Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

Section II. Containing several Propositions demonstrated from the foregoing Foundations.

CHAP. I. Of the Composition and Division of Quantities.

Proposition I.

THE Sum and Difference of two unequal Quantities ad∣ded together, make double of the greatest.

Demonstration.

Suppose a be the greatest, b the least, then will their Sum be

	a+b
And their Difference	a−b
Their Sum	2a,

by Consectary 1. Definition 27, Q. E. D.

CONSECTARY.

HEnce by a bare Subsumption(α) 1.1 you have the truth of Consect. 1. Definit. 8. that 2 unequal contiguous Angles on the same Right Line, ACD and ACE (Fig. 49.(β) 1.2) i. e. if we call the Right Angle BCD or BCE (a) and the dif∣ference between the one and the other (b) a+b and a−b, make 2a, i. e. are equal to 2 right ones.(α) 1.3

Proposition II.

IF the Difference of two unequal Quantities be subtracted from their Sum, the Remainder will be double of the least.

Demonstration.

If from the Aggregate or Sum	a+b
You subtract the Difference	a−b
The Remainder will be	0+2b

by Con∣sectary 2. Definition 27. Q. E. D.

Proposition III.

BUt if the Sum or Aggregate be subtracteed from the Dif∣ference, the remainder is so much less than nothing, as is the double of the last Quantity.

Demonstration.

For if from the Difference	a−b
You subtract the Sum or Aggregate	a+b
The Remainder will be —	0−2b

by Con∣sectary 3 of the aforesaid Definition. Q. E. D.

Proposition IV.

IF a Positive Quantity be multiplied by a Negative one, or contrariwise, the Product will be a Negative Quantity.

Exposition.

If a−b is to be multiplied by c; it is certain, that a multiplied by c, makes ac a Positive Quantity, by Consect. 1. Definit. 28. Moreover b by the same c (a Negative by a Positive) will make −bc; and so the whole Product of a−b by +c, will be ac−bc.

Demonstration.

Suppose a−b=e; therefore ec will be = to the Product of a−b by c: and since a−b is = e by the Hypoth. adding on both sides h, you'l have a=e+b. by Schol. Definit. 26. and multiplying both sides by c, ac=ec+bc, by Consect. 2. Definit. 28. and by further subtracting from each side bc, you'l have ac−bc=ec, that is, to the Product of a−b by c. Q. E D.

CONSECTARY.

SInce ac−bc is the Product of a−b by c, it is manifest also, that if ac−bc be divided by c, you'l have a−b for the Quotient; and so always a Positive Quantity (as ac) di∣vided by a Positive one, c, will give a Positive Quotient; but a Negative Quantity −bc divided by a Positive one, will give a Negative Quotient.

Proposition V.

IF a Negative Quantity be multiplied by a Negative one, the Product will be Positive.

Exposition.

Suppose a−b is to be multiplied by −c; it is certain, that 〈◊〉〈◊〉 multiplied by −c will give the Negative Quantity −ac, by Prop. 4. but −b multiplied by the same −c will produce +bc, and so the whole Product will be −ac+bc.

A Demonstration like the former.

Suppose a−b=e, then will −ec = the Product of a−b by −c: and since a−b is = e, adding b on both sides you'l ••ave a=e+b, by Schol. Definit. 26. and multiplying both sides by −c, you'l have ac=−ec−bc, by Prop. 4. and Consect. ••••. Definit. 28. and by adding bc on both sides, you'l have

−ac+bc=−ec, i. e. to the Product of a−b by −c Q. E. D.

CONSECTARYS.

I. SInce therefore a−c+bc arises from a−b by −c, it is manifest, that if −ac+bc be divided again by −c, you will again have a−b, and consequently a Negative Quantity divided by Negative, will give a Positive Quotient but a Positive Quantity +bc divided by a Negative one, wil•• give a negative Quotient −b.

II. We have therefore the Foundation and Demonstration o•• the Rules of Specious Computation, in the multiplication an•• division of Compounded Quantities, viz. that the same Sig•• multiplied together (as + by + or − by −) give + but different (as + by − or − by +) give the Sign − Which Rules the following Examples will Illustrate, as als•• several other we shall meet with in the following Chapter.

Multiplication.

〈 math 〉〈 math 〉

Division.

〈 math 〉〈 math 〉

CAAP. II. Of the Powers of QUANTITIES.

Containing (after a compendious Way) most part of the 2d Book of Euclid; and the Appendix of Clavius to Lib. 9. Prop. 14.

Proposition VI.

IF any whole Quantity be divided into two parts(α) 1.4 the Rectangle contained under the whole, and one of its parts, is equal to the Square of the same part, and the Rectangle contain'd under both the parts.

Demonstration.

Let	a+b	represent the whole	a+b
	b	one part of it, or	a	the other.
	ab+bb	the Rectangle,	aa+ab	the Rectangle.

(See Fig. 50.) Q. E. D.

Proposition VII.

IF a whole Quantity be divided into two parts(β) 1.5 the Square of the whole is equal to the Squares of both those parts and 2 Rectangles contained under them.

Demonstration.

This is evident from the preceding, and may moreover thus appear further.

Let the Parts be a and b, then will the whole be 〈 math 〉〈 math 〉

Which if you multiply by it self 〈 math 〉〈 math 〉

You have the Square 〈 math 〉〈 math 〉

(See Fig. 51. N. 1.) Q.E.D.

CONSECTARYS.

I. HEnce you have the Original Rule for Extracting of Square Roots, as we have shewn after Definition 30. and here have further Illustrated in Scheme N^o 2.

II. Hence it naturally follows, that the Square of double any Side is Quadruple of the Square of that Side taken singly.

III. Hence also you have the addition of surd Numbers, or in general of surd Quantities, by help of the following Rule (supposing in the mean while their Multiplication:) Suppose these 2 Surds √8 and √18, or more generally √75aa and √27aa, are to be added together; first add their Squares 8 and 18, &c. then double their Rectangle (√144) that is, multiply it by the √4, and then the double of this √576, i. e. having ex∣tracted the Square Root, (24) and added it to the Sum of the first Squares (26) the Root of the whole Summ (50) viz. √50, is the Sum of the two surd Quantities first proposed.

SCHOLIUM.

BUT if it happens that the Root of the double Product can∣not be expressed by a Rational Number (as, when the proposed Quantities are Surds, as √3 and √7, to whose Squares 3+7, i. e. 10, you must add the double Product of √7 by √3, i. e. √84, which cannot be expressed by a Rational Num∣ber) then that double Product must be joined under a Surd Form, or Radical Sign, to the Sum of the Squares (thus, viz. 10+√84) and to this whole Aggregate prefix another Radi∣cal Sign, thus, 〈 math 〉〈 math 〉; or also you may only simply join the Surd Quantities proposed by the Sign + thus, √3+√7. Here also you may note, that the two Surd Quantities proposed

in the first case of Consectary 2. are called Communicants; in the other case of this Scholium, Non-Communicants: For in this case each quantity under the Radical Sign may be divided by some Square, and have the same Quotient (e. g. 8 and 18, may be divided the first by 4, the other by 9, and the Quotient of both will be 2; likewise 75aa and 27aa may be divided, the one by 25aa, the other by 9aa, the Quotient of both being 3; and then if the Quotient on both sides be left under the Radical Sign, and the Root of the dividing Square set before it, the same quantities will be rightly expressed under this form: 2√2 and 3√2, also 5a√3 and 3a√3; and then the addition is easie, viz. only collecting or adding together the Quantities prefixt to the Radical Sign; so that the Sums will be of the one 5√2 and of the other 8a√3, which are indeed the same we have shewn in Consect. 2. For if contrarywise we square the Quantities that stand without, or are prefixt to the Radical Sign, and then set those Squares (25 and 64aa) under the Ra∣dical Sign, multiplying by the Number prefixt to it, you'l have for the one √50, for the other √192aa (Consect. after Schol-Prop. 22.)

Proposition VIII.

IF any whole Quantity (viz. Line or Number) be divided(α) 1.6 into two equal parts, and two unequal ones, the Rectangle of the unequal ones, toge∣ther with the Square of (the intermediate part or) the difference of the equal part from the unequal one, is equal the Square of the half.

An Universal Demonstration.

Suppose the parts to be a and a, and the whole 2a; let one of the unequal Parts be b, the other will be 2a−b, and the dif∣ference between the equal and unequal part a−b.

The equal ones 〈 math 〉〈 math 〉

Rectangle 〈 math 〉〈 math 〉

Difference 〈 math 〉〈 math 〉

The Sum will be aa (the other parts destroying one another) Q.E.D. (Vid. Fig. 52.)

Proposition IX.

IF to any whole Quantity divided into two equal parts(α) 1.7 you add another Quantity of the same kind, the Rect∣angle or Product made of the whole and the part added, mul∣tiplied by that part added, together with th•• square of the half, will be equal to the Square o•• the Quantity compounded of that half, and th•• part added.

Demonstration.

Let the whole be called 2 a, the part added b, then th•• quantity compounded of the whole and the part added will b•• 2a+b; and that compounded of the half and the part added a+b.

The Quantity compounded of the whole, and the part added is, 〈 math 〉〈 math 〉 the half a Comp. 〈 math 〉〈 math 〉

Multip. by the part added 〈 math 〉〈 math 〉

2ab+bb□ of the half aa=□aa+2ab+bb

(Vid. Fig. 53.) Q. E. D.

Proposition X.

IF a Quantity be divided any how into(b) 1.8 two parts, the Square of the whole, together with that of one of its parts, is equal to two Rectangles contained under the whole and the first part, together with the Square of the other part.

[illustration]

Pag. 63.

54

55

56

57

58

59

60

The Universal Demonstration.

Let a be one part and b the other, the whole a+b

a+b	the whole.	The whole	a+b
a	the first part *		a+b
aa+ab			aa+ab
2			ab+bb
2aa+2ab	the double rectangle	□ of the whole	aa+2ab+bb
add bb	the □ of the other part	* add	aa
Sum 2aa+2ab+bb	= to the Sum	....	2aa+2ab+bb

(Vid. Fig. 54. N^o 1.) Q.E.D.

CONSECTARY.

HEnce you have the Subtraction of Surd Numbers, or more generally of Surd Quantities, by help of the following Rule.

Add the Squares of the given Roots according to Consect. 3. Prop. 7. and from their Sum subtract the double Rectangle of their Roots; the Root of the Remainder will be the difference sought of the given Quantities.

As, if the √8(BC) is to be subtracted from √50AC (Fig. 54, N^o 1.) you must add 50, i. e. the whole Square AD) and 8, (i. e. the other Square superadded DE,) and the Sum will be 58, equal to the two Rectangles AF and FE+□GH, by ••his Prop. I find therefore those two Rectangles by multiplying √50 by √8, and then the Product √400 by 2 or √4, there∣by to obtain the double Rectangle √1600, i. e. (having actu∣ally Extracted the Root) 40. This double Rectangle therefore ••or 40, being subtracted out of the superiour Sum, the remain∣der 18 will be the □ GH, and so its Root (viz. √18) gives ••he required Difference between the given Surd Quantities.

SCHOLION.

BUT this Subtraction may be performed yet a shorter way, if each quantity under its Radical can be divided by some

square, so that the same Quotient may come out on both sides that is, if the Surd Quantities are Communicants, as e. g. √5•• (the number 50 being divided by 25) is equal to 5√2 a•• √8 to 2√2; for then the numbers prefixt to the Radical Sig•• being subtracted from one another (viz. 2√2 from 5√2) yo•• have immediately the remainder or difference 3√2, i. e. √1•• But if the proposed Quantities are not Communicants (as if th•• √3 is to be subtracted from √7) the remainder may be brief•• expressed by means of the Sign − thus, 〈 math 〉〈 math 〉, or accordin•• to the foregoing Consectary, thus, 〈 math 〉〈 math 〉.

