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the words Let and Then; and in the Demonstrations the successive steps will be numbered, and their object, in less obvious cases, pointed out. Some few propositions require no construction; and in others the construction is mixed up with the demonstration. Partly with a view to conciseness of expression, and partly as a means of rendering the connection, before each successive steps in the operation, more perspicuous and tangible, the following synıbols have been employed to mark the relation between the several magnitudes, which come under consideration in the

propositions of Euclid, and in those which have been added by way of illustration. The sign +, placed between two magnitudes, indicates

their addition. – , placed between t:vo magnitudes, shows the

remainder, when the latter is subtracted from the former.

denotes equality. > signifies greater; or, placed between two

magnitudes, shows that the former is

greater than the latter. < signifies less; or shows that the former of

two magnitudes is less than the latter.
Il signifies parallel, and om parallelogram.

perpendicular.
angle, and rt. Z right angle.
circle.
circumference.
triangle.

therefore. The square of any magnitude, as the line AB, is marked thus, AB”; and the abbreviations Def., Post., Ax., Prop., Cor., Hyp., Const. imply Definition, Postulate, Axiom, Proposition, Corollary, Hypothesis, and Construction. Also, rad. is for radius, diam. for diameter, st. for straight, cr. for centre, ptfor point, gn. for given, &c. &c. &c.

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PROPOSITION I. PROBLEM.

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GENERAL ENUNCIATION.To describe anequilateral triangle upon a given finite straight line.

PARTICULAR ENUNCIATION.—Let AB be the gn. st. line; then it is required to describe an equilateral a upon it.

CONSTRUCTION.- From the cr. A, with rad.

AB,

describe o BCD; from the cr. B, with rad. BA, describe O ACE (Post. 3); from the pt. c, where the o intersect, draw ca, CB to the pts. A and B (Post. 1); then ABC is an equilateral a.

DEMONSTRATION.-1. Because the pt. A is the cr. of O BCD, .. AC = AB (Def. 15).

2. Because the pt. B is the cr. of O ACE, .: BC = BA (Def. 15).

3. Hence AC and BC are each = to AB.

4. But things which = the same thing, are= to one another (Ax. 1); 1. AC = AB = BC: i. e. the A ABC is equilateral; and it has been described upon the gn. st. line AB.-Q. E. F.

It is clear that, upon the other side of AB, another equilateral A ABF, may in like manner be described; and if CF be joined, the As ACF, BCF are isosceles, having the base, or under side, longer than either of the two equal sides AC, AF, and BC, BF.

A more general proposition than the following will be seen under Prop. III. See also Prop. XXII.

This Problem is practically applied in Fortification.

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А

B

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PROP. A. PROB. Gen. Enun.-To describe an isosceles triangle upon a given finite straight line, which shall be less than either of the equal sides.

Part. Enun.-Let AB be the gn. st. line; then it is required to describe upon it an isosc. K A, of which the two = sides shall each be greater than AB.

CONST.–From crs. A and B, with radii AB and BA, describe OS BCF, ADG (Post. 3); produce AB both ways to c and D (Post. 2); from crs. A and B, with radii ad and BC, describe OS DEK, CEA (Post. 3); and from the pt. E, where these Os intersect, F draw EA, EB to pts. A and B (Post. 1); then ABE is the isosc. A required.

DEMONST.–1. To prove that AD = BC.

By property of the O, AC (AB =) BD (Def. 15). To each of these equals add AB; .. AD = BC (Ax. 2).

2. To prove that AE = BE. Again, by property of O, AE = AD, and BE (Def. 15); .. AE = BE (Ax. 1).

3. Hence the A ABE is isosceles, and it is described upon the gn. line AB, which is > AE or BE.-Q. E. F.

B в

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BC

PROP. II. PROB.

K

H

GEN. ENUN.-From a given point to draw a straight line equal to a given straight line.

PART ENUN.Let A be the gn. pt., Bc the gn. st. line; then it is required to draw from the pt. A, a st. line = BC.

Const.--From the pt. A to B, draw the st. line AB (Post. 1). Upon AB describe the equilateral A DAB. (Prop. I.) Produce

D

A

B

the st. lines DA, DB, to E and F (Post. 2). From cr. B, with rad. BC, describe the o CGH. From cr. D, with rad. DG, describe the O GKL (Post. 3). Then AL is the line required.

DEMONST.-1. By the property of the o, BC = BG (Def. 15). 2. To prove

that BG = AL. For same reason, DG = DL (Def. 15). Now, by the construction, DB = DA ;

.: by subtraction, BG = AL (Ax. 3).

3. .. also al = BC (Ax. 1).

Wherefore, from the gn. pt. A, a st. line AL has been drawn equal to the gn. st. line BC.-Q. E. F.

It should be continually borne in mind by the student, that the propositions of Euclid are of universal application; and that, although the figure may illustrate only a particular case, the demonstration is general. Thus, in the present problem, the gn. pt. might be placed either above or below the gn. line, or anywhere upon the line, whether at the extremity or otherwise ; or, again, beyond the line, so that AB, when joined, would be in the same st. line with Bc. When the pt. is placed at either extremity of the line, as at B, the problem is readily solved by describing one ogch only; and BG is the line required. The problem may also be solved by producing bd upwards, instead of downwards, through o to the Oce; by describing a O from the cr. 1), with the produced line as rad. ; and producing ad through d to the Oce. In this case, the required line is the sum of the rad. of the O, whose cr. is D, and the side of the A;—not their difference. Since then the gn. pt. may be joined to either extremity of the gn. line; since the

lateral A may be described on either side of the line so joined ; and since the problem may be solved in two ways with each A ; it is clear that there are 8 different lines which may be drawn to meet the required conditions. The student will find it a useful

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exercise to work the several cases, of which the figure, wherein the point is placed upon the gn. line, is here subjoined ; and the construction and demonstration will be found to be precisely the same, word for word, and letter for letter, as that which Euclid has given.

It may be supposed that this problem would be sufficiently resolved by ineasuring with the compasses the E distance BC,

and setting off the same distance from A. Practically indeed this may be the case, but the mathematical demonstration would be wanting ; and the connection between this proposition, and those which depend upon it, would not be accurately established. The same remark will apply to many propositions, of which the truth would readily be admitted without proof. A link in the scientific chain is thereby broken ; and the system is of course incomplete.

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PROP. III. PROB.

D

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B

= C.

F

GEN. ENUN.-From the greater of two given straight lines to cut off a part equal to the less.

PART. ENUN.- Let AB and c be the two gn. st. lines, of which AB is >c; then it is required to cut off from AB a part

Const.–From the pt. A, draw AD = to c (Prop. II.); and from the cr. A, with rad. AD, describe the O DEF (Post. 3); then AE is the part required.

DEMONST.–For by the property of the o, AE = AD (Def. 15); and by the construction AD = c; .'. also AE = c (Ax. 1). Hence

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