straction of Problem 17; and it is in the given vanishing line, by the demontration of the present problem.

Cor. 2. And, therefore, if the vanishing lines AB and DE meet in G, the points B, D, and G, will be the vanishing points of the three legs of the solid angle of a cube which are perpendicular to each other; and, drawing DB, BG, GD, and DB, will produce the vanishing lines of the planes that contain that solid angle.

Cor. 3. The distance of the vanishing line DG is equal to the line FP, the point P being the intersection of the line FC, with a circle described on the diameter DG.

Prob. 19. Having the centre and distance of the picture, and the vanishing point of the common intersection of two planes that are inclined to each other in a given angle, and the vanishing line of one of them, to find the vanishing line of the other.

Let C, in fig. 32, be the centre of the picture, BG the vanishing line given of one of the planes, and B the vanishing point of their common intersection, and H the angle of their inclination.

Find the vanishing line GD of planes perpendicular to the original lines which have B for their vanishing point (by Prob. 15), let that vanishing line cut the vanishing line given in G. In GD find the anishing point E, making a given angle H, with lines that have G for their vanishing point (by Prob. 11); that is, in BF produced, take FP, equal to the distance of the vanishing line GD (found by Prob. 15; draw PG and PE, making the angle GPE equal to H, and draw BE, which is the vanishing line required.

Suppose the triangle GPE to be turned up on the line GE, so that the point P may be in the point of sight perpendicular over the centre of the picture C. In this case the planes BPG, GPE, EPB, will be the parallels of three original planes, whose vanishing lines are BG, GE, EB, that whose vanishing line is GE being perpendicular to the other two, by construction, because it is perpendicular to their common section, whose vanishing point is B; there fore the original planes, whose vanishing lines are BG and BE, are inclined to each other in the angle GPE; i. e. in the angle H. For the two planes are always measured by a plane perpendicular to their common intersection. Therefore, BG being the given vanishing line. BE is the vanishing line required. N.B. The centre of the vanishing line BE is found by drawing a line perpendicular to it from C (by Theo. 1), and then its distance is obtained in the same way as the distance PF, in Prob. 19.

Prob. 20. Having the centre and distance of the picture, and the vanishing line of one face of any given solid figure, and the projection of one line in that face, to find the projection of the whole figure.

Having found the projection of the face, whose vanishing line is given, by means of the projection of the line given, find the vanishing lines of the adjacent faces (by Prob. 20), and describe their projections by help of the lines given in the projection of the first face, and so on till the whole projection is completed.

Er. 1. Given the centre and distance of the picture, the inclination of the picture to the horizon, to represent a flight of steps.

Let C, in fig. 34, be the centre of the picture, and let the line L be its distance.

Also, let the line AB represent the base-line of the riser of the first step. Find IK, the vanishing line of the plane of the horizon, its

centre S, and its distance; produce AB to its vanishing point I; then find the vanishing point K; also the vanishing point M of lines whose originals are perpendicular to planes which have IK for their vanishing line; then find the vanishing point N (by Prob. 12,) of the inclination of the inclined plane. Draw AN and BN, then ABN is the inclined plane infinitely extended upwards. By joining IN, IN will be the vanishing line of that inclined plane. Therefore, suppose AD to be the hypothenuse of the triangle which forms the end of the first step. Join AL cutting BN at C. Join IC, and produce IC to D, and CD will form the internal line of the first step. Join AC, and produce it to meet IN in L. Draw DL cutting BN in G; join IG, and produce it to meet AN in H; then GH is the next internal angle. Hence the manner of determining the internal lines of the angles is apparent.

To form any step, suppose the first; by means of the vanishing point K draw DE and CF; and, by the vanishing point M, draw AE and BF; and, by the vanishing point I, draw EF, which completes the first step.

Ex. 2. To put a building with a pediment in perspective, given the centre, distance, and inclination of the picture, and the position of the sides of the building, and the inclination of the pediment to the horizon.

1. By means of the centre and distance of the picture, find the vanishing line of the plane on which the building stands, and the appending lines of operation.

2. Find the vanishing points of the two sides of the building ac cording to the given inclination of its sides.

3. Find the vanishing line of the vertical planes, or sides, which terminate with the pediments and its appending lines.

4. Find the two vanishing points in this last vanishing line of the sloping sides of the pediment, and the picture will be prepared for delineation.

Thus CO is equal to the distance of the picture, and draw DS through C perpendicular to CO; make the angle COS equal to the complement of the angle Q, which is the inclination of the picture to the horizon through S draw AB perpendicular to SD, or parallel to CO. Produce DS to O', and make SO equal to SO. Make the angle AO'B to have the same position to the parallel of the eye through O, as the side of the building has to the intersection corresponding to the vanishing line AB; join AC and produce it to O". Draw BD perpendicular to AC cutting it in S. Find the distance S'O" of the vanishing line BD (that is, make BO" equal to BO'), and join BO", making each of the angles BO"E and BO"F equal to the inclination prr of the inclined line of the pediment with the horizontal line, and let OE and OF meet BD in E and F; then E and F are the vanishing points of the lines which form the representation, of the vertical angle of the roof.

When the picture stands perpendicular to the horizon, the same method will apply, observing that the centre C of the picture, and