draw the line AB, and it will be the perpendicular required.

Another method. 1. From A, fig. 9, plate 4, or any other point in AB, with the radius AC, describe the arc CD. 2. From any other point n, with the radius n C, describe another arc cutting the former in C and D. 3. Join the point CD by a line CGD, and CG will be the perpendicular required. PROBLEM 4. Through a given point C, to draw a line parallel to a given straight line AB, fig. 10, plate 4.

1. On any point D, (within the given line, or without it, and at a convenient distance from C,) describe an arc passing through C, and cutting the given line in A. 2. With the same radius describe another are cutting A B at B. 3. Make B E equal to A C. 4. Draw a line CE through the point C and E, and it will be the required parallel.

This problem answers whether the required line is to be near to, or far from the given line; or whether the point D is situated on AB, or any where between it and the required line.

PROBLEM 5. At the given point D, to make an ́ angle equal to a given angle AB C, fig. 12, plate 4.

1. From B, with any radius, describe the arc nm, cutting the legs BA, BC, in the points n and in. 2. Draw the line Dr, and from the point D, with the same radius as before, describe the arc rs. 3. Take the distance mn, and apply it to the arc rs, from r to s. 4. Through the points D and s draw the line Ds, and the angle r Ds will be equal to the angle m Bn, or ABC as required.

PROBLEM 6. To extend with accuracy a short straight line to any assignable length; or, through two given points at a small distance from each other to draw a straight line.

It frequently happens that a line as short as that

between A and B, fig. 11, plate 4, is required to be extended to a considerable length, which is scarce attainable by the help of a rule alone; but may be performed by means of this problem, without error. Let the given line be A B, or the two points A and B; then from A as a centre, describe an arch C BD; and from the point B, lay off B C equal to BD; and from C and D as centres, with any radius, describe two arcs intersecting at E. From the point A describe the arc FE G, making EF equal to EG; then from F and G as centres, describe two arcs intersecting at H, and so on: then a straight line from B drawn through E will pass in continuation through H, and in a similar manner the line may be extended to any assignable length.

OF THE DIVISION OF STRAIGHT LINES.

PROBLEM 7. To bisect or divide a given straight line A B into two equal parts, fig. 13, plate 4.

1. On A and B as centres, with any radius greater than half AB, describe arcs intersecting each other at C and D. 2. Draw the line CD, and the point F, where it cuts AB, will be the middle of the line.

of

If the line to be bisected be near the extreme edge any plane, describe two pair of arcs of different radii above the given line, as at C and E; then a line C produced, will bisect A B in F.

By the line of lines on the sector. 1. Take A B in the compasses. 2. Open the sector till this extent is a transverse distance between 10 and 10. 3. The extent from 5 to 5 on the same line, set off from A or B, gives the half required: by this means a given be readily divided into 2, 4, 8, 16, 32, 64,

line

may

128, &c. equal parts.

PROBLEM 8. To divide a given straight line AB any number of equal parts, for instance, five.

into

Method 1, fig. 14, plate 4. 1. Through A, one extremity of the line A B, draw A C, making any angle therewith. 2. Set off on this line from A to Has many equal parts of any length as AB is to be divided into. 3. Join HB. 3. Join HB. 4. Parallel to H B draw lines through the points D, E, F, G, and these will divide the line A B into the parts required.

Second method, fig. 15, plate 4. 1. Through B draw LD, forming any angle with A B. 2. Take any point D either above or below A B, and through D, draw DK parallel to A B.. 3. On D set off five equal parts D F, FG, GH, HI, IK. 4. Through A and K draw A K, cutting BD in L. 5. Lines drawn through L, and the points F, G, H, I, K, will divide the line AB into the required number of parts.

Third method, fig. 17, plate 4. 1. From the ends of the line A B, draw two lines A C, B D, parallel to each other. 2. In each of these lines, beginning at A and B, set off as many equal parts less one, as AB is to be divided into, in the present instance four equal parts, A I, IK, KL, LM, on A C; and four, BE, EF, FG, GH, on BD. 3. Draw lines from M to E, from L to F, K to G, I to H, and A B will be divided into five equal parts.

