given dependent only upon the postulates. These postulates are the three following: 1. That a straight line may be drawn from any one point to any other point. 2. That a terminated straight line may be produced to any length in a straight line. 3. That a circle may be described about any point as a centre, and at any distance from that centre. PROBLEM I. Three straight lines, A, B, C, each two of which are greater than the third, being given, to construct a triangle whose sides shall be respectively equal to them. MAKE DE equal to A, and with centres D and E and radii equal to B and C respectively, describe circles intersecting in F, G. Join DF, FE, and likewise DG, GE. Then either of the triangles DFE or DGE will be that required *. E G ABC Scholium. When any two of the three lines are equal, the triangle is isosceles, and when all three are equal, then it is equilateral. These particular cases are, therefore, comprised in the general construction. PROBLEM II. At a given point D in a given line DE to make an angle equal to the given angle BAC. WITH any radius describe arcs from the centres A and D; the first BC meeting AB, AC, in B and C, and the second EH meeting DE in E. With centre E, and radius equal to BC, describe an arc to meet HE in F. The line DF being drawn, will make the angle EDF equal to BAC. XX This is only an application of the last problem, and the equality of the angles will be evident from Geom. th. 8 ↑. PROBLEM III. Through a given point D to draw a line parallel to a given line AB. First method. Draw any line CD through C to meet AB in C, and produce it backwards, till DF is equal to DC. With centre F describe the arcs CKE and DHG, the former meeting AB in E. Join FE meeting the arc DHG in G. Then the line DG being drawn and produced if necessary, will be the parallel required. B * In the drawing, only the small portions of the circles in the estimated regions of F and G need be actually traced. The arc BC need not be drawn at all, it being sufficient that the points B, C, E, be marked, and the circles meeting at F need only be drawn in the estimated region of their intersection. For the line DG divides the sides of the triangle CFE proportionally, (bisects them,) and hence it is parallel to the base CE. Geom. th. 82. Second method. Draw any line DC, meeting AB in C, and with centres C and D, and radius CD, describe arcs DB, CH, the former meeting AB in B. Then with centre C and radius BD describe an arc to meet CH at E. The line CE being drawn will be parallel to AB. E B For by the construction of the second problem, the alternate angles EDC, DCB, are equal; and hence, Geom. th. 13, the line EF is parallel to AB*. Third method. Draw any line CD through D, and with centre D describe the circle CHF cutting AB in H and CD in F, and with centres H and F, and the same radii as before, describe arcs cutting in E. Then DE will be parallel to AB. E H B For, since CDH is isosceles, each of the angles at C and H is half the external angle FDH. But since FD, DE, are equal to HD, DE, and FE equal to HE, the angles FDE, HDE, are equal, and hence each of them is equal to half the angle FDH. Whence FDE is equal to DCH, and therefore the lines CH, DE, are parallel †. Scholium. The frequent occurrence of this, a paper problem, has given rise to the construction of instruments for facilitating the operation. They are, however, reducible, as to general principle of construction, to two,-the parallel ruler and the parallel scales. A brief description of them is annexed. 1. The parallel ruler. It has been proved, Geom. th. 23, that when a quadrilateral has each pair of opposite sides equal, they will also be parallel. Let, then, AB = BD: hence, whatever be the angle BAC, the line CD will be parallel to AB. If now we suppose the lines AB, CD, to be respectively traced on two flat rectangular rulers HK, LM, along the middles of them, and of the same length, and then cross pieces AC, BD, fastened by means of axes to them, R R' H' C L as in the figure, then it is clear that the edges of these rulers will be parallel in all positions which the rulers so united can possibly take. Whence, if the upper edge of the upper ruler were placed along a line A'B', and the lower held firmly whilst the upper is pushed forward to the given point P; then the line RS drawn through P by the edge of the ruler in this position, will be parallel to A'B'. In like manner, if the upper edge be still moved forward to another point P', a line R'S' drawn along the edge of the ruler in its new position, will also be parallel to A'B'. And so on for any number of lines. * None of the intersecting lines or arcs need be drawn, it being sufficient to mark the several points of intersection: and the same remark applies throughout this entire series of problems. The advantage of this method is, that it requires only a single opening of the compasses, and the entire use of it completed without intermediate operations. In the figure the ruler HK adjoining the ruler LM indicates the relative position of the two rulers when the instrument is closed to be laid aside. The descriptive part of the work refers to the upper position of HK. The triangle has no marks except one for the middle of the hypothenuse: and its dimensions are usually the longer and shorter sides about the right angle nine inches and three inches respectively. In using the marquois, the longer leg GH of the triangle is laid along the line to which another is to be drawn parallel, and whilst held in this position, the ruler BE is placed against the hypothenuse GK. The ruler being held in this position, the triangle is slid with the right hand, up or down, as the case may require, till the edge GH passes through the given point, as in the position G'H'. Then both ruler and triangle being held firmly with the left hand in this position, the line is traced along G'H', and this is the parallel required. If there be more parallels through more points required, again hold the ruler firmly, and again slide the triangle up or down till the edge GH passes through a second point, as in the position G"H", and draw the second parallel as before. When the point K" is arrived so far in advance of the end E of the ruler as to give an unsteadiness to the instrument, slide