to this point, and apply them again as at first; perform the operation till the revolution of the compasses coincides with the given line. Thus (fig. 81), suppose it were required to divide the line A B into two equal parts, and the distance A C' was the first guess or opening of the dividers; turning on the point C', the point of the dividers that was at A falls on the point D beyond B, keeping the point of the dividers still on the point C', open them till they embrace the distance Cb, b being at or near as can be judged by the eye the centre of DB; begin again from the point A with the distance C'b contained in the dividers, and apply the distance as at first, dividing the deficit or excess of the two revolutions, till the point of the dividers that was at A falls by revolution on B. The eye, by practice, becomes so accustomed to this means of division, that a plan may be reduced to half scale as quickly and with as little chance of mistake as by the proportional compasses.

To divide a line into any number of equal parts.—If the number is divisible by two, bisect the line, or divide it into two equal parts as above, and continue this as long as the number remaining is divisible by two; but when the number is uneven, measure the given line on the scale, divide numerically the length thus found by the number of parts into which it is to be divided, and take on the scale as accurately as possible the quotient thus obtained; apply this length successively on the line, and if the last distance set off does not agree with the extremity of the line; thus if, as in fig. 81, when the line is to be divided into two parts, the repeated length exceeds the line, divide this excess by the eye into as many parts as the given line is to be divided, and close the dividers so as to include a length less by one of these parts. If the point D should fall inside

A

·D` Fig. 82.

-B

B, divide the deficit as before by the number of parts, but open the dividers by one of these parts.

The above problems may be constructed geometrically as follows:-To bisect or divide into two equal parts a given line A B. From A and B, with any radius

greater than half of A B, describe arcs intersecting each other above

and below the given line, the line C D connecting these intersections will

bisect A B, and also be perpendicular to it.

A÷

PROB. IV.-To divide a given line into a given number of equal parts.

with the line A B, will divide it into four equal parts.

PROB. V.-To draw a perpendicular to a straight line, from a point without it.

1st Method (fig. 84).—From the point A, with a sufficient radius, cut the given line at F and G, and from these points describe arcs cutting at E through E draw A E, which will be the perpendicular required. If there be no room below the line, the intersection may be taken above; that is, between the line and the given point. This mode is not, however, likely to be as exact in practice as the one given.

2d Method (fig. 85).-From any two points B and C, at some distance apart, in the given line, and with radii B A, C A, respectively, describe arcs

cutting at A and D. Draw the perpendicular required, A D. This method is useful where the given point is opposite the end of the line, or nearly so. PROB. VI.-To draw a perpendicular to a straight line from a given point A in that line.

1st Method (fig. 86).—With any radius, from the given point A, in the given line B C, cut the line at B and C; with a greater radius describe arcs from B and C, cutting each other at D, and draw D A the perpendicular.

2d Method (fig. 87).-From any centre F, above B C, describe a circle passing through the given point A, and cutting the given line at D; draw DF, and produce it to cut the circle at E; and draw A E the perpendicular. This method is useful when the point A is at or near one end; and in practice, it is expedient in the first place to strike out a preliminary arc, of any convenient radius, from the point A, as any point in that arc may be chosen for the centre F, with the certainty that the arc from this centre will pass through A, without the delay of adjusting the point of the compass to it. This expedient is of general use where an arc is to be passed through a given point, and particularly if the point of the pencil be round or misshapen, and therefore uncertain.

3d Method (fig. 88).—From A describe an arc E C, and from E, with the

same radius, the arc A C, cutting the other at C; through C draw a line ECD, and set off C D equal to C E, and through D draw A D the perpendicular

required. This method, like the previous one, is useful when the point A is at one end.

4th Method (fig. 89).—From the given point A, set off a distance A E equal to three parts, by any scale; and on the centres A and E, with radii of four and five parts respectively, describe arcs intersecting at C. Draw AC for the perpendicular required. This method is most useful on very large scales, where straight edges are inapplicable, as in laying down perpendiculars or right angles on the ground; as in laying out the corners of houses, beams and girders may be set square with the sides of the houses: columns, and the like, may be set perpendicularly by the same method. numbers 3, 4, 5, are, it is to be observed, taken to measure respectively— the base, the perpendicular, and the slant side of the triangle A E C. Any multiples of these numbers may be used with equal propriety, as 6, 8, 10, ọr 9, 12, 15, whether feet, yards, or any other measure of length. PROB. VII.-To draw a straight line parallel to a given line, at a given distance apart (fig. 90).

C

D

A

B

Fig 90.

The

From the centres A, B, in the given line, with the given distance as radius, describe arcs C, D; and draw the parallel line C D touching the arcs. The method of drawing tangents will be afterwards shown; meantime, in all ordinary cases, the line C D may be drawn by simply applying a straight edge by the eye. PROB. VIII.—To draw a parallel through a given point. 1st Method (fig. 91). With a radius equal to the distance of the given point C from the given line A B, describe the arc D from B, taken considerably distant from C; draw the parallel through C to touch the arc D.

2d Method (fig. 92).-From A, the given point, describe the arc F D, cutting the given line at F; from F, with the same radius, describe the arc E A; and set off F D equal to E A. Draw the parallel through the points A, D.

Fig. 93.

B

PROB. IX.-To construct an angle equal to a given angle (fig. 93).

Thus, on the line ab to construct an angle which shall be equal to the given angle CAB. With the dividers describe the arc CB; from the point a, with the same radius describe cb; with. the dividers measure the length of the arc C B, and on eb lay off this distance; through c draw c a, and we have the required angle. or opening cab, equal to the given angle CA B. PROB. X.-From a point A of a given line D E, to draw a line making an angle of 60° with the given line (fig. 94).

Take any convenient distance in the dividers, and from A describe the arc B C. From B, with the same distance, describe an arc, and mark the point C where the arcs cross. Draw the line A, C. This line will make with the given one the required angle of 60°.

PROB. XI.—From a point B on a given line DE, to draw a line making an angle of 45° with it (fig. 95).

Set off any distance B a, along D E, from B. Construct a perpendicular to D E at a, and set off on this perpendicular a c equal to a B; draw through B c a line, which will make with D E the required angle of 45°.