Proposition XI.

IF any Quantity be divided into two parts,(α) 1.9 the Quadru•• Rectangle contained under the whole and one of its parts, toget•••••• with the Square of its other parts, will be equal to the Square of 〈◊〉〈◊〉 Quantity compounded of the whole and the other part.

Demonstration.

Suppose	a+b	the whole.
	b	one part.
	ab+bb	the Rectangle of these two.
mult. by	4
	4ab+4bb	the Quadruple Rectangle.
Add	aa	the Square of the other part.
Sum	aa+4ab+bb

The Quantity compounded of the whole and the first part 〈 math 〉〈 math 〉

Square of the Compound Quantity Q. E ••

(Vid. Fig. 55.)

Proposition XII.

IF any Quantity be divided into two equal parts(β) 1.10 and into two other unequal ones, the Squares of the unequal parts taken together will be double the Square of half the quantity, and the Square of the difference, viz. of the equal and unequal part, taken to∣gether.

Demonstration.

Suppose the equal parts to be a and a, the difference (b) the greater of the unequal Parts to be a+b, the less a−b.

The greater part 〈 math 〉〈 math 〉	The less 〈 math 〉〈 math 〉	Half 〈 math 〉〈 math 〉	Difference 〈 math 〉〈 math 〉
Sum of these 2aa+2abb	Sum aa+bb

Q. E. D. (Vid. Fig. 56.)

Proposition XIII.

IF to any whole Quantity(α) 1.11 divided into two equal parts there be added another Quantity of the same kind, the Square of the Quantity compounded of the whole, and the quantity added, together with the square of the quantity added, will be double the square of the half the quantity, and the square of the Sum of the half and the part added taken together.

Demonstration.

Suppose the whole to be 2 a, the half parts a and a, the quantity added b; then the quantity compounded of the whole and the quantity added, will be 2a+b, and that of the half and the quantity added a+b.

Comp. of the whole 〈 math 〉〈 math 〉 and quantity added

Sum

Half 〈 math 〉〈 math 〉

Qu. compouded of hal•• 〈 math 〉〈 math 〉 and qu. added,

Sum 2aa+2ab+•••• Manifestly the half of the for¦mer Sum. Q.E.D.

CHAP. III. Of Progression, or Arithmetical Proportionals.

Proposition XIV.

IF there are 3 Quantities in continued Progression, or •• Arithmetical continued Proportion, the Sum of the Extrem•• is double of the middle Term.

Demonstration.

Such are e. g. a, a+x, a+2x ascending, or a, a−x, a−2x descending.

By Definition 32. the Sum of the Extremes in the first 2a+2x, in the latter 2a−2x; in both manifestly double the middle Term Q.E.D.

Proposition XV.

IF there are 4 of these Continued Proportionals, the Sum •• the Extreme Terms is equal to the Sum of the me•••• Terms.

Demonstration.

Such are e. g. a, a+x, a+2x, a+3x, &c. ascending, or a, a−x, a−2x, a−3x, &c descending; in the one the Sum of the Extremes is 2a+3x, in the other 2a−3x; and also of the means 2a+3x and 2a−3x Q. E D.

Proposition XVI.

IF there are never so many of these continued Proportionals, the Sum of the Extremes is always equal to the Sum of any 2 other of them, equally remote from the Ex••remes, or also double of the middle Term, if the number of the Terms is odd.

Demonstration.

Suppose a, a+x, a+2x, a+3x, a+4x, a+5x, a+6x, &c. or a, a−x, a−2x, a−3x, a−4x, a−5x, a−6x, and the Sum of the Extremes, as also of any 2 equally remote from the Extremes, and the double of the middle Term is in the first Series 2a+6x, in the latter 2a−6x, &c. Q.E.D.

SCHOLIUM I.

NOR can we doubt but that this will always be so, how far soever the Progression be continued; if you consider that the last Term contains in it self the first, and moreover the difference so many times taken, as is the number of Terms ex∣cepting one, but that the first has no difference added to it; and therefore tho the last since one contains one difference less than the last; the second on the contrary has one more than the first, and consequently the Sum of the one will necessarily be equal to the Sum of the other; and in like manner the last except two, contains two Differences less than the last; but, on the contrary, the third exceeds the first by a double Diffe∣rence, the double Difference being added to it, &c. as is ob∣vious to the Eye in our first universal Example. Hence you have these

CONSECTARYS.

I. YOU may obtain the Sum of any Terms in Arithmetical Proportion, if the Sum of the Extremes be multiplied by half the number of Terms, or (which is the same thing) half the Sum by the number of Terms,

II. To obtain therefore the Sum of 600, or never so many such Terms, you need only have the Extremes and the number of Terms: So that you have a very compendious Way of pro∣ceeding in Questions that are solvible by these Progressions, if, having the first Term and Difference of the Progression given, you can obtain the last, neglecting the intermediate ones.

III. But you may obtain the last Term, by Multiplying the given given Difference by the given Number of Terms lessened by Unity, and then adding the first Term to the Product; as i•• evident from the preceding Scholium.

IV. Hence we may easily deduce this Theorem, that the Sum of any Arithmetical Progression beginning from o, is subduple of the Sum of so many Terms, equal to the greatest, as is the number of Terms of that Progression. For if the first Term is o and the last x, and the given number of Terms a, the Sum of the Progression will be ½ ax, by Consect. 1. but the Sum of so many Terms equal to the greatest, ax. Q.E.D.

SCHOLIUM II.

NOw if any one would be satisfied of the truth of this last Consect. without the literal or specious Notes, let him consider, that if the first Term be supposed to be o, the last (whatever it is) will be the sum of the Extremes. The last therefore mul∣tiplied by half the number of Terms, gives the Sum of the Pro∣gression, by Consect. 1. and the same last Term multiplied by the whole number of Terms, gives the Sum of so many Terms equal to the greatest. But that this must needs be double of the precedent 'tis evident, because any Multiplicand being multiplied by a double multiplier, must needs give a double Product. Now as this Consectary will be of singular Use to us hereafter for De∣monstrating several Propositions, so the three former are the very same Practical Rules of Arithmetick, which are commonly made

use of in Arithmetical Progressions; for the Illustration whereof Swenterus gives us several Ingenious Examples in his Delic. part 1. Quest. 70. &c.

CHAP. IV. Of Geometrical Proportion in General.

Proposition XVII.

IF there are three Quantities continually (α) Proportional, the Rectangle of the Extremes, is equal to the Square of the mean Term.

Demonstration.

Such are e. g. a, ea, e{powerof2}a,	The mean Term, 〈 math 〉〈 math 〉
The Extremes
Rectangle	Square Q.E.D.

SCHOLIUM.

MOreover if three Quantities on each side are in the same Continual Proportion, as

[illustration]

the Rectangles of the Extremes made Cross-ways, are equal to the Rectangle of the mean Term; being every way e{powerof2}ab.

Whence by the way may appear that Proposition of Archimedes(α) 1.12 (β) 1.13 That the Surface of a Right Cone is equal to the Circle, whose Radius is a mean Proportional between the Side of that Cone and the Semidiameter of the Base. For suppose EF to be a mean Pro∣portional between the side of the Cone BC (Fig. 57.) and the Semidiameter of the Base CD, since an equal number of Peripherys answer to an equal number of Radii in the same Propor∣tion; half the Product of the first Line BC into the last Periphery, ½ e{powerof2}ab (that is, by Consect. 4. Definit. 18. the Surface of the given Cone) will be equal to half Product of the mean Line into the mean Periphery, ½ e{powerof2}ab (i. e. by Consect. 2. Definit. 15.) to the Area of the Circle of the mean Proportional EF. Q. E. D.

The same Proposition of Archimedes may also be Demonstrat∣ed after this Way: If the side of the Cone BC be called b, and the Semidiameter of the Base a) so that the Periphery may, by Consect. 1. Definit. 31. be 2ea, and so the Surface of the Cone, by Consect. 4. Definit. 18. eab) the √ab will be a mean propor∣tional between b and a, by this 17th Proposition; which being taken for Radius, the whole Diameter will be √2ab, and the Pe∣riphery 2e√ab; therefore by Consect. 2. Definit. 15. half the Ra∣dius ½√ab multiplied by the Periphery (since √ab multiplied, by √ab necessarily produces ab) will give you the Area of the Cir∣cle by that mean(a) 1.14 Proportional, equal to the Surface of the given Cone, which before was expressed in the same Terms. Q. E. D.

Hence also naturally flows this other Proposition, That the Surface of the Cone (½e{powerof2}ab) is to its Base (½ ab) as the Side of the Cone (e{powerof2}b) is to the Radius of the Base b, as may appear from the Terms.

Proposition XVIII.

IF(b) 1.15 4. Quantities are Proportional, either continuedly or dircretely, the Product of the Extremes is equal to the Product of the Means.

Demonstration.

Suppose one Continual Proportional, a, ea, e{powerof2}a, e{powerof3}a.

Extremes e{powerof3}a	Means e{powerof2}a
a	ea
Prod. e{powerof3}aa = Prod. e{powerof3}aa. Q.E.D.

SCHOLIUM.

ON this Theorem is founded the Rule of Three in Arithme∣tick; so called because having 3 Numbers, (2. 5. 8.) it finds an unknown fourth Proportional. For altho this fourth be, as we have said, unknown, yets its Product by 2 is known, because the same with the Product of the Means, 5 and 8. Wherefore the Rule directs to multiply the third by the second, that you may thereby obtain the Product of the Extremes: which divided by one of the Extremes, viz. the first, necessa∣rily gives the other, i. e. the fourth sought.

Proposition XIX.

IF 2 Products (on the other side) arising from the Multipli∣cation of 2 Quantities, are equal, those 4 Quantities will be at least directly Proportional.

Demonstration.

Suppose eba be the equal Product of the Extremes, and eab of the Means; the Extremes will either be eb and a, or e and ba, or b and ea, as also the Means. But what way soever either is taken, there can be no other Disposition or placing of them, than one of the following.

1 eb	eb	a	a
—	ee	ab
—	ea	b;	or inversly.
—	a	eb
—	ab	e

—	b	ea
2 e e	e	ba	ba
—	eb	a
—	ea	b;	or inversly.
—	ba	e
—	a	eb
—	b	ea
3 b	b	ea	ea
—	a	eb
—	ba	e;	or inversly.
—	ea	b
—	eb	a
—	e	ba;	or inverting the Order of them all.

In all these Dispositions there may be immediately seen a Geometrical Proportion, by what we have in Definition 3•• and 33.

CONSECTARSY.

I. AS we have shewn one Sign of Proportionality in the Definition of it, viz. That the same Quotient will a∣rise by dividing the Consequents by the Antecedents; so now we have another Sign of it, viz. The Equality of the Products of the Extremes and Means.

II. By a bare Subsumption may hence appear the Truth of Prop. 14. lib. 6. Euclid. at least partly: Which we shall yet more commodiously shew hereafter.

Proposition XX.

IF there are never so many Continual Proportionals, the Pro∣duct of the Extremes is equal to the Product of any 2 of the Means that are equally distant from the Extremes, as also to the Square of the mean or middle Term, if the Terms are odd.

Demonstration.

Such are e. g. a, ea, e{powerof2}a, e{powerof3}a, e{powerof4}a, e{powerof5}a, e{powerof6}a, &c. and the Product of the Extremes, and of any two Terms equally remote

them, and the Square of the mean or middle Term, every where e{powerof6}aa. Q. E. D.

SCHOLIUM I.

