through A and B, will be the perpendicular required; the numbers 6, 8, 10, are used as multiples of 3, 4, 5.

By this method, a perpendicular may be easily and accurately erected on the ground.

Method 5. Let A C. fig. 7, plate 4, be the given line, and A the given point. 1. At any point D, with the radius D A, describe the arc EAB. 2. With a rule on E and D, cross this arc at B. 3. Through A and B draw a right line, and it will be the required perpendicular.

PROBLEM 2. To raise a perpendicular from the middle, or any other point G, of a given line AB, fig. 8, plate 4.

1. On G, with any convenient distance within the limits of the line, mark or set off the points n and m. 2. From n and m with any radius greater than GA, describe two arcs intersecting at C. 3. Join CG by a line, and it will be the perpendicular required.

PROBLEM 3. From a given point C, fig. 8, plate 4, out of a given line A B, to let fall a perpendicular.

When the point is nearly opposite to the middle of the line, this problem is the converse of the preceding one. Therefore, 1. From C, with any radius, describe the arc n m. 2. From mn, with the same, or any other radius, describe two arcs intersecting each other at S. 3. Through the points CS draw the line CS, which will be the required line.

When the point is nearly opposite to the end of the line, it is the converse of Method 5, Problem 1, fig. 7, plate 4.

1. Draw a faint line through B, and any conve nient point E, of the line A C. 2. Bisect this line at D. 3. From D with the radius DE, describe an arc cutting AC at A. 4. Through A and B

draw the line A B, 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 C D 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 are passing through C, and cutting the given line in A. 2. With the same radius describe another arc cutting A B at B. 3. Make BE 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 A B 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 m. 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 rDs will be equal to the angle m Bn, or AB Cas 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 CBD; and from the point B, lay off B C equal to B D; and from C and D as centres, with any radius, describe two arcs intersecting at E. From the point A describe the arc FEG, 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 be extended to any assignable length.

OF THE DIVISION OF STRAIGHT LINES.

may

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.

If the line to be bisected be near the extreme edge of 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 AB in F.

By the line of lines on the sector. 1. Take AB 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 line may be readily divided into 2, 4, 8, 16, 32, 64, 128, &c. equal parts.

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

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 H as many equal parts of any length as AB is to be divided into. 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 AB. 2. Take any point D either above or below A B, and through D, draw D K parallel to A B. 3. On D set off five equal parts DF, 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, B E, EF, FG, GH, on B D. 3. Draw lines from M to E, from L to F, K to G, I to H, and AB 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 NG 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 because they also are the foundation of several mechanical methods of division.

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

1. Draw through either end A, a line A C, forming any angle with AB. 2. Make AC equal to A B. 3. Through C and B draw the line C D. 4. Make B D equal to CB. 5. Bisect AC in a. 6. A rule on a and D will cut off a B equal d of A B.

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 A B. 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, += 4. 1. Now divide AC into four parts, or bisect a C in c, and c C will be the 4th of AC. 2. A rule on c and D cuts off c B 4th of A B. 3. Bisect A c, and the extent c B will divide each half into three equal parts, and consequently the whole line into seven equal parts.

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. 3. Bisect A d. 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 A B 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,