There is no construction in Plane Geometry, i. e. no construction practicable with the assistance of the right line and the circle only, by which a given angle may be divided into three, five, &c. equal parts. It is true that, by the description of other lines which are plane curves, (as the hyperbola, conchoïd, and cissoïd,) any given angle may be trisected, or divided into three equal parts; but such constructions, not being effected by the two species of lines which are the subject of these Elements, are for certain reasons (to be found in the application of Algebra to Geometry) said to be out of the range of Plane Geometry. There are, however, some particular cases in which, by means of the circle, an angle may be divided into three, five, and other numbers of equal parts. Such is the division of a right angle into three equal parts. In this case we are already aware, that, since an equilateral triangle has its three angles equal to one another, and the three taken together equal to two right angles, each of them is two thirds of a right angle. The problem, therefore, will be solved by describing an equilateral triangle upon either of the legs, taken of any length at pleasure. In like manner, by means of an isosceles triangle, each of whose base-angles is double of the vertical angle, (see book iii.) and is, therefore, four-fifths of a right angle, Plane Geometry enables us to divide a right angle into five equal parts. But these are only particular cases, and indicate no general process. PROP. 47. Prob. 6. (Euc. i. 23.) At a given point A, in a given straight line AB, to make an angle equal to the given rectilineal angle Č. From the centre C, with any radius, describe a circle, cutting the sides of the given angle in the points D E, and D join DE: from the centre A, with a radius equal to CE or CD, describe a circle, cutting A B in B, and from the centre B with a radius equal to D E, PROP. 48. Prob. 7. (Euc. i. 31.) Through a given point A, to draw a straight line parallel to a given straight line B C. From A to B C (as directed in Prop. 47., or as in the Therefore, &c. N. B.-In the figure, equal circles are, in the first place, described from the centres A, C, the former cutting BC a second time in B; and the point D is then determined by describing a circle from the centre A with the radius B C. This construction is equally short, in practice, with that of Prop. 47.; and when it is required to obtain the point D at as great a distance as possible from A, may be preferred, the line A C being drawn more obliquely for that purpose. PROP. 49. Prob. 8. (Euc. vi. 9). into any number of equal parts. To divide a given straight line, A B, Let it be required to divide A B into five equal parts. Draw A C, making any angle with A B, and taking any distance M, set it off upon A C five times from A to C: join C B, and through the points of division in A C, viz. D, E, &c. draw (48.) parallels to C B, cutting A B in the points F, G, &c. AB shall be divided at the latter points into five equal parts. Through F and G draw (48.) F H and G K, each of them parallel to A C. Then, because FC is a parallelogram, FH is equal to DC (22.); and for the like reason G K is equal to ED: but ED, DC, are equal to one another; therefore (ax. 1.), FH is equal to G K. But FH, GK, are sides of the E triangles FH B, G K F, which have their angles equal to one another, each to each, because their sides are parallel (18.): therefore (5.), their other sides are also equal to one another, and F B is equal to G F. In like manner it may be shown, that each of the other five parts into which AB is divided, is equal to FB or GF. And it is evident that, by a similar process, A B may be divided into any other number of equal parts. Therefore, &c. It is convenient to draw the parallels DF, EG, &c. by making the angle A B a, equal to B A C (47.), taking B d, de, &c. each equal to M, and joining corresponding points Dd, Ee, &c. For the straight lines which join the extremities of equal and parallel straight lines, are also themselves equal and parallel (21.). A similar process may be recommended, whenever a number of parallels are to be drawn without the assistance of a parallel ruler. PROP. 50. Prob. 9. (Euc. i. 22). To construct a triangle from threeparts given, of which one, at least, is a side. Case 1. Let the given parts be two angles and a side; viz. A the angles A B, and the side CD. If the given side is to be interjacent or 3 to lie between the given angles, make (47.) at the points C, D, the angles DCE, CDE, equal to A, B, respectively: and the triangle ECD will evidently be the triangle required. But if CD is to be opposite to one of the given angles, as B, make at the point C, the angles DCE, ECF, equal to A, B, respectively, and through D, draw DE parallel to C F (48.): then the angle CED is (15. Cor. 2.) equal to the alternate angle E CF, that is, to B, and ECD is the triangle required. Case 2. Let the given parts be two sides and an angle; viz. the sides A. B, and the angle C. If the given angle is to be included, take CD, CE, upon its sides, equal to A, B, respectively, and join DE; the triangle CDE will evidently be the triangle required. B Д But if C is to be opposite to one of the given sides as A, take CE equal to B as before: from the centre E, with a radius equal to A, describe a circle cutting CD in the points D, D; and join ED, ED. Then, if the points D, D fall both on the same side of C with the angle, each of the triangles ECD will satisfy the given conditions; but if they fall upon different sides, only one of them, as ECD, will satisfy those conditions, and therefore that one will be the triangle required. If the points D, D, being