NOR can there be any doubt but this will always be so, how far soever the Progression is continued; if 〈◊〉〈◊〉 con∣sider that the last Term always contains the first, 〈◊〉〈◊〉 way of Reason, so many times multiplied as is the place of that Term in the rank of Terms, excepting one. Altho therefore the last Term but one is in one degree of its Reason less than 〈◊〉〈◊〉 last, ••he second on the contrary, is in one more than the 〈…〉〈…〉 re∣••ore the Product of the one will necessarily 〈…〉〈…〉 ••e Product of the other. Thus also the las•• 〈…〉〈…〉 Degrees of Proportion lower than the 〈…〉〈…〉 being to be multiplied into that, exc••••ds 〈◊〉〈◊〉 fi st 〈…〉〈…〉 of the Proportion, &c. as may be seen thou U••iver••al Ex•••••••••• Hence you have the following

CONSECTARIES.

••. HAving some of the Terms given in a Continual Pro∣portion (e. g. suppose 10) you may easily find any other that shall be required (e. g. the 17th) as the last; If ••he 2 Terms given, being equally remote from the first and ••hat required (as are e. g. the eigth and tenth) be multiplied by one another, and this Product, like that also of the Extremes, be divided by the first.

II. But this may be performed easier, if you moreover take ••n this Observation, That if, e. g. never so many places of pro∣portionals, passing over the the first, be noted or marked by Ordinals or Numbers according to their places (as in this uni∣versal Example)

a,	ea,	e{powerof2}a,	e{powerof3}a,	e{powerof4}a,	e{powerof5}a,	e{powerof6}a,
	I.	II.	III.	IV.	V.	VI.

The place of the 7th Term is (e. g.) VI. (and so the place of any other of them being less by Unity than its number is among the Terms) and also composed of the places of any other equal∣••y distant from the Extremes, e. g. V. and I. IV. and II. or ••wice III. &c.

III. Here you have the Foundation of the Logarithms, i. •• of a Compendious Way of Arithmetick, never enough to b•• praised. For if, e. g. a rank of Numbers from Unity, con••••∣nually Proportional, be signed or noted with their Ordinals, as w•• have said, as Logarithms,

1.	2.	4.	8.	16.	32.	64.	128.	256,	&c.
	I.	II.	III.	IV.	V.	VI.	VII.	VIII.

and any two of them (as 8 and 32) are to be multiplied to¦gether; add their Logarithms III and V, and their Sum VII•• gives you the Logarithm of their Product 256, as the Te•••• equally remote from the 2 given ones and the first, and 〈◊〉〈◊〉 whose Product with the first (which is Unity) i. e. it self w•••••• be equal to the Product of the Numbers to be multiplied: A•••• contrariwise, if, e. g. 128 is to be divided by 4, subtracting t•••• Logarithm of the first II from the Logarithm of the second V•• the remaining Logarithm V points out the number sought 3•• so that after this way the Multiplication of Proportionals 〈◊〉〈◊〉 by a wonderful Compendium, turned into Addition, and the Division into Subtraction, and Extraction of the Square Ro•••• into Bisecting or Halving, (for the Logarithm of the Squa•••• Number 16 being Bisected, the half II gives the Root sough•• 4) of the Cube Root into Trisection (for the Logarithm of th•• Cube 64 being Trisected, the third part gives the Cubi•• Root sought 4).

SCHOLIUM II.

THat we may exhibit the whole Reason of this admirabl•• Artifice (which about 35 years ago was found out b•• the Honourable Lord John Naper Baron of Merchiston in Scotland and published something difficult, but afterwards render'd much easier and brought to perfection by Henry Briggs, the first S••¦vilian Professor of Geometry at Oxford.) I say that we may exhibit the whole Reason of it in a Synopsis, after an easie way when its use appear'd so very Considerable in the great Num∣bers in the Tables of Sines and Tangents, nor yet could they be useful without mixing vulgar Numbers with them, especial∣ly in the Practical Parts of Geometry, the business was to ac∣••ommodate

this Logarithmical Artifice to them both. First ••herefore that Artists might assign Logarithms to all the com∣mon Numbers proceeding from 1 to 1000 and 10000, &c. ••hey first of all pick out those which proceed in continued Geo∣metrical Proportion, and particularly, tho arbitrariously, those which increase in a Decuple Proportion, e. g. 1. 10. 100. 1000. 10000, &c.

But now to fit them according to the Foundation of Consect. 8. a Series of Ordinals in Arithmetical Progression, we do'nt only substitute the simple Number 1, 2, 3, &c. but augmented with several Cyphers after them, that so we may also assign ••heir Logarithms in whole Numbers to the intermediate Num∣••ers between 1 and 10, 10 and 100, &c. Wherefore, by ••his first Supposition, Logarithms in Arithmetical Proportion, ••nswer to those Numbers in Geometrical Proportion, after the ••ay we here see,

1	10	100
Log. 0000000	10000000	20000000
	1000	10000
	30000000	40000000, &c.

As that they also exhibit certain Characteristical initial Notes, whereby you may see, that all the Logarithms between 1 and ••0 begin from 0, the rest between 10 and 100 from 1, the ••ext from 100 to 1000 from 2, &c.

The Logarithms of the Primary Proportional Numbers being ••hus found, there remain'd the Logarithms of the intermediate Numbers between these to be found: For the making of which, ••fter different ways, several Rules might be given drawn from ••he Nature of Logarithms, and already shewn in Consect. 3. See Briggs's Arithmetica Logarithmica, and Gellibrand's Trigonometria Britannica; the first whereof, chap. 5. and the following, shews ••t length both ways delivered by Neper in his Appendix. But ••he business is done more simply by A. Ʋlacq. in his Tables of Sines &c. whose mind we will yet further explain thus: If you are ••o find, e. g. the Logarithm of the Number 9, between 1 and ••0, augmented by as many Cyphers as you added to the Lo∣garithm of 10, or the rest of the Proportionals (h. e. between

10000000 and 100000000) you must find a Geometric•• Mean Proportional, viz. by multiplying these Numbers togeth•••• and extracting the Square Root out of the Product, by Pr•• 17. Now if this Mean Proportional be less than 9 augmen•••• by as many Cyphers, between it and the former Denary Nu••∣ber you must find a second mean Proportional, then betwe•••• this and that same a third; and so a fourth, &c. but if it 〈◊〉〈◊〉 greater, then you must find a mean Proportional between and the next less, &c. till at length after several Operatio•• you obtain the number 9999998, approaching near 90000000. Now if between the Logarithm of Unity a•• Ten (i. e. between 0 and 10000000) you take an Arithm••¦tical Mean Proportional (05000000) by Bisecting their S•• by Prop. 14. and then between this and the same Logarithm Ten, you take another mean, and so a third and a fourth, 〈◊〉〈◊〉 at length you will obtain that which answers to the last abo•• mentioned, viz. 9. See the following Specimen.

A TABLE of the Geometrical Proportionals betwe•••• 1 and 10, augmented by 7 Cyphers, and of t•••• Arithmetical Proportionals between 0 and 10000000 being the Logarithms corresponding to them.

Geometrical Mean Pro∣portionals.		Arithmetical Logar. mean Proportionals
31622777	First,	05000000
56234132	Second,	07500000
74989426	Third,	08750000
86596435	Fourth,	09375000
93057205	Fifth,	09687500
89768698	Sixth,	09531250
91398327	Seventh,	39609375
90579847	Eighth,	09570312
90173360	Ninth,	09550781
89970801	Tenth.	09541015

Which is thus made: In the first Table a Geometrical Mean ••••oportional between 10 000 000 and 100 000 000 the first Number of it; then another Mean between that and ••e same last 100000000, gives the second; and so to the ••••th, 93057205. Which, since it is already greater than the ••ovenary, another Mean between it and the precedent fourth, ••••comes in order a sixth, but sensibly less than the Novenary. ••herefore between it and the fifth you will have a seventh ••ean yet greater than the Novenary; and between the sixth ••••d seventh, an eighth, somwhat nearer to the Novenary, but ••t yet sensibly equal, but somewhat bigger; moreover between ••••e sixth and eighth you will have a ninth, between the ninth ••••d sixth a tenth gradually approaching nearer the Novenary, but ••••t somewhat sensibly differing from it. Now if you con∣••ue this inquiry of a mean Proportional between this tenth, 〈◊〉〈◊〉 somewhat too little, and the precedent ninth as somewhat ••o big, and so onwards, you will at length obtain the Num∣••••r 8999 9998, only differing two in the last place from the ••ovenary Number augmented by seven Cyphers, and conse∣••ently insensibly from the Novenary it self. But for the Lo∣••arithm of this in the second Column, by the same process you ••••e to find Arithmetical Mean Proportionals between every 2 ••ogarithms answering to every two of the superiour ones, till you ••nd, e. g. the Logarithm of the tenth Number 09541015, ••d so at length the Logarithm of the last, not sensibly differ∣••••g from the Novenary, 09542425.

Thus having found, with a great deal of labour, but also ••ith a great deal of advantage to those that make use of them, ••••e Logarithms of some of the numbers between 1 and 10, and ••0 and 100, &c. you may find innumerable ones of the other ••ntermediate Numbers with much less labour, viz. by the help ••f some Rules, which may be thus obtain'd from Consect. 3 of ••e precedent Proposition. The Sum of the Logarithms of the ••umber Multiplying and the Multiplicand, gives the Logarithm of the ••roduct. 2. The Logarithm of the Divisor subtracted from the Lo∣••arithm of the Dividend, leaves the Logarithm of the Quotient: ••he Logarithm of any number doubled, is the Logarithm of the Square, ••ripled of the Cube, &c. 4 The half Logarithm of any number is ••he Logarithm of the Square Root of that number, the third part of 〈◊〉〈◊〉 the Cube Root, &c. Thus, e. g. if you have found the Lo∣garithm

of the number 9, after the way we have shewn, by th•• same reason you may find the Logarithm of the number 5 (vi•• by finding mean Proportionals between the second and the fi•• number of our Table, and between their Logarithms, &c. and by means of these 2 Logarithms you may obtain several o¦thers: First, since 10 divided by 5 gives 2; if the Logarith•• of 5 be subtracted from the Logarithm of 10, you'l have th•• Logarithm of 2, by Rule the second. Secondly, since 10 m••••¦tiplied by 2 makes 20, and by 9 makes 90, by adding th•• Logarithms of 10 and 2, and 10 and 9, you'l have the L••¦garithms of the numbers 90 and 20, by Rule 1. Thirdly Since 9 is a Square, and its Root 3, half the Logarithm of 〈◊〉〈◊〉 gives the Logarithm of 3, by Rule 4. since 90 divided by 〈◊〉〈◊〉 gives 30, the Logarithm of this number may be had by s••••¦tracting the Logarithm of 3 from the Logarithm of 90, b•• Rule the second. Fifthly, 5 and 9 squared make 25 and 8▪ the Logarithms of 5 and 9 doubled, give the Logarithms 〈◊〉〈◊〉 these numbers, by Rule 3. In like manner, sixthly, the Su•• of the Logarithms of 2 and 3, or the Difference of the L••¦garithms of 5 and 30, give the Logarithm of 6, and the Su•• of the Logarithms of 3 and 6, or 2 and 9, gives the Log••¦rithm of 18; the Logarithm of 6 doubled, gives the Loga¦rithm of 36, &c. And after this way you may find and reduce it to Tables, the Logarithms of Vulgar Numbers from 1 to 100•• (as in the Tables of Strauch. p. 182, and the following) or 〈◊〉〈◊〉 100000 (as in the Chiliads of Briggs) But as to the manner ••¦deducing the Tables of Sines and Tangents from these Loga¦rithms of Vulgar Numbers, we will shew it in Schol. of Pr•••• 55, only hinting this one thing before-hand; that this Artifi•• of making Logarithms is elegantly set forth by Pardies in hi•• Elements of Geometry, pt 112. by a certain Curve Line then•••• called the Logarithmical Line; by the help whereof he suppose Logarithms may be easily made; and having found those o•• the numbers between 1000 and 10000, he shews, that all o∣thers may be easily had between 1 and 1000. Wherefore w•• shall Discourse more largely in Schol. Definit. 15. lib. 2.

