to AB as the distance of the picture is to the distance between the point of sight and the original plane.

Theo. 5. The projection of a line is.parallel to its director.

In figure 3, as already explained at the end of the definitions, and in Theo. 3, the lines OF, OG, OD, fg, are all in the same plane; but the directing plane ODE is parallel to the plane of the picture ABIC (Def. 8); consequently, the director OD is parallel to the projection fg (by Prop. XVI. Book II. Euc. Elem.)

Cor. 1. The projections of lines that have the same director, are parallel

to each other.

Cor. 2. When the original line is parallel to the picture its director is parallel to it, and, consequently, is in the parallel of any plane passing through that original line; and, therefore, the vanishing line of that plane, and the projection of the line, are parallel to each other.

Theo. 6. The vanishing line, intersection, and directing line, of any original plane, are parallel to each other.

In the same fig. understood, as has been already explained, the planes OVC and DFH being parallel (Def. 16), as also ODE and CAB (Def. 8), the vanishing line CV, the intersection IB, and the directing line ED, are parallel to each other (by Prop. XVI. Book II. Euc. Elem.) Q. E. D.

Cor. The distance ƒ V between the projection of any point, and the vanishing point V, is to the distance BV, between the intersection B and the said vanishing point V, as the distance OV of the vanishing point V is to the dis tance DF, between the director and the original point F. For OVBD is a parallelogram, consequently, BV is equal to DO, and the triangles fOV and OFD are similar, because their sides Vf and OD are parallel, wherefore ƒV VO :: DO (≈ BV) : DF.

Theo. 7. The vanishing points of all lines in any original plane are in the vanishing line of that plane.

Demon. For, since all the original lines are in the same plane, their parallels, which all pass through the point of sight, will all be in the parallel plane (by Prop. XV. Book II. Euc. Elem.), consequently, all the vanishing points are in the vanishing line.

Cor. 1. Original planes, which are parallel to each other, have the same vanishing line.

Cor. 2. The vanishing point of the common intersection of two original planes, is the intersection of their vanishing lines.

Cor. 3. The vanishing line of a plane perpendicular to the picture, passes through the centre of the picture.

Theo. 8. The intersections of all lines in the same original plane, are in the intersection of that plane.

This needs no demonstration, being so manifest.

Cor. 1. The intersection of the common intersection of two original pianes, is the intersection, or meeting of their intersections.

Cor. 2. Planes, whose common intersection is parallel to the picture, have parallel intersections, and also parallel vanishing lines.

Theo. 9. Any line, in the representation of a figure, parallel to the picture, is to the original line, as the principal distance is to the distance between the spectator's eye and the plane of the original object.

(Fig. 5, No. 2.) Let OG be perpendicular to the original plane, and to the picture cutting them in G and g. The original plane being parallel to the picture; therefore, all the visual rays OČ, OD, OF, &c. are cut in the same proportions by the points c, d, e, &c. as OG by the point g; and de being parallel to DC, the triangle dOc is similar to the triangle DOC; consequently, dc is to DC, as Od is to OD, that is, as Og is to OG, which last is the principal distance.

Theo. 10. The distance between the vanishing point of a line, and the representation of any point in it, is to the distance between the vanishing point and intersection, as the distance between the directing point and the intersection, is to the distance between the directing point and the original point.

(Fig. 3.) Let every thing remain, as explained in the definitions, and f being the representation of F, the line VB being parallel to OD, the triangles OVf, FDO are similar; consequently, Vf: OD : : OV : DF ; but OD=VB, and OV=DB; wherefore, Vƒ : VB : : DB : DF.

Cor. If a plane be supposed to pass through K parallel to the picture, the proportion will be the same as in Theo. 9.

Prob. 1. The centre and distance of the picture being given, it is required to find the projection of a point, whose seat in the picture, as also its distance, are both given.

Let S, in fig. 6, be the centre of the picture, and the given seat of the original point, draw SO at pleasure, equal to the distance of the picture, and parallel to it draw bA, equal to the distance of the original point from its seat; draw Sb and AO cutting in a, which will be the projection sought.

