: Draw VO perpendicular to FD, and equal to the distance given; and draw OD and OE, and make OQ OR c: d. Draw OS parallel to RQ, and AC cutting SD in F. Join FB and DC cutting each other in p. Then draw pS, which will cut CE in I, the point sought. Because of the vanishing points D and F, the figure ABC represents a parallelogram. Wherefore, Cp represents a line equal to the line represented by AB. By the nature of placing the point O, the plane SOD may be considered as the plane generating the vanishing line SD, O being the eye of the spectator. Wherefore, OS being parallel to RQ, D, E, S, are the vanishing points of the sides of a triangle parallel to ROQ. Wherefore, pCI represents that triangle, and, consequently, the line represented by CI is to the line represented by pC, or by AB, as OQ is to OR; that is, as b is to d. When the two given lines terminate at one end in the same point, this problem will be very much simplified, and the process will be shortly described; thus, see fig. 11, CP and CI being the two given lines, and CP being given, it is required to find CI, so that its original of CI may be to the original of CP in a given ratio. Produce CP and CI to their vanishing points D and E. Join DO and EO; make OR to OQ, in the ratio or proportion required; join RQ, and draw OV parallel to RQ, cutting the vanishing line in V; draw PV cutting CE in 1; then the line represented by CI will be to the line represented by CP in the ratio required, that is, as OQ to OR. (Fig. 12.) If one of the lines lay between the common point C and the vanishing point, and the extremity of the other on the contrary side of C, that is, when one of them, when produced, would meet the vanishing line, and the other the intersecting line. Produce the radial DO of PC, which is that next to the intersecting line to R; make OR to OQ in the given proportion, and join RQ; draw OV parallel to RQ, cutting the vanishing line in V; draw PV cutting CE in I; then the line represented by C1 will be to the line represented by CP in the ratio required. N.B. If the original line represented by CI, be equal to the original line represented by CP, then OR will become equal to OQ; and OV being paral· lel to RO, the angle EOD will be bisected; therefore, the point V will also be found by bisecting the angle EOD. The demonstration of these cases is evident from the preceding general demonstration. Prob. 6. Having the vanishing line of a plane, the representation of two lines in that plane; to find a point in one of them, so that it may represent a line divided in the same proportion as the line represented by the other. (Fig. 13.) wk is the vanishing line given, AC and DE being the two representations, it is required to find the point F, so that the parts represented by DF and EF may be in the same proportion as the parts represented by AB, BC. Draw any line sv, parallel to the vanishing line, and, by means of any vanishing point w, transfer the points A, B, C, to s, t, v, and su, will be divided in the same proportion as the original of AC (by Cor. 3, Prop. 1), and because of ws, wt, wv, representing parallels. Whence the point F may be found by Prop. 13. Prob. 7. Given the vanishing line of a plane, and the centre and distance of that vanishing line, as also the projection of a line in that plane; to find the projection of another line making a given angle with the former. (Fig. 14.) Let O be the point of sight, GH the vanishing line, and ab the given projection; it is required to draw ac, so that the original of the angle bac may be equal to the given angle. Produce ab to its vanishing point G; join GO, and draw OI so as to make an angle with OG equal to the given angle, cutting the vanishing line in I; then draw Ica, and ac is the line required. The figure being understood, as in the foregoing problems, let AB be the original of ab, and, consequently, parallel to OG (by Def. 15, and Theo. 3). For the same reason is AC parallel to OI and the original of ac, I being its vanishing point (by Theo. 3). But AB and AC being parallel to OG and OI, the angle BAC is equal to the angle GOI, which is equal to the given angle by the construction. Wherefore, bac representing the angle BAC, represents the given angle. Q. E. D. Scholium Thus it has been seen, that as a geometrical figure can be described by drawing a right line which shall cut a given right line in any proposed point at a given angle, and, by cutting off a given length from a right line, from a given point in that line; so, in like manner, we may describe the re presentation of any right-lined figure, by drawing a right line from a given point in another right line, so that the two right lines shall represent a given