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.

(2) Const.

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 BŽ 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 diculars are proportional.

perpen

B. 5.

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.

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. 6.

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

A A

Fig. 32.

Fig. 33.

portional to the numbers 1, 2, 4 and 6, and let it be required to construct another upon a line equal to the greatest of these; if it be constructed, so that the adjacent angles may be equal to those adjacent to the equal side in the given figure, the quadrilaterals are equal; but if, so that they may be equal to the angles adjacent to sides denoted by 4 or 2 or 1, the figures are to each other as 36 to 16, or as 36 to 4, or as 36 to 1.

PROP. 22. B. VI.

In the demonstration of the second part of this proposition, Euclid assumes that the lines, upon which equal polygons similar and similarly posited are described, are equal; I have preferred demonstrating in B. 5. that the subduplicate ratios of equal ratios are equal, whence the demonstrations of the two parts are similar.

PROP. 23. B. VI.

This demonstration is taken from Candalla. Euclid assumes lines proportional to the sides of the parallelograms, as is done in Cor. 1. which does not seem necessary for the demonstration of the proposition.

PROP. 27. B. VI.

From this proposition is obtained the solution of a problem, which seems to be beyond the limits of elementary Geometry, namely, to find the greatest parallelogram which can be inscribed in a given triangle.

Since the defects of the parallelograms AI, AG, &c. are parallelograms similar to EF, they are about the same diagonal BZ, and are therefore inscribed in the given triangle AZB; of these the greatest is AG described upon half the base, whose sides therefore bisect the other sides AZ and BZ of the triangle.

PROP. 32. B. VI.

The words, and that the sides which are not homolo

gous should form the angle at which they are placed, are added to the enunciation of this proposition, which otherwise is not true; Candalla justly remarks, that the conclusion does not follow from the conditions in the Greek text, as is evident from fig. 33. in which the homologous sides AB and DC, BC and ED are parallel, but the other sides do not form one right line; for this it is necessary that the angles contained by the proportional sides should have a common alternate angle, in order that they may be proved equal. In fact, this proposition contains two conditions, for there are triangles which have two sides of the one proportional to two in the other, and yet which cannot be so placed as to have these sides parallel; that this may be done, the angles contained by them must be equal.

PROP. 33. B. VI.

Euclid demonstrates this proposition differently, from Fig. 34. his own definition of proportionals.-This method is liable to many difficulties, for the given angles may be right or obtuse, whose multiples are therefore not angles, according to Euclid's definition; and if in fig. 34. the angles ACD and ACB be given, one a right angle, the other half a right angle, they can be so multiplied (by 4) that the former becomes the right line CA, the latter the right line AG, and this is one cause for changing the 5. definition of Book V.