Fig. 25.

PROP. 6. B. VI.

If this proposition, and also in the preceding, the construction, which in Euclid is definite, is expressed indefinitely lest any one should imagine there was but one side of the triangle, on which the construction could be made. Euclid says, construct at the side DF, &c. as if it could not be constructed at the other side.

PROP. 11. B. VI.

If the given be a ratio of less inequality LO to LR, the series can be continued until a magnitude be found greater than any assigned.

For let LO, LR, LQ and LI be continually propor-. tional, since LO is to LR as LR to LQ, by conversion, LO is to OR, as LR to RQ; but LR is greater (1)Prop.23. than LO, therefore RQ is greater than OR (1): in the same manner it can be shewn that IQ is greater than RQ; since therefore quantities continually encreasing are added to the first, a magnitude can be attained greater than any assigned.

B. 5.

Fig. 26.

If the given be a ratio of greater inequality, AB to CB, the series can be continued till a magnitude be found less than any assigned.

Let the assigned quantity be OL, as CB is to AB so let OL be to LR, and continue the series till IL be found greater than AB, continue the ratio of AB to CB through as many terms, and let the last be FB, FB is less than OL.

For since there are two series of magnitudes proportional and equal in number, ex æquo AB is to FB as (1) Const. IL to OL, but AB is less than IL (1), therefore FB is (2) Prop.23. less than ÓL (2), and in the same manner a magnitude can be found less than any other given one.

B. 5.

Tacquet has given from Gregorius a S. Vincentio, the following method of finding a series of lines in any given ratio of greater inequality, and of exhibiting the

sum of the series continued through an infinite number of terms.

Take in any right line AZ, parts AB and BC equal Fig. 27. to the terms of the given ratio, draw AL and BO perpendicular to AZ and equal to AB and BC, draw LO meeting AZ in Z, through C draw CQ perpendicular to AZ, and CQ is the third proportional to AB and BC; take CE equal to CQ, the perpendicular ER is the fourth proportional, and so on; and AZ is equal to the sum of the series if it be continued through® an infinite number of terms.

B. 6.

Part 1. AZ is to BZ as AL to BO (1), or as AB to (1) Schol. BC (2), by alternation AZ is to AB as BZ to BC, by Prop. 4. conversion AZ is to BZ as BZ to CZ, but AZ is to BZ (2) Const. as AL to BO (1), and BZ is to CZ as BO to CQ, (1), therefore AL is to BO as BO to CQ (3), and in the (3) Prop.18. same manner it can be proved that the other perpen- B. 5. diculars are proportional.

B. 5.

Part 2. Since AL is to BO as AZ to BZ, and AL is less than AZ, BO is also less than BZ (4), likewise (4) Prop.32. CQ, ER, FS, &c. are less than CZ, EZ, FZ, &c. but AL is equal to AB, BO to BC, CQ to CD, &c. therefore the whole series of proportionals is not greater than AZ. And since AZ, BZ, CZ, &c. are continually proportional, the last term must be less than any assigned magnitude, and therefore the sum of AB, BC, CE, &c. or the whole series of proportionals, if the number of terms be infinite, is not less than AZ.

B. 6.

The difference between the first and second terms, Fig. 27. the first term, and the sum of the series, are continually (1) Cor. 1. proportional. For LX is to XO as LA to AZ (1), but Prop. 4, LX is the difference between LA and OB (2), and XO (2) Const. & is equal to AB, and therefore equal to AL (2), and AZ Prop. 34. is the sum of the series.

PROP. 13. B. VI.

Plato, Philo of Byzantium, and Des Cartes, invented the following methods of finding two mean proportionals, which though not strictly geometrical, are yet deserving of notice.

B. 1.

Fig. 28.

(1) Cor.

Prop. 8.
B. 6.

Fig. 29.

(1) Const.

(2) Cor.
Prop. 36.

B. 3.
(3) Prop.14.

B. 6.

(4) Cor. Prop. 4. B. 6.

(5)Prop. 18.

B. 5.

Plato's method.

Insert in a side of a square a ruler moveable at pleasure along that side and always perpendicular to it; place the given right lines AB and CB at a right angle, and produce them to Z and X, then apply the angle of the square to BX, so that when its side passes through A, and the ruler through C, the vertex of the angle between the ruler and the side of the square in which it is inserted, may be somewhere in the line BZ, then BD and BE are the two required mean proportionals. For in the right angled triangle ADE, DB is drawn from the right angle perpendicular to the opposite side, therefore AB is to DB as DB to BE (1); and for the same reason DB is to BE, as BE to BC.

Philo's method.

Place the given right lines AB and BC at a right angle, and complete the rectangle BADC, circumscribe a circle about it, and produce DC and DA, then apply a ruler at the point B, so that the parts of it FO and BG shall be equal, then AF and CG are the two means sought.

Since GB is equal to FO (1), GO is equal to BF, and therefore the rectangle under OG and GB is equal to the rectangle under BF and FO, but the rectangle under OG and GB is equal to the rectangle under DG and GC (2), and the rectangle under BF and FO is equal to the rectangle under DF and FA (2), therefore the rectangle under DG and GC is equal to the rectangle under DF and FA, and therefore DG is to DF FA to GC (3); but on account of the parallel lines GD and BA, GD is to DF as AB to AF (4), and on account of the parallels DF and CB, GD is to DF as GC to CB (4), and therefore BA is to AF as AF to GC, and AF to GC as GC to CB (5).

as

Des Cartes's method.

Let there be two rulers AZ and AX moveable on a pivot at the point A, and in these let there be alternately inserted perpendicular rulers BC, CD, DE,

&c. so that in opening the first rulers they shall push each other forward; let the given lines be AB and ̃AE, apply the first perpendicular BC at the point B, and open the rulers until the third perpendicular DE pass through the point E, AC and AD are the two means sought.

Since in the triangle ACD, the angle ACD is right, and CB is a perpendicular from it upon the opposite side, AB is to AC as AC to AD (1); likewise in the (7) Cor. right-angled triangle ADE, AC is to AD as AD to Prop. 8. AE; therefore AC and AD are mean proportionals between the given lines AB and AE.

B. 6.

If four mean proportionals are to be found, open the Fig. 31. rulers till FG the fifth perpendicular pass through E, and so on if there be six or eight, &c.

But if the number of means required be odd, the given lines AB and AE must be applied to the same ruler as is evident from fig. 31. where between AB and AE, there are three mean proportionals, AC, AD and AF.

COR. 2. PROP. 16. B. VI.

The second and third corollaries are cases of the same general theorem, that if from the angle of any triangle inscribed in a circle two lines be drawn, one to the opposite side, the other cutting off from the circle a segment containing an angle equal to that made by the first drawn line and the side of the triangle which it meets, the rectangle under the sides of the triangle is equal to the rectangle under these lines.-In Cor. 2. these lines coincide, as is evident.

COR. 1. PROP. 20. B. VI.

That the two polygons should be in the duplicate ratio of the first given line to the second, these lines must be between the equal angles, and the polygons on them be similarly described. Upon a given line as many polygons can be constructed of different magnitudes and yet similar to a given polygon, as there are sides of different lengths in the given figure: let the given figure be a quadrilateral, whose sides are pro

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