horizontal. Find the dividing-points of the relative horizontal, _D_ and _D′_. Through _P_ draw the horizontal line _XY_. With center _P_ and distance _DP_ describe the two arcs _DX_ and _D′Y_, cutting the line _XY_ in _X_ and _Y_. Then _X_ and _Y_ are the dividing-points of the inclined line.[27] _Obs._ The dividing-points found by the above rule, used with the ordinary measuring-line, will lay off distances on the retiring inclined line, as the ordinary dividing-points lay them off on the retiring horizontal line. Another dividing-point, peculiar in its application, is sometimes useful, and is to be found as follows:— [Illustration: Fig. 47.] Let _AB_, Fig. 47., be the given inclined line drawn in perspective, and _Ac_ the relative horizontal. Find the vanishing-points, _V_ and _E_, of _Ac_ and _AB_; _D_, the dividing-point of _Ac_; and the sight-magnitude of _Ac_ on the measuring-line, or _AC_. From _D_ erect the perpendicular _DF_. Join _CB_, and produce it to cut _DE_ in _F_. Join _EF_. Then, by similar triangles, _DF_ is equal to _EV_, and _EF_ is parallel to _DV_.