▲ ACP : ▲ ACB ::m:m+n+p :: AP.cg: AB.CD. Therefore (m+n+p) Ap.cg=m.AB.CD, and cg= m AB CD (m+n+p) AP The line cg being thus known, we soon find Ac; for CD: AC :: cg` : AC = m. AB. AC AC.cg Indeed this expression may be deduced more simply; for, since ACB ACP :: AC. AB : AC.AP :: m+n+p: m, we have (m+n+p) Ac.AP=M.AB.AC, By a like process is obtained, in and AC = m. AB. AC (m+n+p) AP fig. 1, Bd = (m + n)AB. AC (m+n+p) AP° (m+n+p) PB; and, in fig. 2, ad = 4thly. By lines drawn from a given point P within the triangle. Const. Through P and c draw the line crp, and let the triangle be divided into the given ratio by lines pc, pd, drawn from p to intersect AC, BC, or either of them; according to the method described in case 3 of this problem. Through P draw PC, Pd, and respectively parallel to them, from p draw the lines pm, pN: join PM, PN; so shall these lines with PP, divide the triangle in the given ratio. Demon. The triangles CPM, CPP, are manifestly equal, as are also den, dpp; therefore CPM cpc, and CPN = cpd; whence also, in fig. 1, CNPM = cdpc, and, in fig. 2, CBPPN = cepd. = Comput. Since CF. CN cp. cd, we have CN = In like manner CM = cp.cd Remark. It will generally be best to contrive that the smallest share of the triangle shall be laid off nearest the vertex c of the triangle, in order to ensure the possibility of the construction. Even this precaution however may sometimes fail, of ensuring the construction by the method above given: when this happens, proceed thus: By case 1, draw the lines cd, ce, from the vertex c to the opposite side AB, to divide the triangle in the given ratio. Upon AB set off any where MN, so that MN: AB:: Pp (the perp. from P on AB): cp, the altitude of the triangle. If MP and PN are to Md Pee NE ger1 gether to be the least possible, then set off MN on each side the point p: so will the triangle MPN be isosceles, and its perimeter (with the given base and area) a minimum. 5thly. By lines, one of which is drawn from a given angle to a given point, which is also the point of concourse of the other two lines. Const. By case 1st draw the lines ca, cb, dividing the triangle in the given ratio, and so that the smaller portions shall lie nearest the angles A and B (unless the conditions of the division require it to be otherwise). From P and a demit upon AC the perpendiculars Pp, ac; and from P and b, on BC, the perpendiculars pq, bd. Make CM: CA :: ac : Pp, and CN CB bd: Pq. Draw PM, PN, which, with CP, will divide the triangle as required. When the perpendicular from b or from a, upon EC or AC, is longer than the corresponding perpendicular from P, the point N or M will fall further from c than B or A does. Suppose it to be N: then make N'e: eB:: Ne: ep, and draw PN' for the line of division. The demonstration of all this is too obvious to need traċing here. Comput. The perp. ca = aa . sin a; and CM = bd Bb. sin B; and CN= CA. ac Pp CB. bd 6thly. By lines, one of which falls from the given point of concourse of all three, upon a given side, in a given angle. Suppose the given angle to be a right angle, and of the given perpendicular: which will simplify the operation, though the principles of construction will be the same. MA M B Const. Let ca, cb, divide the triangle in the given ratio. Make fN: CB:: bd: pf, and fM: CA :: ac pf; and draw PN, PM, thus forming two triangles PfN, PM, equal to cbв, caA, respectively. If N fall between fand B, and м between A and ƒ, this construction manifestly effects the division. But if one of the points, suppose м, falls beyond the corresponding point A, the line PM intersecting intersecting AC in e: then make me : ea :: em : ep, and draw PM': so shall pf, PM', PN, divide the triangle as required. Comput. Here ca and bd are found as in case 5th; and hence ƒN = and f PM CB.bd CA. = ; and fM. Then PM = √(Mf2+fp2); Pf sin M. Also 180° — (M+A)=MEA. Then sin MCA : sin M: sin A∞ MA(=Mƒ—Aƒ); Ae: Me. Again Pe=PM-Me; Here also the demonstration is manifest. 7thly. By lines drawn from the angles to meet in a determinate point. D H G Construc. On one of the sides, as AC, set off AD, so that AD: AC :: m : m + n + p. And on any other, as AB, set off BE, so that BE: BC::N: m +n+p. Through D draw DG parallel to AB; and through E, EH parallel to BC; to their point of intersection I draw the lines AI, BI, CI, which will divide the triangle ABC into the portions required. Ι A E B Demon. Any triangle whose base is AB, and whose vertex falls in DG parallel to it, will manifestly be to ABC, as AD to AC, or as m to m+n+p: so also, any triangle whose base is BC, and whose vertex falls in EH parallel to it, will be to ABC, as BE to BA, that is, as ʼn to m + n + p. Thus we have AIB ACB:: m : m + n + p, therefore.. AIB BIC :: m : N. And the first two proportions give, by composition, ∞ MN: Comput. BE 2 right angles - ; angle BGI