Take a st. line AB of indefinite length towards B; and let C be the given st. line that is to be cut off from AB: from AB, by 3. I., cut off a part equal to C, as AE; then again from EB another part equal to C, as EF; and so on: the parts in AB are each equal to C and to one another; and AB is a scale of equal parts. (3. I.)

On the same principle the line KL is divided. Each of the parts numbered 1, 2, 3, 4, is representative of 10 equal parts, and those between 0 and K of units; or if the smaller divisions be tenths, the larger will be

units.

By such a scale the comparative lengths of lines are ascertained.

PROB. 18. Given AB the base, angle A the less angle at the base, and AD the difference of the two sides of a triangle, to construct it.

On A D the line forming with AB the angle A, by 3. I., set the difference AD of the two sides; join DB; and, by 10 and 11. I., bisect DB by the perpendicular EC; and produce EC and AD to meet in C: ABC is the triangle required. (19 and 4. I.)

D

B

O

PROB. 19. On a given line AB, to describe an isosceles triangle, having each of the equal sides double of the base.

Produce A B both ways; and by 3. I., make BC, AO each equal to AB: OB and AC each equals twice AB. With AC from A, and with BO from B, describe arcs cutting in C; join C to A and B: ABC is the isosceles triangle with sides each the double of the base. (22. I.)

PROB. 20. To describe an isosceles triangle, the base O ADB C AB being given, and F a line equal to each of the sides.

Bisect the base in D, and raise a perpendicular at D; from A with radius equal to F, draw an arc cutting the perpendicular in C; join CA and CB: the required isosceles triangle is ACB. (22. I.)

PROB. 21. To construct a regular polygon, as ABDEFG. By 32. I., the formula for the angular magnitude of each angle in a regular polygon is,

LA

(S × 180)-360

X

S

two sides; 4 C=

the angle between any

360

for the angle at the cen- G tre between two radii to the angular points. (Note, 32. I., p. 108.)

10. When the side AB is not given. With any radius, as CA, describe a circle; at C the centre, make an angle equal to

360
S'

and pro

F

E

B

D

duce CB to the circumference; join AB: the line AB will, if set round the circumference, divide it into as many parts as there are sides to the polygon; join the points, and the polygon is made.

(S × 180)-360
S

; ascer

2o. When the side AB is given. By formula, A= tain the angles GAB, ABD, and construct them; by 9. I., bisect each of the angles; and the lines of bisection, AC, BC, intersect in C: from C, with radius CA, describe a circle; if now, with AB as radius, successive arcs be cut off from the circumference, there will be as many arcs as sides, and the drawing of the chords to those arcs will complete the polygon. (Use and App. 32. I., p. 108.)

PROB. 22. To divide a finite straight line AL, into any given number of equal parts, as four.

At A, draw a line of indefinite length, AX, making with AL the angle XAL; take a line AB on AX; and, by 3. I., set along A X three other lines BC, CD, DE each equal to AB; join EL; and through the points D, C, and B, by 31. I., draw Dd, Cc, Bb, each parallel to EL: the line AL will be divided into four equal parts in the points b, c, d. (Use and App. 34. I., p. 111.

A

B

EX

D

C

PROB. 23. From a point E in the side AB of a parallelogram, to divide the parallelogram into two equal portions.

Draw the diagonal AD, and, by 10. I., bisect it in F; join EF, and produce EF to G: the line EG makes the portion AEGC equal to the portion EGDB. (Use and App. 6, 34. I., p. 113.)

PROB. 24. To convert a parallelogram, ABDF, into a rectangle, ABCE, of equal area.

Produce indefinitely the parallel DF; C at B and A, the extremities of the other parallel AB, by 11. I., raise the perpendiculars BE and AC to meet the parallel 3 DC; then ABCE is a rectangle equal in area to the parallelogram A BDF. and App. 35. I., p. 115.)

(Use

5

A

E

F

D

PROB. 25. Given a triangle, ABC, and a point, H, in one side, AC; from that point to bisect the triangle.

PROB. 26. On a given line, EK, to draw a parallelogram equal to a given parallelogram, KD.

Produce FK, and by 3. I. make EK equal to the given line; produce indefinitely the sides DF, HK, and DH; and by 31. I. draw through E, AB parallel to HG or DC; draw the diago- E nal AK, and produce it until it cuts DF produced in C; through C draw a parallel to DA or FE, meeting HK and AE produced in G and B: the parallelogram BK will be equal to the given parallelogram KD. (Use and App. 43. Ĭ., p. 127.)