If the angle OSa, and, consequently, Aba, had been a right angle, then, turning the triangles S0a and bAa round the line Sab, as an axis, till SO and bA become perpendicular to the picture, O would be the point of sight, and A the original point, and AO would be the visual ray, cutting the picture in a, which, consequently, would be the projection of the point A, by Theo. 2, but the point a is the same, whether the angle OSb be a right angle or not, because the triangle OSa and Aba are similar, and, therefore, Su : ab :: SOA, which proportion is not affected by altering the angle OSb. Therefore, in all cases the point a, thus found, is the projection of the point sought, whose seat on the picture is b, and its distance from its seat is bA.

Cor. 1. Having drawn Sb, the point a may be found by a scale and compasses, dividing the line Sb in a, so that Su may be to ab as the distance of the picture SO is to the distance bA of the original point from its seat b. Cor. 2. By this proposition the projection of any line may be found, by the projection of any two points in it, and then drawing a line through those two points.

Prob. 2. To find the projection of a line, AC its vanishing point and distance, its seat DE being given, intersection, and the angle EDC it makes with its seat, and the centre S and distance of the picture.

Draw SV parallel to DE, and SO perpendicular to it, and equal ta the distance of the picture. Then draw OV parallel to DC, cutting SV in V, and draw DV,.then will V be the vanishing point, and OV its distance, and DV the indefinite representation.

Imagine the planes OSV and CED to be turned round the lines SV and

DE, as axes, till they become perpendicular to the picture, then will O be the point of sight, and DC the original line; and OV being parallel to it, V is the vanishing point (by Def. 17), and DV is the projection (by Theo. 3). Q. E. D.

Cor. 1. DAC being supposed the original line laid on the picture, by turning the plane CDE round the line ED, the projection of any part of it, AC may be found by drawing the lines AO and CO as visual rays, cutting DV in a and c. For the points a and c depend only on the parallelism of the lines OV and DC and their proportion; aV being to aD as VO is to DA, and CV: CD VO: DC, on account of the similar triangles aVO, and aDA, and cVO, and cDC.

Prob. 3. Having the projection of a line, and the vanishing point, to find the projection of the point that divides the original line in any given proportion.

Let AB be the given projection of the line to be divided (fig. 8), and V its vanishing point; draw VO at pleasure, and ba parallel to it, and through any point O of the line VO, draw CA and CB cutting ba in a and b. Divide ab in c in the proportion given, and draw Oc cutting AB in C. Then will C be the projection sought, the original of BC being to the original of CA as be is to ca.

OV being parallel to ba, ba may be considered as the original line, and OV as its parallel, and, consequently, O as the point of sight, and aO, ¿O, CO, as visual rays projecting the points A, B, C.

Prob. 4. Having the projection of a line, and its vanishing point from a given point in that projection, to cut off a segment that shall be the projection of a given part of the original of the projection given.

Let AB, in fig. 9, be the projection given, V its vanishing point, and C the point from whence the segment is to be cut off. Draw VO at pleasure, and abc parallel to it, and from any point O in VO draw OA, OB, OC, cutting ab in a, b, c. Make cd to ab, as the given part is to the original of AB, and draw Od cutting AB in D, then will CD be the segment required.

If OV be supposed the vanishing line of any plane passing through the original of ABV, abcd being parallel to it, may be conceived as the projection of a line parallel to the picture (Cor. 2; Theo. 5), and, consequently, its parts ab and cd will be in the same proportion to each other as their originals (by Theo. 4). But, because of the vanishing point O, the originals of Oa, Ob, Oc, Od, are parallel (by Cor. 1; Theo. 3.) Wherefore the original of CD is to the original of AB, as cd is to ab (by Proposition II. Book VI. Euclid's Elements.) Q. E. D.

N.B. This last proposition might have been demonstrated as the foregoing, and the foregoing may be considered as a particular case of this, viz. when he point C of this proposition coincides with one of the points A or B.

Prob. 5. Having the vanishing line of a plane, its centre and distance, and the representation of two lines in that plane, from a given point in one of them, to cut off a line that shall represent a line in a given proportion to the line represented by the other.

(Fig. 10.) SD is the given intersecting line, V its centre and AB, CI the representations given, the vanishing points being D and E; it is required to find the point I, so that the line represented by CI may be to the line represented by AB as c is to d.