Fourth method, fig. 16, plate 4. 1. Draw any two lines CE, DF, parallel to each other. 2. Set off on each of these lines, beginning at C and D, any number of equal parts. 3. Join each point in CE with its opposite point in D F. 4. Take the extent of the given line in your compasses. 5. Set one foot of the compasses opened to this extent in D, and move the other about till it crosses N G in I. 6. Join DI, which being equal to AB transfer the divisions of DI to AB, and it will be divided as required. H M is a line of a different length to be divided in the same number of parts.

The foregoing methods are introduced on account

not only of their own peculiar advantages, but be cause they also are the foundation of several mechanical methods of division.

PROBLEM 9. To cut off from a given line AB amy odd part, as d, th, дth, th, &c. of that line, fig. 18, plate 4.

4.

1. Draw through either end A, a line A C, forming any angle with AB. 2. Make AC equal to AB. 3. Through C and B draw the line CD. Make B D equal to C B. 5. Bisect AC in a. 6. A rule on a and D will cut off a B equal d of AB.

If it be required to divide A B into five equal parts. 1. Add unity to the given number, and halve it, 5+1=6, 3. = 2. Divide A Cinto three parts; or, as A B is equal to A C, set off A b equal A a. 3. A rule on D, and b will cut off b B 4th part of AB. 4. Divide A b into four equal parts by two bisections, and AB will be divided into five equal parts. To divide A B into seven equal parts, 7 + 1 = 8, 1. Now divide AC into four parts, or bisect a C in c, and c C will be the 4th of A C. 2. A rule on c and D cuts off c B 4th of A B. 3. Bisect Ac, and the extent c B will divide each half into three equal parts, and consequently the whole line. into seven equal parts.

= 4.

To divide A B into nine equal parts, 9 + 1 = 10 = 5. Here, 1. Make A d equal to A b, and d C will be 4th of A C. 2. A rule on D and d cuts off dB of AC. 14 3. Bisect A d. 4. Halve each of these bisections, and Ad is divided into four equal parts. 5. The extent d B will bisect each of these, and thus divide AB into nine equal parts.

If any odd number can be subdivided, as 9 by 3, then first divide the given line into three parts, and take the third as a new line, and find the third thereof as before, which gives the ninth part required.

Method 2. Let D B, fig. 19, plate 4, be the given line. 1. Make two equilateral triangles ADB,

CDB, one on each side of the line D B. 2. Bisect AB in G. 3. Draw C G, which will cut off HB equal d of DB. 4. Draw DF, and make GF equal to DG. 5. Draw HF which cuts off Bh equal of AB or DB. 6. Ch cuts DB in i one fifth part. 7. Fi cuts A B in k equal 4th of D B. 8. Ck cuts D B in 1 equal 4th of D B. 9. F1 cuts A B in m equal of D B. 10. Cm cuts DB at n equal 4th part thereof.

Method 3. Let A B, fig. 12, plate 5, be the given line to be divided into its aliquot parts, 4, 4. 1. On A B erect the square ABCD. 2. Draw the two diagonals AC, DB, which will cross each other at E. 3. Through E draw FE G parallel to AD, cutting A B in G. 4. Join D G, and the line will cut the diagonal AC at H. 5. Through H draw IHK parallel to AD. 6. Draw DK crossing A C in L. 7. And through L draw MLN parallel to 'AD, and so proceed as far as necessary. AG is, AK, AN of A B.*

Method 4. Let A B, fig. 13, plate 5, be the given line to be subdivided. 1. Through A and B draw CD, FE parallel to each other. 2. Make CA, A D, F B, BE, equal to each other. 3. Draw CE, which shall divide AB into two equal parts at G. 4. Draw A E, D B, intersecting each other at H. 5. Draw CH intersecting A B at I, making A Id of AB. 6. Draw D F cutting A E in K. 7. Join C K, which will cut A B in L, making A L equal + of AB. 8. Then draw Dg, cutting AE in M, and proceed as before.

COROLLARY. Hence a given line may be accurately divided into any prime number whatsoever, by first cutting off the odd part, then dividing the remainder by continual bisections.

PROBLEM 10. An easy, simple, and very useful method of laying down a scale for dividing lines into

*Hooke's Posthumous Works.