Proposition XXI.

IF the first Term of never so many Continual Proportionals, be sub∣tracted from the last, and the Remainder divided by the name of the Reason or Proportion lessen'd by Ʋnity, the Quotient will be equal to the Sum of all except the last.

Demonstration.

	ea
	e{powerof2}a
	e{powerof3}a
	e{powerof4}a
	e{powerof5}a
The last Term less the first	e{powerof6}a−a

Divided by the name of the Reason lessen'd by unity.
e−1
* Quote. e{powerof5}a+e{powerof4}a+e{powerof3}a+e{powerof2}a+ea+a; And it is evident from the Operation, that the same will always happen tho the number of Terms be con∣tinued never so far.
e−1
e−1
	e−1
	e−1
	e−1
	e−1

CONSECTARYS.

I. WHerefore in adding never so great a Series of Geo∣metrical Proportionals, since it is enough that the first and last Term, and the Name of the Reason be known, by this Prop. and having found at least some of the Terms of the Proportion, any other may be afterwards found, whose place will be compounded of the places of the two Antecedent ones, according to Consect. 2. Prop. 20. viz. by Multiplying the Terms answering to the two above-mentioned places, and dividing the Product by the first Term; thence it will be very easie to add a great Series of Proportionals into one Sum, tho the particular separate Terms remain almost all of them unknown.

SCHOLIUM.

THese are the same Practical Arithmetical Rules concerning Geometrical Progressions; for the illustration of whic•• Swenterus in Delic. has given us so many pleasant Examples, li•• 1. Prop. 59. and fol. First of all, that famous Example is of th•• kind which relates to the Chequer-work'd Table or Board t•• fling Dice on, with its 64 little Squares, which Dr. Wa•••• has translated out of the Arabick of Ebn Chalecan, into Latin in Oper. Mathem. part. 1. Chap. 31. for the illustration of whic•• we have heretofore composed an Exercitation, and shall he•• only note these few things: If there are supposed 64 Terms 〈◊〉〈◊〉 double Proportion from Unity, and the first of them, note•• with their local Numbers, are these that follow;

1	2	4	8	16	32	64	128
	I	II	III	IV	V	VI	VII

You may have the Term of the 13th place, 8192, by mu••¦tiplying together the VIth and VIIth place; and the Ter•• of the XXVIth place, by squaring or multiplying this new Pro¦duct again by it self, and moreover the Term of the L•••• place, by multiplying that Product again by itself; and furthe•• more the Term of the LIXth place, by multiplication of the num¦ber last found by the number of the VIIth place, and lastly the Term of the LXIIId place (i. e. the last in the proposed Se¦ries) by multiplying this last of all by the number of the IV•••• place.

II. Moreover you may, by this Art, collect infinite Seri•• of Proportional Terms into one Sum, altho it is impossible 〈◊〉〈◊〉 run over all the Terms separately, because infinite. e. g. in 〈◊〉〈◊〉 continued Series of Fractions, decreasing in a double Propo••¦tion ½ ¼, ⅛ 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, &c. ad infinitum, if you take them bac••¦wards, you may justly reckon a Cypher or 0, for the fi•••• Term (for between ½ and 0 there may be an infinite Numbe•• of such Terms) and the infinite Sum of these Terms will b•• precisely equal to Unity; for subtracting the first 0, from the last ½, and the remainder ½ being divided by the name of th•• Reason lessened by 1, i. e. by I. which divides nothing, th••

Quotient ½ is the Sum of all the Terms excepting the last, by Prop. 21. and so the last ½ being added, the Sum of all in that Series will be I. Now if the last is not ½ but I, the Sum of all will necessarily be 2; if 2 be the last, the Sum of all will be 4; in a word, it will be always double the last Term.

III. And since in this case the Sum of all the precedent Terms is equal to the last Term, the one being subtracted from the other, there will remain nothing, i. e. ½−¼−⅛−〈 math 〉〈 math 〉−〈 math 〉〈 math 〉, &c. in Infinitum, is = 0, and also 1−½−¼, &c. or 2−1−½−¼. &c. = 0.

IV. In like manner the Sum of infini••e Fractions decreasing in triple Reason in an infinite Series (⅓+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉, &c.) will be equ••l to ½: for if from the last ⅓ (again in an inverted Order) you subtract the first 0, and the Remainder ⅓ be divided by the name of the Reason lessen'd by Unit, i. e. by 2, the Quotient ⅙ will be the Sum of all the antecedent Terms, and adding to this last ⅓ or 〈 math 〉〈 math 〉 the Sum of all will be 〈 math 〉〈 math 〉 or ½.

V. Thus an infinite Series of Fractions decreasing from ¼ in a Quadruple Proportion (¼+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉 &c.) is equal to ⅓; for subtracting the first 0 from the last ¼, and the remainder ¼ being divided by the name of the Proportion, i. e. by 3, you will have 〈 math 〉〈 math 〉 the sum of all except the last, and adding also the last ¼ or 〈 math 〉〈 math 〉, you'l have the whole Sum 〈 math 〉〈 math 〉 or ⅓.

VI. Thus also an infinite Series decreasing from ⅕ in a Quin∣tuple Proportion (⅕+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉, &c.) is equal to ¼: ⅙+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉, &c. is equal to ⅕ &c. and so any Series of this kind is equal to a Fraction, whose Denominator is less by an Unit than the Denominator of the last Fraction in that Series.

VII. Generally also, any infinite Series of Fractions decreasing according to the Proportion of the Denominator of the last Term, and having a common Denominator less by an unit than the Denominator of the last Term (e. g. ⅔+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉, &c. or ¾+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉, &c. or ⅘+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉, &c.) is equal to Unity, after the same way as the Series Consect. 2. which may be compre∣hended under this kind, and which may be demonstrated in all its particular cases by the same method we have hitherto made use of, or also barely subsumed from Consect. 4, 5, and 6. For since ⅓+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉, &c. is equal to ½; ⅔+〈 math 〉〈 math 〉+〈 math 〉〈 math 〉 will be equal to 〈 math 〉〈 math 〉, or 1, and so in the rest.

VIII. Particularly the sum of 〈 math 〉〈 math 〉, &c, decreasing in a Quadruple Proportion, is equal to 〈 math 〉〈 math 〉; and the sum of 〈 math 〉〈 math 〉, &c. is equal to 〈 math 〉〈 math 〉; and the sum of 〈 math 〉〈 math 〉, &c. decreasing in Octuple Proportion, is equal to ⅛: For subtracting the first Term 0, and dividing the remainder by the name of the Reason lessen'd by 1, i. e. by 3, the Quotient 〈 math 〉〈 math 〉 gives the sum of all except the last. This therefore (viz. 〈 math 〉〈 math 〉) being ad∣ded, the sum of all will be 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉: In like manner 〈 math 〉〈 math 〉 being divided by the name of the Reason lessen'd by Unity, the Quo∣tient will give 〈 math 〉〈 math 〉, and adding the last, the sum of all will be 〈 math 〉〈 math 〉 i. e. ⅛. So that hence it is evident, that 〈 math 〉〈 math 〉, &c. or −〈 math 〉〈 math 〉, &c. in Infinitum, will be equal to noth∣ing; also ⅛−〈 math 〉〈 math 〉−〈 math 〉〈 math 〉−〈 math 〉〈 math 〉 &c. = 0.

IX. The Sum of a simple Arithmetical Progression (i. e. ascending by the Cardinal Numbers) continued from 1, ad Infinitum, is S••••∣duple of the Sum of the same number of Terms, each of which is equ•••• to the greatest; or on the contrary, this latter Sum is double of the fo••∣mer. We might have subsumed this in Consect. 4. Prop. 16. for, prefixing a Cypher before Unity, it will be a case of that Con∣sectary, the Sum of the Progression remaining still the same. B•••• that this is true, in an infinite Series beginning from Unity (f•••• in a finite or determinate one, the proportion of the Sum is al∣ways less than double, tho it always approaches to it, and come so much the nearer by how much greater the Series is) 〈◊〉〈◊〉 shall now thus Demonstrate: To the Sum of three Terms, 〈◊〉〈◊〉 2, 3, i. e. 6, the sum of as many equal in number to the greatest, i. e. 9, has the same Proportion as 3 to 2; but t•• the sum of six Terms, 1, 2, 3, 4, 5, 6, i. e. 21, the su•• of as many equal to the greatest, i. e. 36, has the same pro∣portion as 3 to 1+¾, that is, as 3 to 2−¼, the decrease be∣ing ¼: but to the sum of 12 Terms, which may be found b•• Consect. 1. Prop. 16.=78, the sum of so many equal to the greatest, viz. 144. has the same proportion (dividing both sid•• by 48) as 3 to 1 〈 math 〉〈 math 〉, i. e. 3 to 1+½+⅛ (for 24 make ½, a•••• the remainder 〈 math 〉〈 math 〉 is the same as ⅛) that is, as 3 to 2−¼−•••• the decrement being now ⅛. Since therefore, by doubling th•• number of Terms onward, you'l find the decrement to be 〈 math 〉〈 math 〉, an•• so onwards in double Proportion; the sum of an infinite Num∣ber of such Terms, in Arithmetical Progression, equal to th•• greatest

will be to the sum of the Progression from 1, ad Infinitum, as 3 to 2−¼−⅛−〈 math 〉〈 math 〉, &c. that is, by Consect. 2 and 3, as 3 to 2−½, that is, as 3 to 1 ½, or as 2 to 1. Q.E.D.

X. The Sum of any Duplicate Arithmetical Progression (i. e. a Progression of Squares of whole numbers ascending) continued from 1 ad Infinitum, is subtriple of the Sum of as many Terms equal to the greatest as is the number of Terms: For any such finite Progression is greater than the subtriple Proportion, but approaches nearer and nearer to it continually, by how much the farther the Series of the Progression is carried on. Thus the Sum of 3 Terms 1, 4, 9=14 is to thrice 9=27 as 1 〈 math 〉〈 math 〉, or 1 〈 math 〉〈 math 〉, or 1+½+〈 math 〉〈 math 〉 to 3 (dividing both sides by 9,) the Sum of six Terms, 1, 4, 9, 16, 25, 36, viz. 91. to six times 36, i. e. to 216 (dividing both sides by 72) is as 1+¼+〈 math 〉〈 math 〉 to 3; and the Sum of 12 Terms 650, to 12 times 144, i. e. to 1728 (dividing both sides by 576) is as 1+⅛+〈 math 〉〈 math 〉 to 3, &c. the Fractions adhering to them thus constantly decrea∣sing, some by their half parts, others by three quarters (for 〈 math 〉〈 math 〉 is 〈 math 〉〈 math 〉; therefore the first decrement is 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉, is 〈 math 〉〈 math 〉; there∣fore the second decrement is 〈 math 〉〈 math 〉, &c.) Wherefore the Sum of the Infinite Progression will be to the Sum of the like number of Terms equal to the greatest, as 〈 math 〉〈 math 〉, &c. to 3, that is, by Consect. 3 and 8, as 1 to 3. Q.E.D.

