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

=

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

▲ 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
CD (m+n+p) AP

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
p. AB. BC

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