Whence EDG = 41° 14′ 53′′, FGD = 39° 45′ 7′′. And DG = DE cos 41°14′53′′ + EF. COS 5°14′53′′ +FG. COS 39° 45′ 7′′ 8812.803. 3dly. In the trapezium GHIA, an exactly similar process gives HGA = 50° 46′ 53′′, IAG = 47° 13′ 7′′, and AG 9780.591. 4thly. In the triangle ADG, the three sides are now known, to find the angles: viz, DAG=60°53′ 26′′, agd=43° 15′ 54′′, ADG = 75° 50′ 40". Hence there results, lastly, = IAB = 47° 13′ 7′′ +60° 53′26′′ +39° 30′ 42′′ = 147° 37′ 15′′, CDE = 32° 29′ 18′′ +70° 50′40′′ +41° 14′53′′ 149° 34′51", FGH=39° 45′ 7"+43° 15' 54" +50° 46′ 53′′ = 133° 47′54′′. Consequently, the required exterior angles are A=32°22′ 45′′, D= 30° 25′ 9′′, G = 46° 12′6′′. Er. 6. Required the area of the hexagon in ex. 1. Ans. 16530191. Ex. 7. In a quadrilateral ABCD, are given AB=24, BC=30, CD = 34; angle ABC = 92°18′, BCD = 97°23'. Required the side AD, and the area. Ex. 8. In prob. 1, suppose PQ = 2538 links, and the angles as below; what is the area of the field ABCDQP? APQ=89°14, BPQ=68°11, CPQ=36 24, DPQ= 19°57'; AQP=25°18′, BQP=69°24′, cap=94° 6', DQP=121°18′. CHAPTER VII. PROBLEMS RELATIVE TO THE DIVISION OF FIELDS OR OTHER SURFACES. PROBLEM I. To Divide a Triangle into Two Parts having a Given Ratio, m: n. 1st. By a line drawn from one angle of the triangle. Make AD AB :: m : m + n; draw CD. So shall ADC, BDC, be the parts required. m Here, evidently, AD = AB, DB=- AB. m+n m+n 2dly. By a line parallel to one of the sides of the triangle. Let ABC be the given triangle, to be divided into two parts, in the ratio of m to n, by a line parallel to the base AB. Make CE to EB as m to n: erect ED perpendicularly to CB, till it meet the semicircle described on CB, as a diameter, in A D. Make CF CD: and draw through F, GF || AB. So shall GF divide the triangle ABC in the given ratio. CD2 CE For, CE: CB= :: CD2(=CF2): CB2. But CE: EB :: m :n, or CE : CB :: m : m+n, by the construction: therefore, CF2: CB2 :: M: m+n. And since A CGF: A CAB :: CF2: CB2; it follows that CGF: CAB:: m : m +n, as required. Computation. Since CB CF2:: m+n: m, therefore, (m + n)cr2 = m. CB2; whence CF/(m + n) = CB1/m, or In like manner, CG = CA CF m m + n CBV 3dly. By a line parallel to a given line. Let HI be the line parallel to which a line is to be drawn, so as to divide the triangle ABC in the ratio of m to n. m m+n By case 2d draw GF parallel to AB, so as to divide ABC in the given ratio. Through F draw FE parallel to HI. On CE as a diameter describe a semicircle; draw GD perp. to AC, to cut the semicircle in D. Make CP = CD: through P, parallel to EF, draw PQ, the line required. H B The demonstration of this follows at once from case 2; because it is only to divide FCE, by a line parallel to FE, into two triangles having the ratio of FCE to FCG, that is, of CE to CG. Computation. Co and CF being computed, as in case 1, the distances CH, CI being given, and CP being to co as CH to CI: the triangles CGF, GPQ, also having a common vertical angle, are to each other, as CG.CF to ca.CP. These products therefore are equal; and since the factors of the former are known, the latter product is known. We have hence given the ratio of the two lines CP(=x) to ca(=y) as CH to ci; say, as p to 9; and their product = CF. CG, sayab: to find x and y. Here we find x = v ✓alp, y = ruby. That is, Ρ N. B. If the line of division were to be perpendicular to one of the sides, as to CA, the construction would be similar: VOL. III. M CP m + n CP would be a geometrical mean between CA and being the foot of t! e perpendicular from в upon AC. 4thly. By a line drawn through a given point P. cỏ, Ꮹ M M A By any of the former cases draw lm (fig. 1) to divide the triangle ABC, in the given ratio of m to n: bisect cl in r, and through r and m let pass the sides of the rhomboid crsm. Make cape, which is given, because the point P is given in position make cd a fourth proportional to ca, cr, cm; that is, make ca: cr:: cm: cd; and let a, and d, be two angles of the rhomboid cabd, figs 1 and 2. Pe, in figure 2, being drawn parallel to ac, describe on ed as a diameter the semicircle efd, on which set off efce = ap: then set off dм or dм' on CA equal to df, and through P and M, P and M', draw the lines LM, L'M', either of which will divide the triangle in the given ratio. The construction is given in 2 figs. merely to avoid complexness in the diagrams. The limitations are obvious from the construction: for, the point L must fall between в and c, and the point M between A and c; ap must also be less than rb, otherwise ef cannot be applied to the