XI. The Sum of a triplicate Arithmetical Progression (i. e. ascending by the Cubes of the Cardinal Numbers) proceeding from 1 thro' 27, 64, &c. ad Infinitum, is Subquadruple of ••he Sum of the like number of Terms equal to the greatest. For the Sum of 4 Terms, 1, 8, 27, 64, i. 100, to 4 times 64, i. e. 256 (dividing both sides by 64) will be found to be as 1+½+〈 math 〉〈 math 〉 to 4; but the Sum of 8 Terms, 1, 8, 27, 64, 125, 216, 343, 582, i. e. 1296 to 8 times 512, that is, 4096 (dividing both Sides by 1024.) will be found to be as 1+¼+〈 math 〉〈 math 〉 to 4, &c. The adhering Fractions thus

constantly decreasing, the one by their ½ part, the others by •••• (for 〈 math 〉〈 math 〉 is 〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉 is 〈 math 〉〈 math 〉, &c. Wherefore the Sum of the In∣finite Progression will be to the Sum of a like (Infinite) num∣ber of Terms, equal to the greatest, as 〈 math 〉〈 math 〉, &c. 〈 math 〉〈 math 〉, &c. to 4; that is, by Consect. 3 and 8, as 1 to 4. Q. E. D.

XII. The Sum of an Infinite Progression, whose greatest Term is a Square Number, the others decreasing according to the odd numbers 1, 3, 5, 7, &c. is in Subsesquialteran Pro∣portion of the Sum of the like number of equal Terms, i. e. as 2 to 3. For the Sum of three such Terms, e. g. 9, 8, 5, i. e. 22 to thrice 9, i. e. 27. is (dividing both sides by 9) 〈◊〉〈◊〉 2 〈 math 〉〈 math 〉, viz. 〈 math 〉〈 math 〉 to 3, or 2+½−〈 math 〉〈 math 〉 to 3. But the Sum of s•••• such Terms, 36, 35, 32, 27, 20, 11, i. e. 161, to six time 36, i. e. 216 (dividing both sides by 72) is as 2+¼−〈 math 〉〈 math 〉, &c. the adhering Fractions thus always decreasing, some by ½, o∣thers by ¾, as above in Consect. 10. Wherefore the Sum of the Infinite Progression will be to the Sum of the like number of Terms equal to the greatest, as 〈 math 〉〈 math 〉, &c. to 3, i. e. by Con¦sect. 3 and 8, as 2 to 3. Q E. D.

SCHOLIUM II.

THus we have, after our method, demonstrated the chie•• Foundations of the Science or Method, or Arithmetick •••• Infinites, first found out by Dr. John Wallis, Savilian Professo•• of Geometry at Oxford, and afterwards carried further by Det••∣lerus Cluverus, and Ismael Bullialdus. And from these Founda∣tions we will in the following Treatise demonstrate, and that directly and à priori, in a few Lines, the chief Propositions o••

Geometry, which the Antients have spent so much labour, and composed such large Volumes to demonstrate, and that but in∣directly neither.

Proposition XXII.

THe Powers of Proportionals whether continuedly or discretely, such as the Squares, Cubes, &c. are also Proportional.

Demonstration.

Continual Proportionals.	Discrete Proportionals.
a ea e{powerof2}a e{powerof3}a	a ea b eb
Squares aa e{powerof2}a{powerof2} e{powerof4}a{powerof2} e{powerof6}a{powerof2}	a{powerof2} e{powerof2}a{powerof2} b{powerof2} e{powerof2}b{powerof2}
Cubes a{powerof3} e{powerof3}a{powerof3} e{powerof6}a{powerof3} e••a{powerof3}	a{powerof3} e{powerof3}a{powerof3} b{powerof3} e{powerof3}b{powerof3}

Q E. D.

SCHOLIUM.

YOu founded in this Truth, 1. the Reason of the Multipli∣cation and Division of Surd Quantities: For since from the Nature and Definition of Multiplication, it is certain, that 1 is to the Multiplier as the Multiplicand to the Product (for the multiplicand being added as many times to it self as there are Units in the Multiplier, makes the Product) if the √5 is to be multiplied by √3, then as 1 to the √3, so the √5 to the Product; and, by the present Proposition, as 1 to 3, so 5 to the □ Product, i. e. to 15. Wherefore the Product is √15; and so the Rule for Multiplying Surd Quantities is this: Multiply the Quantity under the Radical Signs, and prefix a Radical Sign to the Product.(α) 1.16 Likewise since it is certain from the Nature of Division, that the Divisor is to the Dividend as 1 to the Quotient (for the Quotient expresses by its Units how many times the Divisor is contained in the Dividend) if the √15 is to be divided by √5, you'l have √5 to the √15 as 1 to the Quo∣tient, and, by the present Scholium, 5 to 15, as 1 to the □ of the Quotient, i. e, to 3. Therefore the Quotient is the Root of 3, and so the Rule of dividing Surd Quantities this; viz.

Divide the Quantities themselves under the Radical Signs, and pre∣fix the Radical Sign to the Quotient.

II. Hence also flows the usual Reduction in the Arithme∣tick of Surds, of Surd Quantities to others partly Rational, and on the contrary, of those to the form of Surds, e. g. If you would reduce this mixt Quantity 2a√b, i. e. 2a multiplied by the √b, to the form of a Surd Quantity; which shall all be con∣tained under a Radical Sign; The Square of a Rational Quan∣tity without a Sign 4aa, if it be put under a Radical Sign, in this form √4aa, it equivalent to the Rational Quantity 2a; but the √4aa being multiplied by √b makes √4aab. by N•• 1. of this Scholium. Therefore √4aab is also equivalent to the Quantity first proposed 2a√b. Reciprocally therefore, if th•• form of a meer Surd Quantity √4aab, is to be reduced to on•• more Simple, which may contain without the Radical Sig•• whatever is therein Rational, by dividing the Quantity com∣prehended under the sign √ by some Square or Cube, &c. as here by 4aa, (i. e. √4aab by √4aa, i. e. 2a) the Quotient wil•• be √b, which multiplied by the Divisor 2a, will rightly ex∣press the proposed Quantity under this more simple Form 2a√•• Which may also serve further to illustrate the Scholia of Prop. 7. and 10.

Proposition XXIII.

IF there are four Quantities Proportional, (a, ea, b, eb) they will be also Proportional,

• 1. Inversly. ea to a as eb to b.
• 2. Alternatively,(α) 1.17 a to b as ea to eb.
• 3. Compoundedly,(β) 1.18 a+ea to ea, so b+eb to eb.
• 4. Conversly, a+ea to a as b+eb to b.
• 5. Dividedly,(γ) 1.19 a−ea to
  • ea as b−eb to
  • or a
    • ...eb
    • or 〈◊〉〈◊〉
• 6.(α) 1.20 By a Syllepsis, a to ea as a+b to ea+eb.
• 7. By a Dialepsis, a to ea as a−b to ea−eb.

Which are all manifest, by comparing the Rectangles of the Means and Extremes according to to Prop. 19. and its Consect. 1.

or by dividing any of the Consequents by their Antecedents, ac∣cording to Def. 31.

Proposition XXIV.

IF in a(β) 1.21 double Rank of Quantities you have

• as a to ea,
• so b to eb,

and also

• as ea to oa,
• so eb to ob,

then you'l have also by proportion of Equality orderly pla∣ced,

• as the first a, to the last oa, in the first Series;
• so the first b, to the last ob, in the second Series.

Which is manifest from the Terms themselves.

Proposition XXV.

BUt(γ) 1.22 if they are disorderly plac'd

• as oa to ea
• † so eob to ob

* as ea to a † so ob to eb, * you'l have here again by proportion of Equality,

• as the first oa to the last a, in the first Series;
• so the first eob to the last eb, in the second Series.

As is evident from the Rectangles of the Extremes and Means, as also from the very Terms.

Proposition XXVI.

IF(α) 1.23 as the whole ea to the whole a, so the part eb to the part b; then also will

the Remainder	Remainder	Whole	Whole
ea−eb to the	a−b, as the	ea	to the a.

This is evident from the Rectangle of the Extremes and Means, both which are eaa−eab. Q.E.D.

Proposition XXVII.

REctangles or Products having one common Efficient or Side, are one to another as the other Efficients or Sides.

Demonstration.

Suppose the Products to be ab and ac, having the common Efficient a; I say they are

• as b to c, so ab to ac.

Which is evident at first sight, by comparing the Products of the Extremes and Means, and also fully shews, that other way of proving Proportionality, whereby by dividing the Conse∣quents by their Antecedents, the identity or sameness of the Quotients are wont to be demonstrated.

SCHOLIUM I.

I. THe Reduction of Fractions either to more compounded or more simple ones is founded on this Theorem; on the one hand by multiplying, on the other by dividing, by the same quantity, both the Numerator and the Denominator, as, e. g, 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉, ⅓, 〈 math 〉〈 math 〉, 〈 math 〉〈 math 〉, &c. are in reality the same Fractions. And

II. The Reduction of Fractions to the same Denomination, as if 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 are to be changed into two o∣thers that shall have same Denominator;(α) 1.24 this is to be done by multiplying the Denominators together for a new Denominator,(β) 1.25 and each Numerator by the Denominator of the other for a new Numerator, and you'l have for the two Fractions above — 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉

SCHOLIUM II.

WE will here for a conclusion of Proportionals, shew the way of cutting or dividing any Quantity in Mean and Extreme Reason, viz. if for the greater Part you put x, the less

will be a−x; and so by Hypoth. these three, a, x and a−x, will be proportional, by Def. 34. Therefore by Prop. 17. the Product of the Extremes aa−ax = to the Square of the Mean xx, and (adding on both sides ax) aa=xx+ax; and more∣over adding on both sides ¼aa, you'l have 〈 math 〉〈 math 〉aa=xx+ax+¼aa. Now this last Quantity, since it is an exact Square, whose Root is x+½a, you'l have √〈 math 〉〈 math 〉aa=x+½a, and (subtracting from both sides 〈 math 〉〈 math 〉 a) √〈 math 〉〈 math 〉aa−½a=x.

Now therefore we have a Rule to determine the greater part of a given Quantity to be divided in Mean and Extreme Rea∣son, viz. if the given Quantity be a Line, e. g. AB=a (Fig. 58.) join to it(α) 1.26 at Right Angles AC=½a: Wherefore by the Theorem of Pythagoras from Schol. Definit. 13. the Hypo∣thenuse CB, or, which is equal to it, CD=√¾aa; and conse∣quently AC=½a being taken out of CD, the Remainder AD, or AE, which is equal to it, will be = x, the greatest part sought; according to Euclid, whose Invention this first Specimen of Analysis, by way of Anticipa∣tion, reduces to its original Fountain. As for Numbers (tho none accurately admits of this Section) the sense of the Rule, or which is all one as to the thing it self, is this: Add the Squares of a whole Num∣ber and its half, and subtract the said half from the Root of the Sum (which can't be had exactly, since it is √〈 math 〉〈 math 〉.

CAAP. V. Of the Proportion or Reasons of Magnitudes of the same kind in particular.

Proposition XXVIII.

TRiangles and Parallelograms, also Pyramids and Prisms and Paral∣lelepipeds, lastly Cones and Cylinders, each kind compared among themselves, if they have the same Altitude, are in the same Proportio•• to one another as their Bases.

Demonstration.

This and the following Proposition might have been by •• bare Subsumption added, as Consectarys to the precedent; fo•• the Altitudes in the one, and Bases in the other, may be looke•• on as common Efficients, and the Magnitudes mentioned as their Products: But for the greater distinction sake, we will thus De∣monstrate them more particularly.

I. If the equal Altitudes of two Triangles, o••(α) 1.27 two Parallelograms, are called b and 〈◊〉〈◊〉 Base of the one a, and of the other ea; these Products will be ba and bea, the other ½ ba and ½ bea, by Def. 28. Schol. 2.