the ruler down before moving the triangle, and then proceed with the triangle as before *. PROBLEM IV. To bisect a given angle ABC, First method. Take any equal distances BA, BC, in the sides containing the angle, and with A, C, as centres, and any equal radii describe circles cutting in D. Then BD being drawn will bisect the angle ABC. For if AD, CD be drawn, the three sides of the triangle ABC are equal each to each to those of the triangle CBD. Hence, Geom. th. 8, the angles ABD, CBD, are equal. Second method. Produce one of the sides AB to D, and take BD equal to BC. Then a line BH through B, parallel to CD will bisect the angle ABC. For by construction BD is equal to BC, hence the angles BDC, BCD, are equal (th. 3). But by the parallels DC, BH, the angles ABH, ADC, are equal, and the angles. HBC, BCD, are equal (th. 12): whence the angles ABH, HBC, are equal †. B 1 C * The use of this instrument is easily acquired, though it requires a little more practice than the common parallel ruler to use it with complete facility. On the whole, especially in respect of its use in drawing perpendiculars, it is a more convenient instrument than the old parallel ruler and it is matter of surprise that it has not obtained more attention from architectural and mechanical draughtsmen than it has yet done. Military draughtsmen, to whom time, as well as accuracy, is an important object, seldom use any other; and this is a good proof of its value as a parallel ruler. : + This latter method is the preferable one in drawings, where a parallel ruler is admitted, as the line DC need not then be drawn at all. PROBLEM V. Through a given point C in a given line AB to draw a perpendicular. First method. Take equal distances CD, CE, on each side of the given point, and with any convenient equal radii describe arcs meeting in F. Join FC; it is the perpendicular required. For, conceive DF, EF, joined: then the equality of the sides of the triangles DCF, ECF, give the angles at C equal (th. 8), and hence CF perpendicular to AB (def. 25)*. Second method. With centre C and any radius describe a circle cutting the line AB in G. Set off the arcs GD, GE, equal to one another; and through C draw CF parallel to DE. For DE is perpendicular to AB, and CF parallel to ED t. Third method. Take any point D with which as a centre, and DC as radius, describe a circle, cutting AB in E. Draw ED to meet this circle in F, and join FC. This will be the perpendicular required. For the angle ECF being in a semicircle is a right angle (th. 52)‡. Scholium. These methods are often, in practice, superseded by the following one. In every case of instruments there is a rectangular parallelogram of wood or ivory, DGHE called a scale, having, amongst other lines, one CF drawn from C, the middle of DE, at right angles to DE, or parallel to the ends DG, HE. The edge of the ruler is placed to coincide with the given line AB, and its middle point with the point C: then the opposite end of CF being marked on the paper, the scale is removed and CF drawn. A D C EB It is also often effected by means of the triangle and ruler described in the Scholium to Prob. 3. This method is applicable where there is sufficient space obtainable in the drawing on both sides for setting off CD and CE. + This is adapted to the case where AB does not admit of prolongation sufficiently beyond C, as near the edge of the drawing. This method is adapted to the case where AB does not admit of sufficient prolongation to apply the first method, as when near the edge of the drawing. When C is near a corner, this is the only method applicable. It may, indeed, be generally advantageous to the student to practise each of these problems and modes of construction in all the corners and edges of the drawing, as he will then see at once the circumstances which determine the applicability of each method. Place the ruler DEHK to coincide with the given line AB, and holding it in this position, place the right angle of the triangle at C. Hold the triangle firm, (the ruler being either retained or removed,) and draw the line FC by its edge. If there be several perpendiculars to be drawn at equal distances, this method is convenient, if HD be a graduated scale adapted to those distances. However, in most cases, if the divisions be already made on AB, it will be more convenient to draw one perpendicular to AB, and draw parallels to it through the given point by the parallel ruler. PROBLEM VI. From a given point A without a given line BC to draw a perpendicular to BC. First method. With centre A, and any convenient radius AD, describe a circle cutting BC in two points D and E; and from D and E with any equal radii describe circles intersecting at F. Then FA being joined, cutting BC in G, will be the perpendicular required. For conceiving the lines DA, AE, DF, FE, to be drawn, it may be proved, as before, that the angle DAE of the isosceles triangle is bisected by AG; and hence AG is perpendicular to DE *. Second method. With any two centres H and D in BC describe circles passing through A and intersecting again in F. Join AF, cutting BC in G. It will be the perpendicular required. For, as before, HD bisects the angle AHF of the isosceles triangle AHF, and is therefore perpendicular to AF; that is, AF is perpendicular to BC †. Third method. Draw any line AE nearly perpendicular to BC by estimation, and produce it till EF EA. With centres A and radii AF, AE, describe circles, the former meeting BC in H, and the B latter to intersect the line drawn from A to H. With centre D and radius DA describe a semicircle cutting BC in G: then AG being drawn, will be the perpendicular required. E G B C H B H D F D E G C For since AE = EF by construction, and that AH = AF; and AE being * This method can only be used when A is not near the edge of the drawing. When it is near the top or bottom of the drawing, (BC supposed horizontal,) the two triangles DAE, DFE, may be drawn on the same side of AB, but not with the sides of the one equal to those of the other. This method is very convenient when the point A is near the margin, but not near a corner of the drawing. The points H and D should be taken as remote as circumstances will allow, to prevent the arcs intersecting under too acute an angle. |