angle, and, by finding a limited distance from the point of intersection upon either of these lines, and by continuing the operation until we construct a figure which shall have any given original that we imagine. This will enable us to make the perspective representation of any object, of which we have a perfect idea, without having recourse to geometrical draw ings of that object. Prob. 8. Given the vanishing line of a plane, the centre and distance of that vanishing line, and the projection of one side of a trian gle of a given species in that plane, to find the projection of the whole triangle. The projections of the sides, in fig. 15, are to be found by the foregoing problem, the angles of the triangle being given. Thus the projection ab of the side AB of the triangle ABC is found, by making the angle IOG equal to the angle CĂB, and the vanishing point H of the side ab is found by making the angle IOH equal to the angle ACB. Prob. 9. Given the vanishing line of a plane, the centre and distance of that vanishing line, and the projection of one side, or principal line of any figure in that plane, to find the projection of the entire figure. Resolve the figure into triangles, and find the projections of those triangles, one after another (by Prob. 8), beginning with those that have the line given for two of their sides. The same thing may be done several ways, by the application of the foregoing problems, as in every case is most convenient. This will be better understood by a few examples. Ex. 1. (Fig. 16.) In this example, IK is the vanishing line, S its centre, and SO its distance, and AB parallel to IK is the given projection of one side of a regular hexagon. Having drawn OG parallel to IK (by Def. 15), the original of AB M m being parallel to the picture (by Cor. 2, Theo. 5), the vanishing points H, I, K, of the sides and diagonals BC, FE, AD; af, be, CD: AC are found by Prob. 2, making the angles HOG 60 degrees, IOG 120 degrees, KÖG 30 degrees, then drawing AK and BH will find C; drawing AH and CI will find D; drawing DE parallel to IK, and joining AS, will find E (for S is the vanishing point of AE, the original of the angle. EAB being a right angle, as is GOS); lastly, drawing EH and AI, the point P is found, which completes the figure required. Prob. 10. Given the vanishing line of a plane, its centre, and distance, and the representation of one radius of a circle to find the representation of the circumference of that circle. (Fig. 17.) This is only an application of Prob. 5, which directs as to find the length of the representation of a line, so that the original line may be equal to the original of another given representation at the same point; and repeating the process for as many radii as we please; then, drawing a curve, which will represent the circumference through the extremities of these radii. Thus CA is the given representation of a radius; it is required, in the direction CP, to represent another radius. Produce CA and PC to their vanishing points E and D, and draw OE and OD; bisect the angle EOD by the right line OF, cutting ED in F; draw FA, and produce it to meet CP in B; then CB is the representation required; this direction is only a repetition of the simplest case of Prob. 5. Prob. 11. Having the vanishing line of a plane, its centre, and distance, to draw the representation of a circle through three given points. Let the three given points be A, B, C, in fig. 18, and let the vanishing line be Df, and the eye of the spectator the point O above the vanishing line. Draw AC and BC meeting the vanishing line in D and E, which will, therefore, be their vanishing points, and join OD and OE. Draw any line do, and make the angle dOe equal to DOE, and produce Oe to meet/D in f. Join dA and B cutting each other in p, which will be the representation of a fourth point in the circumference. NB. q is another point found in the same manner. This construction is aken from the equality of angles in the circle insisting upon the same base, whence the vanishing points are found by Prop. 13. Prob. 12. The same things being given, from a given point in the picture, to draw a line that shall touch the representation of a circle. (Fig. 19.) Let DE be the vanishing line, O the eye of the spectator, C the representation of the centre of the circle, and CA the representation of the radius; it is required to draw a line from a given point P to touch the representation of the circumference. Draw CP cutting the vanishing line in E, and (by Example 3), find the radius CB. Any where apart draw bp, and make be: cp :: BCX PE: CPX BE; that is, make the line bcp like the original of BCP. With the centre e, and radius cb, describe a circle, a |