II. Likewise the equal Altitudes of two Priso••(β) 1.28 or Pyramids, may be called b, and the Pro∣portion of their Bases expressed by a and ea; a•••• the Prisms will be among themselves as ba to be•• and the Pyramids as ⅓ba to ⅓bea,(γ) 1.29 by the sai•• Schol. Num. 3.(δ) 1.30

III. There is also the same Proportion of Cylin∣ders and Cones as of Pyramids and Prisms, by Consect. 4. Definit. 17. But,

as a to ea so is ba to bea.

−½ba to ½bea.

−⅓ba to ⅓bea.

Q.E.D.

CONSECTARY.

THerefore Magnitudes of the same kind upon the same or equal Bases (dgr;) and of the same heighth, are equal among themselves, and the contrary.

Proposition XXIX.

TRiangles and Parallelograms, Pyramids and Prisms and Parallele∣pipeds, Cones and Cylinders, being on equal Bases, are in the same Proportion as their heighths.(*) 1.31

Demonstration.

Let all their Bases be called a, and the Proportions of their Heighths be as b to eb: Therefore, 1. the Parallelograms, Pa∣rallelepipeds and Cylinders, are one to the other of the same kind, as ba to eba; the Triangles as ½ ba to ½ eba; the Pyra∣mids and Cones as ⅓ba to 3 eba, by Def. 28. Schol. 2. But,

as b to eb,	so is ba to eba.
	and ½ ba to ½ eba.
	and ⅓ ba to ⅓ eba.

Q.E.D.

Proposition XXX.

EQual‖ 1.32 Triangles, Parallelograms, Prisms, Parallelepipeds, also equal Pyramids, Cones, and Cylinders, have their Bases and Heighths reciprocally Proportional.

Demonstration.

For if for the equal Triangles you put ½ab, for the Cones and Pyramids ⅓ ab, and for the rest ab;

whether the Bases of the equal Quan∣tities are supposed to be a, and so the Altitudes

on both sides b; or if the Base of the one be a and b the Al∣titude, but the Base of the other b and the Altitude a, you'l certainly have eitherways,

as a to a, so Reciprocally b to b; the Base of the former to the Base of the latter, as the Altitude of the latter to the Altitude of the former, or,

as a to b, so Reciprocally a to b. Q.E.D.

CONSECTARY.

AND those Magnitudes of the same kind, whose Bases and Altitudes are thus Reciprocal, are equal by Prop. 18. for the Product or Rectangle of the Extremes is ab, and that of the Means ba.

Proposition XXXI.

TRiangles, Parallelograms, Prisms, Parallelepipeds, Pyramids, Co•••••• and Cylinders, each kind compared among themselves, are in the Proportion compounded of the Proportion of their Altitudes and Bases.(α) 1.33

Demonstration.

Suppose the Base of the one to be a, and the other ea, and the Altitude of the one b, of the other ib; therefore the one will be to the other,

as a b to eiab,

or ½ ab to ½ eiab,

or ⅓ ab to ⅓ eiab; i. e. every where as a to ei•• i. e. in Proportion compounded of a to ea, and of b to ib, by Consect. 2. Def. 34. Q.E.D.

SCHOLIUM.

FRom what we have hitherto Demonstrated, we may not only make an estimate of Magnitudes of the same kind compared together, which is easie to any one who attentively considers them; but also with F. Morgues, deduce a General Rule of expressing the Proportions of any Rectilinear Planes or Solids, contained under

Plane Surfaces, by the proportion of one Right Line to ano∣ther. For since the one may be resolved into Triangles, and the other into Pyramids, having first two Rectilinear Planes given and thus resolved, upon a Right Line I make the ▵ abc (Fig. 49.) Equal to one of the Triangles of either of the Planes e. g. to ABC; then having drawn the Parallel cm, if the ▵ BCD has the same Altitude with the former, you need only joyn the Base BC to the Base ab. But if the Altitude DS is greater than the Altitude of the other e. g. by ⅕, then you must make b f equal to the Base bc augmented by a fifth part, and the Trian∣gle bcf will = BCD, and the whole acf = to the Rectilinear Figure ABCD. If now therefore I likewise make another Tri∣angle ghi equal to another Rectilinear Figure between the same Parallels, then will the ▵ acf be to the ▵ ghi, that is, the Right Lined Figure ABCD to the Right Lined Figure FGHIK, as af to gh, by Prop. 28. 2. Having 2 Right Lined Solids gi∣ven, and having resolved them into Triangular Pyramids, they may be transferr'd between 2 parallel Planes, viz. by augment∣ing or diminshing their Triangular Bases reciprocally, ac∣cording to the excess or defect of their Altitudes, as was done above with the Linear Bases; then those Triangular Bases on both sides may be converted into one Triangular Base, and con∣sequently each Solid into a Pyramid equal to it self; which two Pyramids will be one to the other as their Triangular Bases. And because the Proportions of these Bases may be reduced to the Proportion of two Lines each to the other, by N^o 1. of this; therefore also the Reason or Proportion of the two Solids may be expressed by the Proportion of two Lines. Q.E.D.

Proposition XXXII.

CIrcles(β) 1.34 are in the same Proportion to one another as the Squares of their Diameters.

Demonstration.

Suppose a to be the Diameter of one Circle, and b of another; then by Definit. 31. Consect. 1. the Area of the one will be ¼ eaa, and that of the other ¼ ebb. But as aa to bb so is ¼ eaa to ¼ ebb by Consect. 1. Prop. 19. Q E. D.

CONSECTARY I.

THe same will in like manner be manifest of like Sectors Circles, while for the parts of the Periphery you put and ib, as for the wholes we put ea and eb: for thus the A•• of the one will be ¼ iaa, and of the other ¼ ibb.

CONSECTARY II.

CYlinders whose Altitudes are equal to the Diameters of th•• Bases, are in proportion to one another as the Cubes their Diameters; for the Cylinders will be ¼ea{powerof3} and ¼eb{powerof3}, Cubes a{powerof3} and b{powerof3}.

CONSECTARY III.

HEnce also (whatever the Reason of the Sphere is to the Cylinder of the same Diameter and Heighth; which will hereafter Demonstrate, and which in the mean while will denote by the name of the Reason y) I say, hence Sphe•••• which have the same Proportion to one another as these Cyli••¦ders (viz. as ¼ ea{powerof3} to ¼eb{powerof3}, so ¼ yea{powerof3} to ¼ yeb{powerof3}) will also (by C•••• sect. 1.) be in the same proportion as the Cubes, a{powerof3} to b{powerof3} is also evident from these Terms themselves.

Proposition XXXIII.

THE Angle(α) 1.35 (β) 1.36 at the Center of any Circle ACB (Fig. 60) to an Angle at the Circumference which has the same Arch its Base ADB, as 2 to 1.

Demonstration.

The truth of this has already appear'd fro•• Schol. Definit. 10. N^o 3. but here we will demo••¦strate it otherwise in its three Cases, after E••¦clids way. In the first Case DE being conceive Parallel to CB, by Def. 11. Consect. 1 and 2. th•• External Angle ACB is = to the Internal A••¦gle

[illustration]

Pag. 95.

61

62

63

64

65

66

67

68

69

70

71

72

ADE, and the Angle BDE, is equal to the alternate Angle BCD, i. e. to the other at the Base CDB, by Consect. 2. Definit. 13. Therefore BDE is as 1, and CDE, i. e. ACB as 2.

In the second Case the whole ECB is double of the whole EDB, and the subtracted Angle ECA is double of the sub∣tracted Angle EDA, by Case 1. Therefore the Remainder ACB is also double of the Remainder ADB, by Prop. 26. In the third Case the part ECA is double of the part EDA, and also the part ECB is double of the part EDB, by Case 1. Therefore the whole ACB is double the whole ADB. Q.E.D.

CONSECTARYS.

I. HEnce all Angles ADB(α) 1.37 in the same Segment are equal, and the Angle ADB (Fig. 61.) in a Semicircle is a Right one; because the Aperture at the Center answering to it, ACB contains two Right Angles: The Angle in a less Segment than a Semicircle EDF, is greater than a Right one; because the Aperture at the Center EGHFC answering to it, comprehends more than two Right Angles. An Angle, last∣ly, in a Segment greater than a Semicircle GDH, is less than a Right one; because its double at the Center GCH is less than two Right ones. All which we have already otherwise demon∣strated in Schol. Def. 10. N^o 6.

II. All the three Angles(β) 1.38 of any Triangle ABD taken together, are equal to two Right ones; because they are the half of the three at the Center C, which always make 4 Right ones, by Definit. 8. Consect. 2.

III. Therefore any external Angle IAB, is e∣qual to the two Internal opposite ones at B and D; because that, as well as they with the other contiguous to them BAD, make two Right ones, by Consect. 1. of the same Definit.

IV. And the greatest Side of a Triangle, because it insists on a(α) 1.39 greater Arch of a Circumscribed Circle, does also ne∣cessarily subtend a greater Angle, by vertue of Consect. 1. hereof.

Proposition XXXIV.

IN Equiangular Triangles (ACB and abc, Fig. 62.) the Sides ••¦bout the equal Angles are Proportional, viz. as AB to BC, so ab to bc, and as BC to CA so is bc to ca. &c. (β)

Demonstration.

For having described Circles thro' the Vertex of each Trian∣gle, according to Consect. 6. Definit. 8. by reason of the supp••¦sed equality of the Angles A and a, B and b, C and c, th•• Arches also AB and ab, &c. will necessarily agree in the num¦ber of Degrees and Minutes, by the foregoing 33 Prop, a•••• so also the Chords AB and ab, BC and bc, &c. will agree in th•••• number of Parts of the Radius or whole Sine ZA and za, 〈◊〉〈◊〉 Consect. 2. Definit. 10. Wherefore as many such Parts as A•••• has, whereof az has also 10000000, so many such also will a•• have, whereof az has also 10000000, &c. Therefore AC•• to CB as ac to cb, &c. Q. E. D.

CONSECTAYS.

I. WHerefore by the same necessity the Bases of such T••••¦angles AB and ab, will be proportional to their A••¦titudes CD and cd, as being Right Sines of the like Arches (〈◊〉〈◊〉 and cb, or rather CE and c e; and so for similar or like Tria••¦gles (and consequently also Parallelograms) we may rightly sup¦pose that their Bases are as a to ea, and their Heighths as b 〈◊〉〈◊〉 eb; tho we must not immediately conclude on the contrary, tha•• because their Bases and Altitudes are so, therefore they are S••¦milar.

II. As also in Similar Parallelepipeds it will be manifest 〈◊〉〈◊〉 any attentive Person, that the Bases are in a duplicate Propo••¦tion of the Altitudes. For since the Planes of Similar Solids a•••• equal in number, and Similar each to the other, if for A•• (Fig. 63.) we put a, and for BC b, AB will = ea and BC=eb; and so that Basis will be to this as ab to eeab. Moreover having let fall the Perpendiculars EH and EH, the Triangle

••BH and EBH are similar, and by putting c for BE, BE will be ec, putting also d for EH, EH will consequently be ed. But the Reason of the Base ab to the Base eeab, is du∣plicate of the Reason of d to e d, by Def. 34. Wherefore in Similar Parallelepipeds we may rightly suppose, that their Bases are as a b to eeab, or as a to eea, and their Altitudes as d to ed.

SCHOLIUM I.