upon the same side with the angle, coincide with one another, which will happen when EDC is a right angle (12. Cor. 3.), there is only one triangle EDC, which is the triangle required. Case 3. Let the given parts be three sides, viz. A, B, and C D. AFrom the cen- Btres C, D,with radii equal to A, B, respectively, describe circles intersecting in E, and join E C, ED: then ECD is evidently the triangle required. Therefore, &c. Scholium. In each of these three cases we have supposed the data, or given parts, to be such that the problem is possible: and the same will be supposed in all future problems. Many of them, however, will be found to be of that description, that, if the relations of the data be not confined within certain limits, the required solutions will be impossible. In problems of a more complex nature than the present, the determination of these limits is facilitated by the consideration of loci, to be noticed hereafter (in Book III. Sect. 6), and which may here be explained as circumscribing the range, if it be limited, within which every particular datum confines the solution of the problem; for it is obvious that, if a second datum cannot be satisfied within that range, the two will be inconsistent, and the solution impossible. In the mean time, it may be observed, the limits of the data are in many cases readily suggested by the known properties of the figure, or by the necessary construction of the problem. It is evident, for instance, that in every case of the problem before us, there are certain limits, beyond which the solution will be impossible. This will happen in the first case if the two given. angles should be together equal to or greater than two right angles, for there PROP. 51. Prob. 10. Given the three angles of a triangle (together equal to two right angles), and the perimeter A B, to construct the triangle. At the points A, B, make (47.) the angles BAC, ABC, equal to two of the given angles, each to each; bisect (46.) the angles at A, B, α by the straight lines A c, B c, meeting in c, and through c draw (48.) ca, cb, parallel to CA, CB, respectively: the triangle ABC shall have its angles equal to the three given angles, and its perimeter equal to the given perimeter A B. The first is evident, for its sides are parallel to the sides of the triangle CAB, which has its angles at A and B, and therefore at C, equal to the given angles (18.). Again, because ca is parallel to CA, the angle acA is equal (15.) to the alternate angle CAc, that is to the angle a A c, and therefore (6.) ac is equal to aA; and in the same manner it may be shown that bc is equal to B; therefore, the three sides of the triangle ab c are together equal to A B, that is, to the given perimeter. Therefore, &c. PROP. 52. Prob. 11. Given two sides and the included angle of a parallelogram, to construct the parallelogram. Let A C, B C, be the two given sides, and C the given angle. From the centres A, B, with radii equal to B C, AC, respectively, describe circles intersecting in D, and join D A, D B. has its opposite sides equal to one ano- N. B. The same end will be obtained by drawing parallels to C B, CA through the points A, B. It may be observed that if AC be equal to BC, the parallelogram is a rhombus; if AC be at right angles to B C, it is a rectangle; and if AC be both equal to B C, and at right angles to it, it is a square. This problem, therefore, includes the following as particular cases; 1st, to describe a rhombus with a given side and angle; 2d, to deand 3d, to describe a square upon a scribe a rectangle with two given sides; given finite straight line. PROP. 53. Prob. 12. To describe a triangle which shall be equal to a given quadrilateral A B C D, and shall have a side and angle adjacent to it, the same with a given side A B and adjacent angle B of the quadrilateral Join AC: through is parallel to AC, A E the triangle ACE is equal to ACD PROP. 54. Prob. 13. be equal to a given rectilineal figure To describe a triangle which shall ABCDEF, and shall have a side and adjacent angle the same with a given side AB and adjacent angle B of the figure. FG parallel to AE (48.), to meet DE Join A C, AD, AE: through F draw produced in G: through G draw G H parallel to AD, to meet CD produced in H; through H draw HK parallel to HC, to meet B C produced in K, and join A K, AH, A G.* Then because the triangle ADG is AH is not joined in the figure, to avoid confusion for the same reason, A M and AL are not Then because the quadrilateral AD joined in the corresponding figure of Prop. 55. A B K, is equal to the figure ABCD EF, and it has the same side A B, and adjacent angle B with the figure. Therefore, &c. PROP. 55. Prob. 14. To bisect a given triangle or rectilineal figure by a straight line drawn from a given angle. First, let it be required to bisect the triangle A B C by a straight line to be drawn from the angle A. Bisect (43.) B C in D, and join AD. Then because B D is equal to D C, the triangle ABD is (27.) equal to ADC, and the triangle ABC is bisected by the straight line A D. B Next, let it be required to bisect the rectilineal figure ABCDEF by a straight line, to be drawn from the angle A. With the same side A B and angle B, describe (54.) the triangle A BK, equal to the given rectilineal figure: bisect (43.) BK in L: through L draw (48.) LM parallel to A C, to meet C D produced in M: through M draw M N parallel to AD, to meet D E in N, and join AN, AM, AL. Then, it may be shown, as in the last proposition, that the triangle A BL is equal to the figure ABCDN; but A B L is equal to half the triangle ABK, that is to half of the given figure; therefore the figure ABCDN, is also equal to half of the given