When, therefore, the line _AB_ is horizontal (or marks the water surface), its direction is the direction of the inclined plane, and the inclination of the line _DC_ is the inclination of the inclined plane. In beds of rock geologists call the inclination of the line _DC_ their “dip.” To fix the position of an inclined plane, therefore, is to determine the direction of any two lines in the plane, _AB_ and _CD_, of which one shall be horizontal and the other at right angles to it. Then any lines drawn in the inclined plane, parallel to _AB_, will be horizontal; and lines drawn parallel to _CD_ will be as steep as _CD_, and are spoken of in the text as the “steepest lines” in the plane. But farther, whatever the direction of a plane may be, if it be extended indefinitely, it will be terminated, to the eye of the observer, by a boundary line, which, in a horizontal plane, is horizontal (coinciding nearly with the visible horizon);—in a vertical plane, is vertical;—and, in an inclined plane, is inclined. This line is properly, in each case, called the “sight-line” of such plane; but it is only properly called the “horizon” in the case of a horizontal plane: and I have preferred using always the term “sight-line,” not only because more comprehensive, but more accurate; for though the curvature of the earth’s surface is so slight that practically its visible limit always coincides with the sight-line of a horizontal plane, it does not mathematically coincide with it, and the two lines ought not to be considered as theoretically identical, though they are so in practice. It is evident that all vanishing-points of lines in any plane must be found on its sight-line, and, therefore, that the sight-line of any plane may be found by joining any two of such vanishing-points. Hence the construction of Problem XVIII. II. DEMONSTRATIONS WHICH COULD NOT CONVENIENTLY BE INCLUDED IN THE TEXT. I. THE SECOND COROLLARY, PROBLEM II. In Fig. 8. omit the lines _CD_, _C′D′_, and _DS_; and, as here in Fig. 75., from _a_ draw _ad_ parallel to _AB_, cutting _BT_ in _d_; and from _d_ draw _de_ parallel to _BC′_. [Illustration: Fig. 75.] Now as _ad_ is parallel to _AB_— _AC_ ∶ _ac_ ∷ _BC′_ ∶ _de_; but _AC_ is equal to _BC′_— ∴ _ac_ = _de_. Now because the triangles _acV_, _bc′V_, are similar— _ac_ ∶ _bc′_ ∷ _aV_ ∶ _bV_; and because the triangles _deT_, _bc′T_ are similar— _de_ ∶ _bc′_ ∷ _dT_ ∶ _bT_. But _ac_ is equal to _de_— ∴ _aV_ ∶ _bV_ ∷ _dT_ ∶ _bT_; ∴ the two triangles _abd_, _bTV_, are similar, and their angles are alternate; ∴ _TV_ is parallel to _ad_. But _ad_ is parallel to _AB_— ∴ _TV_ is parallel to _AB_. II. THE THIRD COROLLARY, PROBLEM III. In Fig. 13., since _aR_ is by construction parallel to _AB_ in Fig. 12., and _TV_ is by construction in Problem III. also parallel to _AB_—