semicircle on ed. = = ef Demon. Because crcl, the rhomboid crsm= triangle clm, and because ca: cr: cm: cd, we have ca.cd=cm.cr, therefore rhomboid cabd = rhomboid crsm triangle clm. By reason of the parallels CB, bd, and CA, ab, the triangles. aĹp, dGм, bGp, are similar, and are to each other as the squares of their homologous sides ap, dм, bp: now ed2 +df, by construction; and ed = rb, ef ap, df = dм; therefore pb ap2 + dm2, or, the triangle PbG taken away from the rhomboid, is equal to the sum of the triangles aPL, dMG, added to the part capGd: consequently CLM = cabd, as required. By a like process, it may be shown that aL ́P, dG'M', pbG, are similar, and aL'P + do'm' pbG'; whence rbdм'= aL'P, and CL'M' cabd, as required. = Com Computation. cl, cm, being known, as well as ca, ap, or œe, ep, cr = cl, is known; and hence cd may be found by the proportion ca: cr:: cm: cd. Then cd-ce= ed, and √ ed2 — ef2 =√ ed2 — ap2 — dƒ — dû = d'. Thus cM is determined. Then we have cl. cm CM = CL. N. B. When the point is in one of the sides, as at M; then make CL. CM. (m+n) = CA. CB. m, or, CL: CA:: M. CB : (m+n)cм, and the thing is done. 5thly. By the shortest line possible. MA Draw any line PQ dividing the triangle in the given ratio, and so that the summit of the triangle cra shall be c the most acute of the three angles of the triangle. Make CмCN, a geometrical mean proportional between CP and co; so shall MN be the shortest line pos- A sible dividing the triangle in the given ratio.The computation is evident. Demons. Suppose MN to be the shortest line cutting off the given triangle CMN, and CG MN. MN MG + GNCG. cot M+ CG. cot N = CG (cot M + cot N). But, cot м + COS N sin (M + N) + = cot N = COS M sin M sin N sin M.sin N And (equa. P M G N B XVIII, ch. iii) sin м.. sin N = COS (M — N) — — COS (M+N)= sin (M+N) = COS (MN) + cos c. Theref. MN=CG. which expression is a minimum when its denominator is a maximum; that is, when cos (M-N) is the greatest possible, which is manifestly when M-N=0, or M⇒ N, or when the triangle CMN is isosceles. That the isosceles triangle must have the most acute angle for its summit, is evident from the consideration, that since 2 A CMN CG. MN, MN varies inversely as CG; and consequently MN is shortest when CG is longest, that is, when the angle c is the most acute. N. B. A very simple and elegant demonstration to this case is given in Simpson's Geometry: vide the book on Max. and Min. See also another demonstration at case 2d prob. 6th, below. PROBLEM II. To Divide a Triangle into Three Parts, having the Ratio of the quantities m, n, p. 1st. By lines drawn from one angle of the triangle to the opposite side. ་ Divide the side AB, opposite the angle c from whence the lines are to proceed, in the given ratio at D, E; join CD, CE; and ACD, DCE, ECB, are the three triangles required. The demonstration is manifest; as is also, the computation. If it be wished that the lines of division be the shortest the nature of the case will admit of, let them be drawn from the most obtuse angle, to the opposite or longest side. E 2dly. By lines parallel to one of the sides of the triangle. Make CD: DH: HB::m:n: p. Erect DE, HI, perpendicularly to CB, till they meet the semicircle described on the diameter CB, in E and I. Make CF CE, and CK= CI. Draw GF through F, and LK through K, L parallel to AB; so shall the lines GF and LK, divide the triangle ABC as required. G A H The demonstration and computation will be similar to those in the second case of prob. 1. 3dly. By lines drawn from a given point on one of the sides. Let (fig. 1) be the given point, a and b the points which divide the side AB in the given ratio of m, n, p: the point P falling between a and b. Join PC, parallel to which draw ac, bd, to meet the sides A€, BC, in the points c and d join PC, rd, so shall the lines CP, Pd, divide the triangle in the given ratio. : In fig. 2, where P falls nearer one of the extremities of AB than both a and b, the construction is essentially the same; the sole difference in the result is, that the points c, and d, both fall on one side AC of the triangle. Demon. The lines ca, cb, divide the triangle into the given ratio, by case 1st. But by reason of the parallel lines ac, PC, bd, ▲ acc▲ acr, and ▲ bdc = bdr. Therefore, in fig. 1, A Aac + acpAac + acc, that is, ACP = Aac: and вbd+bdr Bbdbdc, that is, вdPBbc. Consequently, the remainder ccrd cab.-In fig. 2, ACP = Aac, and Adp = Acb; therefore cpd AdP ACB =aCP; and ACB Acb, that is, CBPd CBb. Computation. The perpendiculars cg, CD being demitted, ACP |