figure, and the latter is bisected by the straight line A N. Therefore, &c. For Cor. 1. By a similar construction, any part required, for example a fifth, may be cut off from a given triangle or rectilineal figure, by a straight line drawn from one of its angles. in the case of the triangle, if the base BC be divided into five equal parts in the points D, &c. the triangles A BD, &c. will be equal to one another (27.); and therefore any one of them, as A B D, will be equal to one fifth of the given triangle. And hence the passage is easy to the division of the rectilineal figure; for it is evident (53.) that, whatever be the point L taken in AK, the figure ABCD N, constructed as above, will be equal to the triangle A B L, and will therefore be a fifth part of the given figure, if A B L be a fifth of the triangle ABK. It is manifest in this case, that, if the given fractional part be such, that B L is less than BC, the problem will be solved at once by joining A L. Cor. 2. And hence it appears in what manner a triangle or rectilineal figure may be divided into any number of equal parts by straight lines drawn from one of its angles. PROP. 56. Prob. 15. To bisect a given triangle or rectilineal figure by a straight line drawn from a given point in one of its sides. First, let it be required to bisect the triangle ABC, by a straight line to be drawn from the point D in the side A B. F Join D C bisect (43.) AB in E: through E draw (48.) E F, parallel to DC, and join DF, EC. Then, because DC is parallel to E F, the triangle D E F is 27.) equal to CEF; therefore DEF, BEF together, are equal to CEF, BEF together, that is, the triangle DB F, is equal to the triangle C E B, or (27.) to half of the tri. angle ABC. Therefore, ABC is bisected by the straight line D F. Next, let it be required to bisect the figure A B C D E F, by a straight line, to be drawn from the point G in the side AB M Describe the triangle ABK (54.) equal to the given figure, and, as already shewn, bisect it by the straight line GL, drawn from the point G: join GC, GD, &c.; through L draw (48.) LM, parallel to G C, to meet CD produced in M: through M draw M N parallel to GD to meet DE in N, and join GN, GM, GL.* Then, as in the last proposition, it may be shown that the figure GB CDN is equal to the triangle G B L, that is to half of the given figure; therefore the latter is bisected by the straight line GN. Therefore, &c. Cor. 1. By a similar construction, any part required may be cut off from a given triangle or rectilineal figure, by a straight line drawn from a given point in one of its sides. For we have but to make CBE the same part of C BA in the first case, and G BL the same part of ABK in the case of the rectilineal figure, and proceed as before. If the fractional part be such that BL is less than BC, the problem will be solved at once by joining G L. Cor. 2. And hence a given triangle or rectilineal figure may be divided into any number of equal parts, by straight lines drawn from a given point in one of its sides. PROP. 57. Prob. 16. (Euc. i. 44.) Upon a given base BD to describe a rectangle which shall be equal to a given triangle ABC. From the point D draw DE, at right angles to BD (44.) : through A draw A E parallel to BD, to meet DE in E (48.): join M LA EB: bisect BC in F (43.): through F draw FG, parallel to DE, to meet BE in G: through G draw H K parallel to BD, and complete the rectangle DH: DH shall be the rectangle required. Complete the rectangles DL, DM. Then, because GM, GD are complements of the rectangles F H, K L, which are about the diagonal of the rectangle DM, GD is (23.) equal to G M. Therefore, adding F H to each, the whole HD is equal to the whole M F, that is, (26.) to twice the triangle AB F, or to the triangle ABC; and HD is described upon the given base B D. Therefore, &c. GD and GL do not necessarily coincide, as in the figure. Cor. 1. Hence, upon a given base, a rectangle may be described which shall be equal to a given rectilineal figure: for a triangle (54.) may be described equal to the figure, and a rectangle equal to the triangle. Cor. 2. It is evident that, with a like construction, a parallelogram may be described upon a given base BD which shall be equal to a given triangle A B C, and shall have one of its angles equal to a given angle B D E. PROP. 58. Prob. 17. (Euc. ii. 14). To describe a square which shall be equal to a given rectangle A B C D. Produce A B to E, so that BE may be equal to B C: bisect the centre F, with AE in F (43.): from the radius FA or D FE, describe a circle, and produce CB to meet the circumference in G: the square of BG shall be equal to the rectangle ABCD. to FE, AB is equal to the sum, and Join FG, then, because FA is equal BE to the difference, of FE, FB: that is, the rectangle ABC D, is equal therefore (34.) the rectangle AB, BE, to the difference of the squares of FE, F B, that is to the difference of the squares of F G, FB, or (36, Cor. 1.) to the square of B G. Therefore, &c Cor. Hence a square may be described which shall be equal to a given rectilineal figure (57. Cor. 1.). PROP. 59. Prob. 18. To describe a square which shall be equal to the difference of two given squares, viz. the squares of AB, A C From the point C draw AB, and from the centre (44.) CD perpendicular to A with the radius A B, describe the circle B D cut- A ting CD in D: the square of CD shall be equal to the difference of the squares of AB, A C. For (36. Cor. 1.) the square of CD is the difference of the squares of AD, A C, of which A D is equal to A B. Therefore, &c. Cor. Hence a square may be deference of two given rectilineal figures scribed, which shall be equal to the dif(58. Cor.). PROP. 60. Prob. 19. |