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Note. By the assistance of this proposition, we may divide a given ▲ into any number of As, which shall be to one another in a given ratio, by dividing the base of the given ▲ so that the parts will be to one another in that ratio; if then the extremities of those parts be connected with the opposite, the several ▲s on those parts will be as the parts (Prop. 1. 6).

PROP. XI. PROB.

To find a third proportional to two given right lines.

Draw two right lines making any with one another; on those lines from the vertex of this, assume portions respectively to the given antecedent and consequent ; connect their extremities; from the connected extremity of the part to the antecedent, assume in directum with it a part to the consequent; draw through the extremity of this a line par. to the connecting line, and it will cut from the other leg of the a segment (between the parallels) which shall be a third proportional to the given lines.

For the assumed antecedent: the consequent, assumed in directum with it, the first assumed consequent is to this intercept. (7. 6); .., &c.

Note. If a ratio of lesser inequality is given, it may be continued till we come to a magnitude greater than any assignable.

It is thus demonstrated. Assume on an indefinite right line, from one and the same point in it, parts continually proportional, which shall be the successive terms of the series (suppose them to be to 1, 3, 9, 27, &c.); then since the first three of those are in continued proportion, (viz. 1:3:3:9); .. convertendo, the first is to the dif. between the first and second, as the second is to the dif. between the second and third, i. e. 1:2::3:6; ·. permutando 1:3:: 2:6, i. e. the first term of the series is to the second as the dif. between the first and second is to the difference between the second and third; and, since the second term is greater than the first, it follows, that the difference between the second and third is greater than the difference between the first and second; since therefore magnitudes continually increasing are added

to the first, we will at length come to a magnitude greater than any assigned.

It is necessary to prove that the increments are continually increasing; for, if they were decreasing, the sum might never exceed a certain quantity.

Note 2. But, if the given ratio be of greater inequality, we may at length come to a magnitude less than any assignable.

It is thus demonstrated. Suppose the given terms of the series are assumed on a right line from one and the same point in it, and that they are 16:8; then assume any other right line, ever so small, suppose; then, as the second term of the given ratio is to the first, i. e. as 8: 16, so let this assumed line, be to another part assumed in the same right line, and from the same point in it, i. e. ; for :::8:16; let this series of be continued till a magnitude is found greater than (16) the first term of the given series (by preceding note), i. e. let it be continued through the terms 4, 1, 1, 2, 4, 8, 16, 32; then, let the given series of 16:8 be continued down through as many terms as there are up from to 32, i. e. let it be continued through the terms 16, 8, 4, 2, 1, 1, 4, 1, and let the least term be; this is less than the assigned quantity; and so on we may continue the series through ever so many terms, without ever coming to an end, and the sum of all those terms, though always coming nearer to the first (15), can never be equal to it, even though continued to ; and ... they can never exceed the assigned quantity, though they will approach nearer to it than by any assignable difference.

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Proof. Since there are two series of magnitudes continually proportional and equal in number, viz. 16, 8, 4, 2, 1,,,, and 32, 16, 8, 4, 2, 1, 1, 4; ex æquo, the first term is to the last in the first series, as the first is to the last in the second series, i. e. 16:¦::32 : ; but the first term of the first series is less than the first term of the second; .. the last term of the first series is less than that of the second; but this last term of the second series is presumed a very small quantity; yet we have found a quantity smaller than this; in like manner a quantity can be found less than any assumed quantity, be it ever so small.

Tacquet gives from Gregorius à S. Vincentio, a method

of finding a series of right lines in any given ratio of greater inequality, and of exhibiting the sum of the series continued through an infinite number of terms; assuming those premises proved in the last two notes.

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On any right line assume a part to the antecedent, or to the first term of the series; from the extremities of this assumed antecedent raise perpendiculars respectively

to it, and the consequent. Connect the extremities of those perpendiculars, and produce the connecting line to meet the line, on which the part is assumed. That latter line will be the base of a right angled ; assume on it from the point, in which the second perpendicular meets it, but remotely from the first perpendicular, another part

to the second perpendicular (or to the given consequent); and, from the extremity of this perpendicular, raise a perpendicular to meet the connecting line (or the hypotenuse); that perpendicular will be the third term of the series; and so on the series may be continued through any number of terms, by assuming on the base, from the extremity of the last perpendicular, a segment to it, and again raising from the extremity of this segment another perpendicular.

