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When A+B 180°, sin. (A + B) is negative, and therefore the fraction in which it occurs becomes positive.

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(505) Since similar triangles are as the squares of their homologous sides,


BDE: BFG :: BD2: BF2; whence BF = BD
3D (BDE).

1. BF × FG =†. BF × BF. tan. B;

(506) BFG=†. BF × FG

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(511) The final formula results from the proportion

FAE: CDE:: AE2 : ED2.

(512) Since triangles which have an angle in each equal, are as the products of the sides about the equal angles, we have

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Substituting these values in the preceding proportion, cancelling the common factors, observing that sin. (A + B)=sin. E, multiplying extremes and means, and

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(515) The first formula is a consequence of the expression for the area of a triangle, given in the first paragraph of the Note to Art. (65).

(517) The reasons for the operations in this article (which are of very frequent occurrence), are self-evident.

(518) The expression for DZ follows from Art. (65), Note. The proportion in the next paragraph exists because triangles having the same altitude are as their bases.

(519) By construction, GPC = the required content. Now, GPC GDC, since they have the same base and equal altitudes. We have now to prove that LMC=GDC. These two triangles have a common angle at C. Hence, they are to each other as the rectangles of the adjacent sides; e.,


Here CM is unknown, and must be eliminated. We obtain an expression for it by means of the similar triangles LCM and LEP, which give


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Substituting this value of CM in the first proportion,

and cancelling CD in the last two terms, we get


GDC: LMC:: GC: ; or GDC: LMC:: GC X LE : LC2.

LC2=(LH + HC)2= LH2 + 2 LH × HC + HC2.

But, by construction,



HKa— HE’— EK2 = HE’— EC2 — (HE+EC) (HE-EC)=HC (HE—EC).
GC2 HC; and LE LH + HE.

Substituting these values in the last proportion, it becomes


:: 2 LH + 2 HE

: HE-EC + 2 LH + HC.

: HE EC + 2 LH + HE + EC.

: 2 HE+ 2 LH.

The last two terms of this proportion are thus proved to be equal. Therefore, the first two terms are also equal; i. e., LMC=GDC= the required content.

Since HK √ (HE2 — EK2), it will have a negative as well as a positive value. It may therefore be set off in the contrary direction from L, i. e., to L'. The line drawn from L' through P, and meeting CB produced beyond B, will part off another triangle of the required content.

(520) Suppose the line LM drawn. Then, by Art. (65), Note, the required content, c. CLX CM. sin. LCM. This content will also equal the sum of the two triangles LCP and MCP; i. e., c = † • CL X p + † • CM × q. The first of

these equations gives CM= tion, we have


2 c
CL. sin, LCM®

Substituting this in the second equacq

c =†. CLXp + CL. sin. LCM

1 p. CL2. sin. LCM + cq = c. CL. sin. LCM.

Transposing and dividing by the coefficient of CL2, we get

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If the given point is outside of the lines CL and CM, conceive the desired line to be drawn from it, and another line to join the given point to the corner of the field. Then, as above, get expressions for the two triangles thus formed, and put their sum equal to the expression for the triangle which comprehends them both, and thence deduce the desired distance, nearly as above.

(522) The difference d, between the areas parted off by the guess line AB, and the required line CD, is equal to the difference between the triangles APC and BPD sin. A. sin. P sin. (A + P)

By Art. (65), Note, the triangle APC=}·AP2.

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By the expression for sin. (a + b) [Trigonometry, Art. (8)], we have


sin. A. sin. P
sin. A. cos. P+ sin. P. cos. A


sin. B. sin. P

sin. B. cos. P+ sin. P. cos. B

Dividing each fraction by its numerator, and remembering that


+ AP9

cot. P+cot. A cot. P+cot. B'

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For convenience, let p

cot. P; a cot. A; and b = cot. B. The above equation

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2 dp2 + 2 dap + 2 dbp + 2 dab=p. AP2+b. AP2 — p. BP2 — a. Transposing, dividing through by 2 d, and separating into factors, we get

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b. AP2
2 d

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If A = 90°, cot. Aa=0; and the expression reduces to the simpler form given in the article.

(523) Conceive a perpendicular, BF, to be let fall from B to the required line DE. Let B represent the angle DBE, and ẞ the unknown angle DBF. The angle BDF = —ẞ; 90° - ß; and the angle BEF= 90° — (B — ß) 90° — (B — ß) — 90° — B+ ß. By Art. sin. BDE. sin. BED

(65), Note, the area of the triangle DBE = DE2

sin. (90° — ß) sin. (90° — B+ ß)

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Hence, DE2:

sin. B

sin. (BDE+BED)

2 X DBE X sin. B
cos. ß. cos. (B—ß)'

2 X DBE X sin. B sin. (90°-6). sin. (90° - B+B) Now in order that DE may be the least possible, the denominator of the last fraction must be the greatest possible. It may be transformed, by the formula, cos. a. cos. b =† cos. (a + b) + ‡. cos. (a - b) [Trigonometry, Art. (8)], into † cos. B + † . cos. (B — 2 ß). Since B is constant, the value of this expression depends on its second term, and that will be the greatest possible when B-2 ẞß = 0, in which case ẞ = B.