FRrom this Proposition flows first of all the chiefest part of Trigonometry for the Resolution of Right Angled ▵▵: For since in any Right Angled Triangle, if one side, e. g. AB (Fig. 64.) be put for the whole Sine, the other BC will be the Tangent of the opposite Angle at A (and in like manner if CB be the whole Sine, BA will be the Tangent of the Angle C;) but if the Hypothenuse AC be made Radius or whole Sine, then the Side BC will be the Right Sine of the Angle A, or the Arch CD described from the Center A, and AB the Right Sine of the Angle C, or the Arch AE, described from the Center C, (we will omit mentioning the Secants, because the business may be done without them) which all follow from Def. 10. Wherefore you may find,

• I. The Angles.
  • 1. From the Sides by inferring As one leg to the other, so the whole Sine to the Tangent of the Angle opposite to the other Leg.
  • 2. From the Hy∣poth. & one side, by inferring As the Hyp. to the W.S. (whole sine) so the given leg to the S. of the opp. angle
• II. The Sides.
  • 1. From the Hy∣poth. and Angles: As the W. S. to the Hypoth. so the Sine of the Angle, opposite to the Leg sought, to the Leg it self.
  • 2. From one Leg and the Angles: As the W. S. to the given Leg, so the Tan∣gent of the Angle adjacent to it, to the Leg sought.
  • 3. From the Hypoth. and one of the Sides: Having first found the Angles, it's done by the 2, 1. or by the Pythagorick Theorem.

• III. The Hypothenuse.
  • 1. From the An∣gles and one of the Legs. As the S. of the Angle, opposite to the gi¦ven Leg, to that Leg, so the W. S. to the Hypoth.
  • 2. From the Legs given; Having first found the Angles its done by the 1. or by the Pythagorick Theorem.

III. Inversly also, if two Triangles ABC and ABC (〈◊〉〈◊〉 the Figure of the present Proposition) have one Angle of o•••• equal to one Angle of the other (e. g. B and B) and the Sid•••• that contain these equal Angles proportional (viz. as AB to B•• so AB to BC) then the other Angles (A and A, C and C will be also equal, and the Triangles similar(α) 1.40 for to 〈◊〉〈◊〉 like Chords AB and AB, BC and BC, there answer by t•••• Hypoth. like or similar Arches, i. e. equal in the number 〈◊〉〈◊〉 Degrees and Minutes; and to these also there answer equal A••¦gles both at the Periphery and Center.

IV. (Fig. 65. N^o 1.) If(β) 1.41 the Sides of the Angle BA•• are cut by a Line DE, parallel to the Base BC, the Segments 〈◊〉〈◊〉 those, Sides will be proportional, viz. AE to EC as AD to BD•• for by reason of the Parallelism of the Lines BE and BC, th•• Triangles ADE and ABC are Equiangular: Therefore as th•• whole BA to the whole AC, so the part AD to the part A•• and consequently also the remainder EC to the remainder D•• as the part EA to the part AD, by Prop. 26. and alternative•••• by Prop. 24. EC will be to EA as BD to AD.

SCHOLIUM II.

THere are several useful Geometrical Practices depend 〈◊〉〈◊〉 this Consectary and its Proposition. 1. That(γ) 1.42 where•• we are taught to cut off any part required, e. •• ⅓ from a given Line AB, and so generally to 〈◊〉〈◊〉 or divide any given Line AC, in the same pr••¦portion as any other given Line, is supposed 〈◊〉〈◊〉 be divided in D, (and consequently into as ma•••• equal parts as you please;) viz. if in the fi•••• Case, having drawn any Line AF, you take AD•• 1, and make DB 2, and having joined CB, dra••••

the Parallel LE: for as AD to DB so is AE to AC; that is, as 1 to 2, by this 4th Consect. therefore AE is one third of the whole AC, &c.

II. A Rule(α) 1.43 to find a third Proportional to the 2 Right Lines given AB and BC (N^o 2. Fig. 65.) (or a fourth to three given;) if, viz. having drawn AF at pleasure, you make AD equal to BC, and Joining DB, draw the Parallel EC: For as AB to BC, so AD i. e. BC) to DE. Now if AD be not equal to BC but to another (viz. a) third Proportional, then by the same Reason DE will be a fourth Proportional.

III. Another Rule(β) 1.44 to find a mean Proportional between two Right Lines given AC and CB; which is done by join∣••ng both the Lines together, and from the middle of the whole AB describing a Semicircle, and from C erecting the Perpendi∣cular CD: For since the Angle ADB is a Right one, by Con∣sect. 1. of the preceding Proposition, and the two Angles at C are Right ones, and those at A and B common to the whole Triangle ADB, and to the two partial ones ACD and BCD, ••hese two will be Equiangular and Similar to the great one, and consequently to one another: Therefore by the present Proposition, as AC to CD, so CD to CB, Q. E. D. and also as AB to BD so BD to BC, and as AB to AD so AD to AC, &c.

IV. The Analytical Praxis of multiplying and dividing Lines ••y Lines, so that the Product or Quotient may be a Line; and ••lso the way of Extracting Roots out of Lines: Which Des Cartes, gives us, p. 2. of his Geom. viz. assuming a certain Line ••or Unity, e. g. AB (in Fig. 65. N^o 2.) if AC is to be multi∣••lied by AD, having joined BD, and drawn the Parallel CE, ••he Product will be AE; for it will be as 1 to ••he Multiplier AD, so the Multiplicand AC to ••he Product AE; or if AE is to be divided ••y AC, having joined EC and drawn the Pa∣••allel BD, the Quotient will be AD; (for AC ••he Divisor, will be to AE the Dividend, as an ••nit AB to the Quotient AD;) all which are e∣••ident from the Nature of Multiplication and Division, and the

Precedent Praxes. As also taking CB (in the same Fig. N^o 3.) for Unity, if the Square Root is to be extracted out of any o∣ther Line AC, this being joined to your Unity in one Line AB▪ and having described thereon a Semicircle, the Perpendicular CD will be the Root sought, as being a Mean Proportional be∣tween the two Extremes CB and AC, according to Prop. 17.

V. A Right Line AG which divides(α) 1.45 any given Angle A into two equal Parts (Fig. 66.) being prolonged, divides the Base BC proportionally to the Legs of the Angle AB and AC For having prolonged CA to E, so that AE shall be = to AB•• the Angles ABE and AEB will be equal, by Consect. 2. Def. 13. and consequently also equal to each of the halves of the exter∣nal Angle CAB, by Consect. 3. of the antecedent Proposition Therefore the lines AG and EB will be parallel, by Cons. 1. Def. 11. Therefore as AC to AE, i. e. to AB, so GC to GB, by Co••¦sect. 3. of this Proposition. Q.E.D.

VI. Hence also there follows further, by conversion of th•• last inference, as AC+AB to AC, so GC+GB (i. e. BC) 〈◊〉〈◊〉 GC; and inversly GC to BC as AC to AC+AB; and lastly alternatively, GC to AC as BC to AC+AB.

N. B. This last Inference follows also immediately from the preceding Consectary. For by reason of the Similitude of the ▵▵ ACG and ECB, as GC to AC so BC to CE, i. e. to AC+AB.

SCHOLIUM III.

FRom these two last Consectarys there an•• these or two or three Practical Rules, t•••• first whereof shews, how having the two Legs A•• and AC given, and also the Base BC, to find the Se••∣ments GC and GB, made by the Bisection of the Intercrural Angle•• (viz. by this inference, according to Consect. 6: As the Sum 〈◊〉〈◊〉 the Sides to one Side (e. g.) AC;) so the Sum of the Segmen•• of the Base, i. e. the whole Base to one of the Segments, vi•• that next the said Side GC. 2. It shews on the contrary, how having the Base and one of its Segments given, and moreover the S•••• of the Sides, to find separately the Side AC next the known Segment

by inferring as the Sum of the Segments, or the Base BC to the Sum of the Sides, so the given Segment GC to the sought AC: or also, 3dly, Having only the Base and Sum of the Sides given, but not the Segment GC, yet to express its Proportion to the next side AC, — viz. in the Quantities of the given Terms, by putting (by Consect. 6.) for GC the value of the Base BC, and for AC the value of the Sum AB+AC; the great use of which last Rule will appear hereafter in the Cy∣clometry (or Quadrature of the Circle) of Archimedes.

VII. In any Triangle ABC (Fig. of the present Proposition) the Sides are to one another as the Sines of their opposite An∣gles: For they are as the Chords of the double Angles at the Center, by Prop. 33. therefore they are also one to another as half those Chords, i. e. by Definit. 10. as the Sines of the half Angles.

SCHOLIUM IV.

HEnce flow two new Rules of Plane Trigonometry, for Oblique-angled Triangles to find, viz.

• 1. The other Angles:
  • From 2 gi∣ven Sides, & an Angle op∣posite to one of them: by inferring As the Side opposite to the given Angle to the other Side, so is the Sine of the given Angle to the sine of the angle opposite to the other Side; which being given, the third is easily found.
• II. The other Sides:
  • From one side and the angles given, As the Sine of the Angle opposite to the given side, to that side; so is the Sine of the Angle op∣posite to the side sought to the side sought.

So that this way we have reduced all the Cases excepting one of Plane Trigonometry, and consequently all Euthymetry to their original Foundations (for in that Case of having two Sides, and the included Angle given, we may find the rest by

the Resolution of the Obliqueangled Triangle into two Right Angled ones; and so it's done by the Rules we have deduc'd in (Schol. 1.) I say, excepting one, in which from the three sides of an Obliqueangled Triangle given, you are required to find the Angles: the Rule to resolve which we will hereafter deduce in the 2d Consect. of Prop. 45. from that Theorem which Euclid gives us, lib. 2. Prop. 13.

VIII. Because in in the Right Angled ▵ BAC (Fig. 67) BC is to CA as CA to CD, by N^o 3. of the 2d Schol. of this Prop. the □ of CA will be = ▭ CE, by Prop. 17. In like manner because as CB to BA so is BA to BD; the □ of BA will be = to ▭ BE: Wherefore the two Rectangles BE and CE taken together, that is, the □ of the Hypothenuse BC, will be = to the two □'s BA and CA taken together: Which is the very Theorem of Pythagoras demonstrated two other ways in Schol. of Definit. 13.

SCHOLIUM V.

THis Theorem of Pythagoras as it furnishes us with Rules of adding Squares into one Sum, or subtracting one Square from another; so likewise it helps us to some Foundations where∣on, among the rest, the structure of the Tables of Sines relies, &c. Whose use we have already partly shewn in Schol. 1 and 4. 1. If several Squares are to be collected into one Sum, having joined the Sides of two of them so as to form a Right Angle, e. g. AB and BC (Fig. 68. N^o 1.) the Hypothenuse AC being drawn, is the Side of a Square equal to them both; and if this Hypothenuse AC be removed from B to D, and the Side of the third Square from B to E, the new Hypothenuse DE will be the Side of a Square equal to the three former taken together. 2. If the Square of the side MN (N^o 2.) is to be subtracted from the Square of the side LM. Having described a Semicircle upon LM, and placed the other MN within that Semicircle, then draw the Line LN and that will be the Side of the remain∣ing Square. 3. Having the Right Sine EG of any Arch ED gi∣ven (but how to find the Primary Sines we will shew in another place), you may obtain the Sine Complement CG or EF, by the preceding

Numb. viz. by subtracting the □ of the given Sine from the □ of the Radius; and moreover the versed Sine GD by subtract∣ing the Sine Complement CG from the Radius CD. 4. The Squares of the versed Sine GD, and of the Right sine EG being added together, give the □ of the Chord ED of the same Arch, (which all are evident from the Pythagorick Theorem) and half of that EH gives the Right Sine of half that Arch. 5. From the Right Sine EG you have the Tangent of that Arch, if you make, as the Sine Complement CG to the Right Sine GE, so the whole Sine CD to the Tangent G I. 6. Lastly, From these Data you may also have the Secants (if required) thus, as the Sine Complement CG to the W. S. CE, so the W. S. CD to the Secant CI; or as the Right Sine EG to the W. S. E.C. so the Tangent ID to the Secant IC; both which are e∣vident by our 34th Proposition.

