2

USE OF INSTRUMENTS.

Parallel lines should be drawn by means of the set squares; (they are far better than parallel rulers). The edge of one must be adjusted in the required direction and held firmly on the paper, the other should be placed in contact with a second edge of the first and held in that position, and the first may then be made to slide along the second till it comes into the position of the required parallel line. A line perpendicular to another and passing through a given point should be drawn by adjusting an edge containing the right angle of one of the squares to the given line, placing the second square in contact with the hypotenuse of the first and sliding the first along the second until its third side passes through the given point, when the required perpendicular can be drawn.

If a line is to be drawn through two given points, the point of the pencil should first be placed on one of the points, the square can then be brought up to the pencil and worked against it as a centre till it coincides with the other, when the line can be drawn, and care must be taken that the line passes accurately through both points, as owing to the thickness of the edge of the square it is quite possible to make a slight but quite appreciable error. This is particularly the case if the pencil is cut to a chisel edge instead of to a circular point, and the author would express his decided conviction as to the superiority of the circular point. It is of course quite impossible to draw accurately unless a good sharp point to the pencil be constantly maintained.

Lines whether straight or circular should be bisected, trisected, &c. by trial, mechanical methods however good in theory being unnecessary and indeed objectionable in practice. A very little practice in handling a pair of dividers will enable this to be done with great ease and with all attainable accuracy, if the amount by which the first shot exceeds or falls short of the desired result is noted and the legs of the dividers closed up or extended to the necessary estimated fraction of this amount. If the required number of parts admits of division, the line should first be divided in the smaller number of parts necessary, i. e. if it is to be divided into six parts, it should be first bisected, and

then each half trisected; if into nine parts it should be first trisected, and so on. Care must be taken by a light handling of the instruments, not to damage the paper, until it is found that it can be marked with the points in the right places, and when a point is being marked on a line with the dividers, special care should be taken to press in the point on the line and not merely somewhere in its neighbourhood. In handling the instruments they should be constantly kept in as nearly vertical planes as possible. A point when found should be marked by a light pencil ring round it and not by a smudge made with a blunt pointed pencil, which entirely obscures the exact position of the point.

PROBLEM 1. (Figs. 1, 2.) To draw a line bisecting the angle between two given lines.

It is frequently necessary to do this when the lines are so nearly parallel or are otherwise so situated that their point of intersection does not fall within the limits of the sheet of paper, or drawing-board, and since the method of proceeding in this case includes the ordinary simple case, it is the one chosen as

Draw GH

an example. Let AB, CD (fig. 1) be the given lines. parallel to AB at any convenient distance (BE) from it, and draw GK parallel to CD at a distance DF equal to BE from it. This can be done by drawing BE perpendicular to AB from any point B on it, and DF perpendicular to CD from any point D on it and making_BE = DF, and then using two set squares in the

4

BISECTION OF AN ANGLE.

way referred to in the introduction. The distance BE should be so chosen as to bring the point G about as in the figure, i.e. BE should be somewhat greater than half the least distance between the given lines. If the angle EGF be now bisected, its bisector will obviously also by symmetry bisect the angle between AB and CD. Take any equal distances GH, GK on GE, GF respectively or, what comes to the same thing, with centre G and any radius describe an arc HK, and with centres H and K, and with any (the same) radius describe arcs intersecting in L. Then GL will be the required bisector. For the triangle GHL is obviously equal and similar in all respects to the triangle GKL.

This method is scarcely satisfactory when the lines are nearly parallel, on account of the smallness of the angle EGF and the difficulty of determining accurately the point of intersection G of two nearly coincident lines, and an alternative method evading this difficulty is shewn in fig. 2. As before, let AB, CD be the

two given lines. At any point B of the one line draw a line as BF, and at any point D of the other construct an angle GDH equal to the angle EBF. The exact size of this angle is immaterial but preferably it should not differ much from half a right angle. [An angle (GDH) can be constructed equal to a given angle (EBF) by describing arcs EF, GH, with the angular points B, D as centres and with any (equal) radius, and then

making the chord GH equal to the chord EF by means of a pair of dividers.]

Let BF and DH intersect in K. The bisector of the angle BKD will, by symmetry, be parallel to the required bisector, i, e. bisecting the angle BKD by the line KL, the direction of the required bisector is known. To find its position, draw any line AC perpendicular to KL meeting the given lines in A and C. The required bisector must evidently pass through M the centre point of AC. It can therefore be drawn through this point parallel to KL.

PROBLEM 2. (Fig. 3.) To find a fourth proportional to three given lines AB, CD, EF, or to find a line of such length (1) that

or that the rectangle contained by the two lines AB and I shall be equal in area to the rectangle contained by CD and EF.

All questions involving proportionals depend on the construction of similar triangles. Draw any two lines OK, OL meeting

in O and containing any angle. From O along one line set off OG = AB, the first term of the proportion, and OK = EF, the third term of the proportion. From O along the other line set off OH = CD, the second term of the proportion, then through K draw KL parallel to GH meeting OH in L. OL will be the required fourth term. For obviously by the similar triangles OGH,

6

SIMILAR DIVISION OF TWO LINES.

A similar construction will obviously give a third proportional to two given lines AB, CD; i.e. a line of length (1) such that AB CD CD : 1,

or that the rectangle contained by AB and 7 shall be equal in area to the square on CD; the only difference being that in this case the lengths OH and OK will be equal to each other.

PROBLEM 3. (Fig. 4.) To divide a line of given length (AB) similarly to a given line CD divided in any manner as at E, F.............. (There may be any manner of points of division.)

Draw any two lines as OG and OH. Make OG = AB, OH=CD, OK=CE, OL= CF... and draw KM, LN... parallel to HG. The

line OG, i.e. AB will be divided in M, N... similarly to CD in E, F....

PROBLEM 4. (Fig. 1.) To draw a line through a given point and through the intersection of two given lines.

It is of course the simplest possible thing to do this when the actual point of intersection of the two lines is available. As in Problem 1 however it is frequently necessary to draw a line the direction of which depends on an inaccessible point. Let AB, CD be the two given lines and M the given point. (M may be between the lines as in fig. 1 or on the farther side of either with regard to the other.) Draw any line through M meeting the given lines in N and O, and at any convenient distance from M