[Geometric diagram]
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.[Footnote 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:—

[p56] [Geometric diagram]
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.

Hence it follows that if from D, the dividing-point of Ac, we raise a perpendicular and make DF equal to EV, a line CF, drawn from any point C on the measuring-line to F, will mark the distance AB on the inclined line, AB being the portion of the given inclined line which forms the diagonal of the vertical rectangle of which AC is the base.