B. Hence, in the triangle BGI, there are known two sides and the included angle, to find the third side BI. E Remark. When m=n=p, the construction becomes simpler. Thus: from the vertex draw CD to bisect AB; and from в draw BE in like manner to the middle of AC: the point of inter- A section 1 of the lines CD, BE, will be the point sought. For, on BE and BE produced, demit, from the angles cand A, the perpendiculars CI, AK: then the triangles CEI, AEK, are equal in all respects, because AE = CE, KAE = ICE, and the the angles at E are equal. Hence AK = CI. But these are the perpendicular altitudes of the triangles BPC, BPA, which have the common base BP. Consequently those two triangles are equal in area. In a similar manner it may be proved, that APC = APB or CPB: therefore these three triangles are equal to each other, and the lines PA, PB, PC, trişect the A ABC. PROBLEM III. To Divide a Triangle into Four Parts, having the Proportion of the Quantities m, n, p, q. This, like the former problems, might be divided into several cases, the consideration of all which would draw us to a very great length, and which is in great measure unnecessary, because the method will in general be suggested immediately on contemplating the method of proceeding in the analogous case of the preceding problem. We shall therefore only take one case, namely, that in which the lines of division must all be drawn from a given point in one of the sides. Let p be the given point in the side AB. Let the points l, m, n, divide the base AB in the given proportion; so will the lines cl, cm, cn, divide the surface of the triangle in the same proportion. Join CP, and parallel to it draw, from l, m, n, the lines lL, mм, nn, to cut the other two sides of the triangle in L, M, N. Draw PL, PM, PN, which will divide the triangle as required. M J The demonstration is too obvious to need tracing throughout for the triangles LlP, L/C, having the same base Ll, and lying between the same two parallels Ll, CP, are equal; to each of these adding the triangle ALl, there results ALP = Acl. And in like manner the truth of the whole construction may be shown. The computation may be conducted after the manner of that in case 3d prob. 2. PROBLEM IV. To Divide a Quadrilateral into Two Parts having a Given Ratio, m: n. 1st. By a line drawn from any point in the perimeter of the figure. Construc. From P draw lines PA, PB, to the opposite angles A, B. Through n draw DF parallel to PA, to meet BA produced in F: and through c draw CE parallel to PB to meet AB produced in E. DPC RE Divide FE in M, in the given ratio of m to n: join F, M; so shall the line PM divide the quadrilateral as required. Demon. That the triangle FPE is equal to the quadrangle ABCD, may be shown by the same process as is used to demonstrate the construction of prob. 36, Geometry; of which, in fact, this is only a modification. And the line PM evidently divides FPE in the given ratio. But FPM = ADPM, and EPM= BCPM: therefore PM divides the quadrangle also in the given ratio. Remark 1. If the line PM cut either of the sides AD, BC, then its position must be changed by a process similar to that described in the 5th and 6th cases of the last problem. Remark 2. The quadrilateral may be divided into three, four, or more parts, by a similar method, being subject however to the restriction mentioned in the preceding remark. Remark 3. The same method may obviously be used when the given point P is in one of the angles of the figure. Comput. Suppose I to be the point of intersection of the sides DC and AB, produced; and let the part of the quadrilateral laid off towards 1, be to the other, as n to m. Then we n(ID. IA-IB.IC) As to the distances DI, AI, (since have IM (m + n) IP the angles at A and D, and consequently that at I, are known), they are easily found from the proportionality of the sides of triangles to the sines of their opposite angles. 2dly. By a line drawn parallel to a given line. Construc. Produce DC, AB, till they meet, as at I. Join DB, parallel to which draw CF. Divide AF in the given ratio in H. Through D draw DG parallel to the given line. Make IP a mean proportional between IH, IG; through P draw PM parallel to GD: so shall PM divide the quadrilateral ABCD as required. = case 1. Demon. It is evident, from the transformation of figures, so often resorted to in these problems, that the triangle ADF quadrilateral ABCD (th. 36 Geom.): and that DH divides. the triangle ADF in the given ratio, is evident from prob. 1 We have only then to demonstrate that the triangle IHD is equal to the triangle LPM, for in that case HDF will manifestly be equal to BCMP. Now, by construction, IH : IP:: IP: IG:: (by the parallels) IM: ID; whence, by making the products of the means and extremes equal, we have ID. IH = IP, IM; but when the products of the sides about the |