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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 fM. Then PM = √(Mf2+fp2);
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.
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.
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,
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.
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 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.
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.
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.
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.
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.
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 equal angles of two triangles having a common angle are equal, those triangles are equal; therefore ▲ IHD = ▲ IPM.
Q. E. D.
Comput. In the triangles ADI, ADG, are given all the angles, and the side AD; whence AI, AG, DI, and IC, = DI DC, become known. In the triangle IFC, all the angles and the side Ic are known; whence IF becomes known, as well as FH, since AH : HF :: mn. Lastly, IP = √(IH. IG), and IG:
ID: IP: IM.
Cor. 1. When the line of division PM is to be perpendicu lar to a side, or parallel to a given side; we have only to draw DG accordingly so that those two cases are included in this.
Cor. 2. When the line PM is to be the shortest possible, it must cut off an isosceles triangle towards the acutest angle; and in that case IG must evidently be equal to ID.
3dly. By a line drawn through a given point.
The method will be the same as that to case 4th prob. 1, and therefore need not be repeated here.
Scholium. If a quadrilateral were to be divided into four parts in a given proportion, m, n, p, q: we must first divide it into two parts having the ratio of m+n, to p + q; and then each of the quadrangles so formed into their respective ratios, of m to n, and p to q.
To Divide a Pentagon into Two Parts having a Given Ratio, from a Given Point in one of the Sides.
Reduce the pentagon to a triangle by prob. 37, Geometry, and divide this triangle in the given ratio by case 1 prob. 1.
To Divide any Polygon into Two Parts having a Given Ratio.
1st. From a given point in the perimeter of the polygon. Construc. Join any two opposite angles A, D, of the polygon, by the line
Reduce the part ABCD into an equivalent triangle NPs, whose vertex shall be the given point P, and base AD produced an operation which may be performed at once, if the portion ABCD be quadrangular; or by several operations (as from 8 sides to 6, from 6 to 4, &c,) if the sides be more than four. Divide the triangle NPS into two parts having the given ratio, by the line PH. In like manner, reduce
ADEFGA into an equivalent triangle having H for its vertex, and FE produced for its base; and divide this triangle into the given ratio by a line from н, as HK. The compound line PHK will manifestly divide the whole polygon into two parts having the given ratio. To reduce this to a right line, join PK, and through H draw Hм parallel to it; join PM; so will the right line PM divide the polygon as required, provided м fall between F and E. If it do not, the reduction may be completed by the process described in cases 5th and 6th prob. 2d. All this is too evident to need demonstration. Remark. There is a direct method of solving this blem, without subdividing the figure; but as it requires the computation of the area, it is not given here. 2dly. By the shortest line possible. Construc. From any point P', in one of those two sides of the polygon which, when produced, meet in the most acute angle 1, draw a line P'M', to the other of those sides (EF), dividing the polygon in the given ratio. Find
the points P and M, so that IP or IM shall be a mean proportional between IP', IM'; then will PM be the line of division required.
The demonstration of this is the same as has been already given, at case 5 prob. 1. Those, however, who wish for a proof, independent of the arithmetic of sines, will not be displeased to have the additional demonstration below.
The shortest line which, with two other lines given in position, includes a given area, will make equal angles with those two lines, or with the segments of them it cuts off from an . isosceles triangle.
Let the two triangles ABC, AEF, having the common angle A, be equal in surface, and let the former triangle be isosceles, or have AB = AC; then is BC shorter than EF.
First, the oblique base Er cannot pass through D, the middle point of BC, as in the annexed figure. For, drawing CG parallel to AB, to meet EF produced in. Then the two triangles DBE, DCG are identical, or mutually equal in all respects. Consequently the triangle DCF is less than DBE, and therefore ABC less than AEF.
EF must therefore cut BC in some point н between в and D, and cutting the perp. AD in some point I above D, as in the