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PARTI. Those perpendiculars, or the segments into which the base is divided, are in continued proportion; for the whole base the difference between it and the first perpendicular, i. e. the second base :: the first perpendicular the second (by similar As), or as the first assumed part the second; and.. permutando, the whole base: the first assumed part:: as the second base: the second assumed part; and convertendo, the whole base the second base the second base the third base; but the first base the second base :: the first perpendicular : the second; and the second base; the third base :: the second perpendicular the third perpendicular; and thence the first perpendicular the second: the second the third. In like manner it can be shewn, that the other perpendiculars are proportional.

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PART II. Since the first perpendicular: the second:: the first base the second base; and as the first perpendicular is less than the first base, the second perpendicular will be less than the second base; in like manner the third, fourth, fifth, &c. perpendiculars will be respectively less than the third, fourth, fifth, &c. bases; but the first perpendicular is to the first assumed part, the second

perpendicular to the second assumed part, and the third to the third, &c.; therefore, the whole series of propor tionals is not greater than the first base; and since the bases themselves are continually proportional, the last term may be less than any assignable magnitude; and therefore the sum of all the assumed parts, or of the entire series of proportionals, if the number of terms be infinite, is not less than the entire base, as the diff. is unassignable.

The difference between the first and second terms, the first term, and the sum of the entire series, are continually proportional.

For this difference : to a line to the first term :: the first perpendicular: the sum; but the first perpendicular is also to the first term; .., &c.

PROP. XII. PROB.

To find a fourth proportional to three given right lines.

Form a rectilineal Z, and on one of its legs from the vertex, assume a part to the first antecedent; and, in continuum with this, assume another part to the given consequent; on the other leg assume from the vertex a part to the second antecedent, and connect its extremity with the extremity of the first assumed antecedent; from the extremity of the assumed consequent draw a line parallel to this connecting line; the intercept of the other leg (between those parallels,) shall be the fourth proportional.

For (by Prop. 2. 6.) the first assumed antecedent: assumed consequent :: second assumed antecedent : intercept;., &c.

PROP. XIII. PROB.

To find a mean proportional between two given right lines. Draw any right line, and on it take, one after another, parts respectively to the given extremes. Bisect the sum of those parts; and, with the point of bisection as centre, and half sum as radius, describe on the sum a semicircle; from the common extremity of the parts erect a perpendicular to meet the circumference; it will be the mean proportional required.

For, draw chords from the extremity of this perpendicular at the circumference, to the extremities of the diameter; the , contained by those chords, will be a right (Prop. 31, 3); and the perpendicular, from this right on the hypotenuse, is a mean proportional between the segments of the bypotenuse (Cor. 8, 6.).

Cor. 1. In like manner mean proportionals can be found between the given lines and this mean; from whence a series of five right lines continually proportional would arise; and again, mean proportionals being found between the adjacent terms in this last series, a series of nine proportionals arises, and so on. But the number of proportions in any series will be always one less than double the number of terms in the preceding series.

Cor. 2. Given, of three proportionals, the sum of the extremes, and the mean, the extremes themselves can be found.

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On the given sum describe a semicircle; from either extremity of the diameter raise a perpendicular to the given mean; through the extremity of this draw a line parallel to the diameter to meet the circumference; and from the point, in'which it meets the circumference, let fall a perpendicular on the given sum; this perpendicular will divide the given sum into the required segments. For, it is a mean proportional between the segments (by prop.) and to the perpendicular raised from the extremity, i. e. to the given mean.

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NOTE. This proposition, with the 14th of second, proves, that the of the mean is to the rectangle under the extremes. Thus, if we are given two numbers, suppose 4 and 9, we can find a mean proportional between them, by extracting the√ of their product, i. e. ✓ 36 = 6; then 6 is a mean proportional between 4 and 9. NOTE 2. In Cor. 2, the par. to the diameter will meet the periphery; for it is evident, that the given mean cannot be greater than half the sum of the extremes.

NOTE 3. If we have the sum of two right lines and the rectangle under them, we can find the lines; for the of the rectang., or the side of a 2 the rectangle, is = the mean proportional between them. Thus, in numbers; let 30 be the sum of the numbers and 9 the mean; then the

of half the sum - the of the mean the 2 of in

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