Fig. 50.

As the inclined plane is given, one of its steepest lines must be given, or may be ascertained.

Let A B, [Fig. 50.], be a portion of a steepest line in the [p62] ]given plane, and V the vanishing-point of its relative horizontal.

Through V draw the vertical G F upwards and downwards.

From A set off any portion of the relative horizontal A C, and on A C describe a semicircle in a vertical plane, A D C, cutting A B in E.

Join E C, and produce it to cut G F in F.

Then F is the vanishing-point required.

For, because A E C is an angle in a semicircle, it is a right angle; and therefore the line E F is at right angles to the line A B; and similarly all lines drawn to F, and therefore parallel to E F, are at right angles with any line which cuts them, drawn to the vanishing-point of A B.

And because the semicircle A D C is in a vertical plane, and its diameter A C is at right angles to the horizontal lines traversing the surface of the inclined plane, the line E C, being in this semicircle, is also at right angles to such traversing lines. And therefore the line E C, being at right angles to the steepest lines in the plane, and to the horizontal lines in it, is perpendicular to its surface.