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of the two unknown angles A and C: now IG being parallel to AC; AH:1H::CH : GH. (by cor. 1. theo. 20. sect. 4.) But the wholes are as their halves, that is, ÁH: IH:: CE:ED, that is as the sum of the two sides AB and BC, is to their difference; so is the tangent of half the sum of the two unknown angles A and C, to the tangent of half their difference. Q. E. D.

THEO. II.

Fig. 12

In any right-lined plane triangle ABD; the base AD will be to the sum of the other sides, AB, BD, as the difference of those sides is to the difference of the segments of the base, made by the perpendicular BE; viz. the difference between AE and ED.

Produce BD, till BG=AB the lesser leg; and on B as a centre, with the distance BG or BA, describe a circle AGHF; which will cut BD, and AD in the points H and F; then it is plain, that GD will be the sum, and HD the difference of the sides AB and BD; also since AE-EF (by theo. 8. sect. 4.) therefore FD is the difference of AE, ED, the segments of the base; but(by theo. 17. sect. 4.) AD: GD:: HD: FD; that is, the base is to the sum of the other sides, as the difference of those sides is to the difference of the segments of the base. Q. E. D.

Cor. 1. In the above triangle the longest side is made the base, and then the perpendicular falls within the triangle; but if DF (the same construction remaining as in the above, only joining BF, (Fig. 3. Pl. 14.) be considered the base of the triangle BDF, then BE is a perpendicular on the base produced; GD is equal to the sum of the sides BF, BD; HD is equal to their differ'.

ence; also AD is equal to the sum of the segments

DE, EF. But (by theo. 17. sect. 4.) FD X AD = GD x HD, hence FD: GD :: HD: AD. That is, as the base, is to the sum of the two sides; so is the difference of the sides, to the sum of the segments of the base, Q. E. D.

Cor. 2. Hence, (by calling any side the base) the base, is to the sum of the sides; so is the difference of the sides, to the difference or sum of the segments of base, according as the perpenelicular falls within or without the triangle.*

THEO. IV.

Fig. 13.

If to half the sum of two quantities, be added half their difference ; the sum will be the greatest of them ; and if from half the sum be subtracted half their difference ; the remainder will be the least of them.

Let the two quantities be represented by AB and BC: (making one continued line ;) whereof AB is the greatest, and BC the least; bisect the whole line AC in É; and make AD=BC; then it is plain, that AC is the sum, and DB the difference of the two quantities; and AE or EC, their half sum, and DE or EB their half difference. Now if to AE we add EB, we shall have AB the greatest quantity; and if from EC we take EB, , we shall have BC the least quantity. Q. E. D.

Cor. Hence, if from the greatest of two quantities, we take half the difference of them, the remainder will be half their sum; or if to half their

* The perpendicular falls within or without the triangle, according as the square of the greater side is less or greater than the sum of the squares of the lesser side and the base. For a demonstration of which the reader is refer red to (Prop. 12. 13. B. 2.) Simpson's Euclid.

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difference be added the least quantity, their sum will be half the sum of the two quantities.

THEO. V.

PL. 14. fig. 4.

In any triangle, the rectangle under two sides, is to the rectangle under the semiperimeter, and its excess above the base, as the square of the radius, to the square of the co-sine of half the contained angle.

In the triangle CBE, the perimeter being denoted by P, CB X CE: PCP-BE):: R’: cos. C? Produce EC to A, making CA = CB; draw BD perpendicular to CE, bisect CEin H,and join AB. Let CB be greater than EB, then, (by theo. 3.

CB 2.BE fig. 12.) CE: CB+BE :: CB-BE:

CE = 2. HD, by adding half this to half the base =

CB BE2+ CE2 CH. The segment CD= this adding CA, or CB, gives

AD= CBBE +CE?+2 CE. CB CB +CE- BE 2. CE

2 CE CB+CE+BEX CB+CE-BE

2 CE AC+CD=CB+CD; hence ADP = CB2 + 2 CB. CD+CD2 = 2 CB. AD; also, BD2 = CB2 CD?; hence AB2= ADP + BD2 =2. CBP + 2 CB. CD = 2 CBX CB + CD = 2 CB. AD; therefore

P(; P-BE) AD. ABP = 2 CB. AD, or

AB

СЕ = 2 CB. AD?, or P ( P--BE). ABP = CE. 2 CB. AD, dividing both sides by 2; CE, CB.

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2. CE

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Again, AD

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AD? = P (P--BE). AB?, consequently CE. CB: 1P (I.P--BE) :: AB?: AD2. That is, CE XCB:HP P-BE:: rad.? : (cos. BCE)?* Q. E. D.

OBLIQUE ANGLED TRIANGLES.

CASE I.

TWO sides, and an angle opposite to one of them given ; to find

the other angles and side.

/

PL. 5. Ag. 14.

In the triangle ABC, there is given AB 240, the angle A 46° 30',

and BC 200 ; to find the angle C, being acute, the angle B, and the side AC

1st. By Construction. Draw a blank line, on which set AB 240, from a scale of equal parts; at the point A, of the line AB, make an angle of 46° 30', by an indefinite blank line; with BC 200, from a like scale of equal parts that AB was taken, and one foot in B, describe the arc DC to cut the last blank line in the points D and C. Now if the angle Chad been required obtuse, lines from D to B, and to A, would constitute the triangle; but as it is required acute, draw the lines from C to B and to A, and the triangle ABC is constructed. From a line of chords let the angles B and C be measured; and AC from the same scale of equal parts that AB and BC were taken; and you will have the answers required.

* For a different method of demonstrating this Theorem, as well as the demonstration of other useful Theorens, the reader is referred to Leslie's Geometry (page 372 and 373).

2d. By Calculation.

This is performed by theo. 1. of this sect. thus;

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180°—the sum of the angles A and C, will give the angle B, by cor. 1. theo. 5. sect. 4.

A 46°. 30'
C 60. 31

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Extend from 200 to 240, on the line of numbers ; that distance will reach from 46° 30' on the line of sines, to 60° 31' for the angle C.

Extend from 46° 30', to 72° 59', on the line of sines; that distance will reach from 200 to 263.7 on the line of numbers, for AC.

NOTE. The method by Natural Sines will be obyious from the foregoing analogies,

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