B

A

H

Ꭰ

K

F

G

PROB. 27. To change any right-lined figure, ABCDE, first into a triangle, and then into a rectangle of equal area.

Divide the figure into triangles by joining DA, DB; produce AB indefinitely; through E, by 31. I., draw EH parallel to DA, and through C, CF par. to DB; join DH and DF: F then the triangle DHF is equal in area to ABCDE.

Next, draw DN parallel to HF; by 10. I. bisect HF in L; at L, 11. I., raise the perpendicular LM, and draw FN parallel to LM: the H A

fig. LMNF is a rectangle equal in area to

DHF, which is equal to ABCDE. (Use and App. 2, 45. I., p. 132.)

PROB. 28. To make straight a crooked boundary, ABCDE, between two fields, M and N.

Draw AC, the subtend to angle B; and through B, 31. I., FB parallel to AC, and join CF: the crooked boundary AB, BC is now converted into the single boundary CF. In a similar way FC and CD will be converted into one boundary; and this last and DE into a single boundary: and thus the crooked boundaries AB, BC, CD, DE, will be changed into one straight boundary without affecting the areas of the two fields. (Use and App. 3, 45. I., p. 132.)

M

D

C

N

B

A

PROB. 29. Given the diagonal, AB, to construct a square.

At A and B, by 11. I., draw the perpendiculars AE, BF; by 9. I., bisect the right angles by AC and BC meeting in C; and by 31. I., draw through A and B, parallels, AD to BC, and BD to AČ: then the figure ABCD is the square required. (46. I., p. 133.)

E

C

PROB. 30. To find a square equal to the sum of any number of given squares, as on AB, BC, CD, DE; or a square that is a multiple of any given square, AB; or a square that is equal to the difference of two squares.

10. Set the lines AB, BC, representative of the two squares on AB, BC, at right angles, and join G AC; then A C2=AB2+BC2; at C place CD, representative of the square on CD, at right angles to AC; and AD2=AC2 +CD2: and at D, forming a right angle with AD, place DE representative of the square on DE; join AE: and the square on AE equals the squares on DE, DC, CB, and BA. (Cor. 3, 47. I., p. 137.)

20. Supposing AB to be representative of the line on which the given square is constructed, its multiple square will be obtained in a similar way;

B

D

for in this case, BC, CD, DE being each equal to AB, the square on AE is the multiple of the square on AB. (Cor. 3, 47. I., p. 137.)

30. Let A B be the less line, and AC the greater; at B, the extremity of the less, raise a perpendicular BG, and from A at the other extremity, with AC as radius, inflect on BG the greater line: the square of the intercept CB will equal the difference of the squares on AC and AB. (Cor. 3, 47. Î., p. 137.)

PROB. 31. To make a square that shall be the half, fourth, &c., of a given square on AB.

At A and B make the angles each equal to half a right angle; C being a right angle, the square on A C will be one-half of the square on AB. Again, at A and C make the angles CAD, ACD each equal to half a right angle: then the square on CD will be half that on AC, or one-fourth of the square on AB. By a similar process a square may be obtained that is 1, 16, 32, &c., of the original square. (Cor. 3, 47. I., p. 138.)

1

B

C

E

1/4

D

1/2

A

II. THE PROBLEMS IN BOOKS 3, 4, AND 6, WHICH THOUGH DEMONSTRATED UNDER THEIR RESPECTIVE PROPOSITIONS, ARE, FOR THEIR CONSTRUCTION AND SOLUTION, MOST INTIMATELY CONNECTED WITH THE FIRST AND SECOND BOOKS.

Book III.

PROB. 1. In a given circle, ABC, to find the centre.

Draw any chord AB; and, 10. I., bisect it at D; from D, 11. I., raise a perpendicular produced both ways to meet the circumference in C and E; bisect CE in F; and F is the centre of the circle. (1. III.)

B

E

PROB. 2. To draw a tangent to a given circle BCD, from a given point, either without as A, or in the circumference as D.

Find the centre E of the circle BCD, and join AE; from E with EA describe the circle AFG; from D, 11. I., draw DF at right angles to EA, and join EBF and AB: then AB shall touch the circle in B, and DF in D. (17. III.)