It hence appears that the required line DE is perpendicular to the line, BF, which bisects the given angle B. This gives the direction in which DE is to be run. Its starting point, D or E, is found thus. The area of the triangle DBE=BD. BE. sin. B. Since the triangle is isosceles, this becomes

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DE is obtained from the expression for DE2, which becomes, making ß = †B,

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(524) Let a=value per acre of one portion of the land, and b that of the other portion. Let x = the width required, BC or AD. Then the value of xX BE BCFE a X 10

and the value of ADFE=bx


Putting the sum of these equal to the value required to be parted off, we obtain value required × 10

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ax BE+bx AE

(525) All the constructions of this article depend on the equivalency of triangles which have equal bases, and lie between parallels. The length of AD is derived from the area of a triangle being equal to its base by half its altitude. (527) Since similar triangles are to each other as the squares of their homologous sides,


ABC': DBE :: AB3: BD2; whence BD=AB ABC

The construction of Fig. 363 is founded on the proportion

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BF: BG :: BG: BA; when BD=BG=√✓ (BA × BF)=BA

m + n

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m + n m + n The similar triangles AEF and ABD give AD:DB:: AE: EF =

(528) By hypothesis, AEF: EFBC::m:n; whence AEF: ABC::m:m+n; AC X DB m and AEF-ABC. Also, AEF = } · AE × EF 2 DB X AE AD

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(530) In Fig. 366, the triangles ABD, DBC, having the same altitude, are to each other as their bases.

In the next paragraph, we have ABD: DBC::AD: DC :: m:n; whence AD: AC::m:m+n; and AC: DC ::m+n:n; whence the expressions for AD and DC.

In Fig. 367, the expression for AD is given by the proportion AD: AC::m:m+n. Similarly for DE, and EC.

(531) In Fig. 368, conceive the line EB to be drawn.

The triangle

AEB = ABC, having the same altitude and half the base; and AFD = AEB, because of the equivalency of the triangles EFD and EFB, which, with AEF, make up AFD and AEB.

The point F is fixed by the similar triangles ADB and AEF

The expression for AF, in the last paragraph, is given by the proportion,

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(532) The areas of triangles being equal to the product of their altitudes by half their bases, the constructions in Fig. 369 and Fig. 370 follow therefrom.

(533) In Fig. 371, conceive the line BL to be drawn. The triangle ABL will be a third of ABC, having the same altitude and one-third the base; and AED is equivalent to ABL, because ELB = ELD, and AEL is common to both. A similar proof applies to DCG.

(534) In Fig. 372, the four smaller triangles are mutually equivalent, because of their equal bases and altitudes, two pairs of them lying between parallels.

(535) In Fig. 373, conceive AE to be drawn. The triangle AEC —†. ABC, having the same altitude and half the base; and EDFC=AEC, because of the common part FEC and the equivalency of FED and FEA.

(536) In Fig. 374, in addition to the lines used in the problem, draw BF and DG. The triangle BFC = ABC, having the same altitude and half the base. Also, the triangle DFG = DFB, because of the parallels DF and BG. Adding DFC to each of these triangles, we have DCG BFC = ABC. We have then to † prove LMC — DCG. This is done precisely as in the demonstration of Art. (519), page 402.

(537) Let AE=x, ED=y, AH=x', HF=y', AK=a, KB=b.

The quadrilateral AFDE, equivalent to ABC, but which we will represent, generally, by m2, is made up of the triangle AFH and the trapezoid FHED.

AFH=1.x'y'. FHED=(x-x') (y+y').

.. AFDE=m2 = 4. x'y' + } (x − x') (y+y') = x (y + y) − x'y. The similar triangles, AHF and AKB, give

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Substituting this value of y' in the expression for m2, we have

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The formula is general, whatever may be the ratio of the area m2 to that of the triangle ABC.

(538) In Fig. 376, FD is a line of division, because BF the triangle BDF divided by half its altitude, which gives its base. So for the other triangles.

(539) In Fig. 377, DG is a second line of division, because, drawing BL, the triangle BLC=} ABC; and BDGC is equivalent to BLC, because of the common part BCLD, and the equivalency of the triangles DLG and DLB.

To prove that DF is a third line of division, join MD and MA. Then BMA BGA. From BMA take MFA and add its equivalent MFD, and we have MDFB == 1 BGA =† (ABDG (ABDG — BDG)=( ABC — BDG) = ABC — † BDG.

To MDFB add MDB, and add its equivalent, tion, and we have

BDG, to the other side of the equa

ABC — BDG +1 BDG; or, BDF


(540) In Fig. 378, the triangle AFC = } ABC, having the same base and onethird the altitude. The triangles AFB and BFC are equivalent to each other, each being composed of two triangles of equal bases and altitudes; and each is therefore one-third of ABC.

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