15. TO BISECT LINES, or to divide them into two equal parts.

Strictly speaking, this rule forms part of the general rule in Sect. 18 post for the dividing a line into any number of equal or proportional parts; but as the bisection of a line is so often required in geometrical operations, it is thought better to give separately the rule for the division into two equal parts.

In general the mode of bisection taught is by cross arcs, but such mode is not so neat as that by parallels and proportionals, and involves the use of another tool.

The mode here given and recommended for use is based on the demonstration in Sect. 11 of Div. F post, that angles cut by parallels are divided proportionably.

Thus in Fig. B 12, to bisect or divide into two equal parts the line ab.

From either end (say a), and at any angle, draw the indefinite line ax.

On line ax, with any scale, set off two equal parts, 1 and 2.

Through point 2 and 6 draw parallel 1, and through point 1 and parallel to 26 draw parallel 1', cutting line ab at c. Then c will be the point of bisection.

16. TO BISECT AN ANGLE (Fig. 13).

[NOTE.-In considering an angle the lengths of the including sides will be immaterial.]

The mode proposed is to convert the angle into a rhombus, and then bisect the angle by a diagonal.

On each side of the angle mark with equal lengths ad and af.

From d parallel to af draw the indefinite line dx, and from ƒ parallel to ad draw fg, cutting dx at g, and from a through g draw line ah, which will bisect the angle as required.

For demonstration see Div. C post, Sect. 13.

17. TO BISECT A CHORD OR AN ARC (Fig. B 14).

Let ab mark the chord and acb the arc. By Sect. 15 bisect the chord (at d), and through d (Sect. 12) draw the perpendicular xy in both directions, and it will bisect the arc (C, Sect. 37).

In this instance (one of the very few instances in which it is so) the bisecting by cross arcs will give the operation in certain cases more directly than by parallels, as by the cross arcs shown in the figure, with centre successively at a and at b and the same radius, the direction of the perpendicular xy which bisects the chord and the arc is at once given.

18. TO DIVIDE A LINE into any number of equal or proportional parts.

For rule and examples see Div. F, Sect. 12.

For the division of lines into any parts or proportions very rapidly by an equilateral triangle, see F, Sect. 27.

19. TO DRAW a line which will be a THIRD PROPORTIONAL to any two given right lines (a: b::b: x).

By this proportion the second side of a rectangle which will be equivalent in area with a given square may be

found.

Rule and example in F, Sect. 13, and Fig. 2.

20. TO DRAW a line which shall be a FOURTH PROPORTIONAL to three given right lines (a: b:: c: x).

Rule, &c., in F, Sect. 14, and Fig. F 3.

21. TO DRAW a line which shall be a MEAN PROPORTIONAL between two given right lines (ab × bc = x2).

This shows the line a square of which will be equivalent in area with a rectangle of which the length and altitude are given.

In this case the extremes are given, and the mean term has to be found.

See F, Sect. 15, and Fig. F 4.

For constructional purposes the mean proportional may be found very easily by drawing a line on which is marked off the lengths of the two lines, and at the point of junction (b) drawing an indefinite perpendicular bx (Fig. F 4), taking a strip of thin paper with length fg = ac (diameter),. double it in half fd for radius.

Place one end of the strip at f, and move the other till it cuts the perpendicular at d.

Then bd will be the mean proportional required.

22. TO DIVIDE A LINE into mean and extreme ratio, giving the proportion, as the whole line is to the greater part, so is the greater part to the less; or as in Fig. F 5, ab ac ac: cb.

This proportion also shows that the square of the greater part is equivalent with the rectangle of the whole line and the smaller part.

See Rule, &c., in Div. F, Sect. 16, and Fig. F 5.

23. To describe Geometrically the Arithmetic, Geometric, and Harmonic mean.

See F, Sect. 17, and Fig. F 6.

Sectns. 18 and 19 of Div. F, and Figs. 7 and 8, contain two interesting operations, but not necessarily forming part of it, though having an affinity for it.

24. TO DIVIDE the Base of a Triangle into two Segments bearing the same proportion as the other sides of the Triangle. Rule and demonstration in F, Sect. 20, and Fig. 9.

25. From a given point (a) to draw AN ANGLE equal to a given Angle. Fig. 15 B.

This may be effected by arcs or by perpendiculars.

Case 1. The common mode by arcs, a being the given point and hed the given angle.

With centre c and any radius (say cd) draw the arc df. From a draw the line ai, and with centre a and radius cd draw the indefinite arc ix and set off ij = df.

From a through j draw line dk.

Then jai will be the angle required.

Case 2. By Perpendiculars (Fig. 15 B).

On cd set off any length, cg or cd (selected as the inner giving the sine, and the outer the tangent, of the arc, but the length is not material).

From g or d draw a perpendicular af or dh. Through point a draw any line (ai), and set off al = cg, or ai = cd, and draw the indefinite perpendicular ly or ix.

Set off ljgf, or in dh, and from point a through either point draw line ak, which will complete the required angle.

In large operations the perpendiculars will be found far better than the arcs.

26. To ascertain THE ANGLE where only parts of the sides are given and the apex of the Triangle is inaccessible, and also to bisect such Angle.

Rule and demonstration in Div. C, Sect. 32, and Fig. C 27.

The following Sectns. 27 to 34 show the construction of Triangles.

27. AN EQUILATERAL TRIANGLE. (Fig. B 16.)

Base taken to equal line ab.

With radius ab and successively with a, and with b as centre, draw arcs cutting at c, and complete the triangle by joining ca and cb.

Sides respectively equal to base therefore equal (Axiom 1). The like, with given altitude. (Fig. 25, ah.)

The readiest mode is to draw an equilateral triangle with a greater altitude (akc), and then from the required altitude at h to draw hi and hj parallel to ka and kc.

The triangles will be similar and equiangular, the sides being parallel (F, Sect. 29).

The sides will also be equal by the proportion ka: hike: hj (F, Sect. 11).

On the ground on a large scale.

In such case the triangle will be best constructed by drawing at each end of the base a small triangle (Fig. B 16) with equal sides, and make ad and bf covering-points to complete the triangle at c. The same course might be adopted with other triangles on a large scale.