Consect. 9. If the Quadrant of a Circle (CBEG, Fig. 70.) be inclined to another Quadrant (CADG) and two other Per∣pendicular Quadrants cut both of them, viz. FBAG and FEDG, and the latter do so in the extremities of them both) having let fall Perpendiculars from the common Sections E and B, thro' the Planes of the Perpendicular Quadrants, and the inclined Quadrant, (viz. on the one side EG and BH, as Right Sines of the Segments EC and BC; on the other EI and BK, as Right Sines of the Segments ED and BA) you'l have 2 Tri∣angles EIG and BKH Right Angled at I and K, Equiangular at G and H (by reason of the same inclination of the Plane CBEGC) and consequently similar, by our 34th Proposition; wherefore as the Sine EG to the Sine EI, so the Sine BH to the sine BK, or as EG to BH so EI to BK, and contrariwise.

SCHOLIUM VI.

HEnce you have several Rules of Spherical Trigonometry for resolving Right Angled ▵▵(α) 1.46 1. Having given in the Rightangled ▵ ABC the Hypothenuse BC and the Oblique Angle ACB, for the Leg AB op∣posite to this Angle, make: as the sine T (EG) to the sine of the Hypoth. (BH) so the sine of the given Angle (EI) to the sine of the Leg sought (BK). 2. Having given the Hypothe∣nuse

BC and the Leg AB for the opposite Angle ACB make as the sine Hypoth. (BH) to S. T. (EC) so the sine of the given Leg (BK) to the sine of the Angle sought (EI) 3. Having given the side AB and the Angle opposite to it ACB, for the Hypothenuse BC (supposing you know whether it be greater or less than a Quadrant) make as the sine of the given Angle EI to the sine T. (EG) so the sine of the given Leg (BK) to the sine of the Hypoth. BH). 4. Having given in the Right Angled ▵ EBF (which we take instead of ABC that so we may not be obliged to change the Figure) one Leg EB and the Hypothenuse BF for the other Leg EF, you may find its complement, if you make as the sine Complement o•• the given side (BH) to S. T. (EG) so the sine Compleme•••• of the Hypothenuse (BK) to the sine Compl. of the side sought (EI) 5. Having both Legs EB and EF given, for the Hy∣pothenuse BF its Compl. BA may be found thus: as S. T. (EG) is to the sine Cpmpl. (BH) of one side EB, so the sine Compl. (EI) of the other side (EF) to (BK) the sine Compl. of the Hypothenuse.

6. Having given in the same Right Angled Triangle EBF one Leg EF, and the Angle adjacent to it ETB, first prolong into whole Quadrants BA to f, that A f may = BF Hypoth. & BC to e that Ce may = EB, and AC to d that Cd may = D•• the measure of the given Angle EFB: secondly from d thro' e and f let fall a Quadrant thro' the extremities of the Quadrant Bf and Be, that so the ▵ C de may be Right Angled, in which there are given the Hypoth. Cd = to the given Angle, and the Angles C = to the Compl. of the given Leg (viz. to the Arch ED) and so, thirdly, there is sought the side de, as the Complement of the Arch ef, or of the Angle sought ABC, or EBF; viz. by the first case of this, by inferring, as S. T. to the sine Hypoth. cd (i. e. of the given Angle EFB;) so the An∣gle dce (i. e. DE the Compl. of the given Leg RF) to the sine de (as the Compl. of the Angle fBe or EBF.

7. Having given, in the same Triangle, the side EF and the opposite Angle EBF (i. e. the Arch ef) for the other Angle EFB (that is the Hypoth. cd in the ▵ cde) make by the third of this:

As the Sine of the Angle dce (i. e. the sine Compl. of the gi∣ven Leg DE) to the S. T. so the sine of the Leg de (i. e. the

fine Compl. of the Angle EBF) to the Hypoth. cd (i. e. the sine of the Arch DA or Angle EFB.).

8. Having the Oblique Angles given to find either of the sides, viz. EF; which may be done thus by the second of this:

As the sine of the Hypoth. cd (i. e. the sine of the Angle at F) to the W. S. so the sine de (i. e. the sine Compl. of the Angle at B) to the sine of the Angle dce (i. e. the sine Compl. of the side sought EF.)

Consect. 10. The same being given as in Consect. 7. if instead of the Right Sines EI and BK, you erect Perpendicularly DL and AM(Fig. 71) because of the similitude of the Triangles DGL and AHM, you'l have, as DG sine T. to DL the Tangent of the Arch DE, so AH the Right Sine of the Arch AC to AM the Tangent of the Arch AB; or as DG to AH, so DL to AM, and contrariwise.

SCHOLIUM VII.

HEnce flow the other Rules of Spherical Trigonometry for Re∣solving Right Angled Triangles, viz. 9. Having given the side AC in the ▵ ABC, and the adjacent Angle ACB, for the other side AB, make as the W. S. (DG) to the sine of the given side (AH) so the Tangent of the given Angle (ACB) to the Tangent of the Angle sought (AB.) 10. Having given the side (AB) and the opposite Angle (at C) for the other side (AC, so you know whether it be greater or less than a Quadrant) make as the Tangent of the given Angle (DL) to the Tangent of the given Leg (AB) so the whole S. (DG) to the sine of the Leg sought (viz. at AH.) 11. Both sides being given, for the Angles, make, as the sine of one Leg (AH) to the W. S. (DG) so the T. of the other Leg (AM) to the Tangent of the Angle opposite to the same (at C.) 12. Having given moreover in the Right Angled Triangle EBF the Hypothenuse (BF) and the Angle (EFB) for the adjacent side EF, make, as the sine Compl. of the given Angle (AH) to the W. S. so the Tangent Compl. of the Hypoth. (AM) to the Tang. Compl. of the Leg sought (DL) 13. Having given the side (EF) and the adjacent Angle F for the Hypoth. BF make; as the W. S. to the sine Compl. of the given Angle (AH) so the Tangent Complement of the given Leg (DL)

to the Tangent Compl. of the Hypoth. (AM.) 14. Having given the Hypoth. (BF) and one side EF for the adjacent An∣gle (F) make as the Tang. Compl. of the given Leg (D. L.) to the W. S. so the Tang. Compl. of the Hypoth. (AM) to the Sine Compl. of the Angle sought (AH). 15. Having gi∣ven the Hypoth. (BF, i. e. the arch Af, or the angle at d) and either of the oblique angles (at F) for the other angle (EBF) make by help of the new Triangle cde, by the 12th of this.

As the Sine Compl. of the angle cde (i. e. the Sine Compl. of the Hypoth. (AH) to the W. S. so the Tang. Compl. of the Hypoth. cd (i. e. Tang. Compl. of the given angle) to the Tang. Compl. of the Side de (i. e. to the Tang of the angle sought ABC or EBF.)

16. Having given the oblique Angles to find the Hypoth. (BF, or the arch Af, or the angle cde) it is done by the 14 of this Schol.

As the Tang. Compl. of the Leg de (i. e. the Tangent of the angle ABC or EBF) to the W. S. so the Tangent Com∣plement of the Hypoth. cd (i. e. the Tangent Complement of the other angle EFB) to the Sine Compl. of the angle cde (i. e. the Sine Compl. of the Hypoth. BC sought.)

So that now we have with Lansbergius (but much more com∣pendiously) Scientifically Resolved all the Cases of Rigt-angled Triangles; the Resolution of Oblique-angled ones only now remaining.

Consect. 11. In Oblique-angled Spherical Triangles, as well as Right-angled ones, the Sines of the angles are directly proportio∣nal to the Sines of the opposite Sines. 1. Of the Right-angled ones this is evident from N^o 3. Schol. 6. and from the 9th Consect. For as the Sine of the angle A (Fig. 72.) to the Sine of BD, so the W. S. (i. e. of the angle D) to the Sine of AB. 2. The same is immediately evident of an Oblique-angled Tri∣angle ABC, resolved into 2 Right-angled ones. For,

The Sine of the angle C is to the Sine of BD as the sine of the gle D to the Sine of AB; and also,

The Sine of the angle C to the Sine of BD as the Sine of the angle D to the sine of BC, by the 1.

In each Proportionality the means are the Sines of BD and D; therefore the Rectangles of the Extremes of the Sines of AB in∣to the Sine of A, and the Sine of BC into the Sine of C, will be equal among themselves, since the Rectangles of the same Means are equal, by Prop. 18. therefore by Prop. 19. as the Sine of A to the Sine of BC, so the Sine of C to the Sine of AB, Q. E. D.

SCHOLIUM VIII.

THe latter may appear of Oblique-angled Triangles after this way also; since the Sine of the angle A is to the Sine of BD as the Sine of the angle D to the sine of AB, call the first a, the second ea, the third b, the fourth eb; and because the Sine of the angle C (which we call c) is likewise to the Sine of BD (i. e. to ea) as the Sine of D (i. e. b) to the Sine of BC (which will consequently be 〈 math 〉〈 math 〉) it will be manifest, that the Sine of the angle

A is to the sine of BC as the sine of the angle C to the sine of AB, i. e. as a to .. 〈 math 〉〈 math 〉, i. e... c to .. eb. by multiplying the Means and Extremes, whose Rectangles are on both sides eab. Therefore as by the present and precedent Consectary 7, it is universally true, That in any Triangle whether Right Lined or Spherical, Right-Angled or Oblique-angled, the Sides or their Sines, are to one another, as the Sines of their opposite Angles (which therefore is commonly called a Common Theorem:) so also hence flow 2 new Rules of Spherical Trigonometry for Oblique-angled Triangles, like those we found in Schol. 4.

To find

• I. The other Angles.
  • From 2 sides given of an an∣gle opposite to one of them, by inferring As the sine of the side opposite to the gi∣ven angle to the sine of the other side, so the sine of the given angle to the sine of the angle sought.

• II. The other Sides.
  • From one side and the angles given, by inferring As the sine of the angle opposite to the given side to the sine of that side, so the sine of the angle opposite to the side sought to the sine of the side sought.

And thus we have reduced all the Cases and Rules of Sphe∣rical Trigonometry to their original Fountains (for from 2 Sides given and the interjacent Angle, or 2 Angles and their adjacent side, we may find the rest in Oblique-angled Triangles, by resolving them into 2 Right-angled ones; and so by the Rules we have deduc'd in Schol. 6 and 7) excepting two Cases, viz. when from 3 sides given, the Angles, or from 3 Angles the Sides are sought; to resolve which, we are supplied with Rules from the following

Cons. 12. In the given Oblique angled Spherical Triangle ABC (Fig. 73.) whose Sides are unequal and each less than a Quadrant, having produced the sides AB and AC to the Quadrant AD and AE, and effected besides what the Figure directs, then will

The Arch DE be the Mea∣sure of the angle A, AF=AC, and so FB the difference of the Sides AB and AC.

BC=BG, and so GF the difference of the third side, and the differences of the rest FB.

But now, 1. As EH or DH to CM or FM so will PH be to NM (by reason of the Equi∣angular Triangles EPH and CNM;) therefore by Prop. 26. so will also DP be to FN. Make therefore DH=a FM=ea, DP=b FN=eb.

AI the R. Sine of the side AB. GM the Sine of the Side AC GL the Sine of GB, or of the side BC.

FK the R. Sine of the Arch FB. BI the versed Sine of AB.

BL the vers. Sine of GB or BC. BK the versed Sine of FB. KL or NO the difference of the versed Sines we have now men∣tioned.

EP the Right Sine and DP the versed Sine of the arch DE.

CN the R. Sine and FN the ver∣sed Sine of the arch FC.

[illustration]

Pag. 108.

73

74

75

76 n. 1.

76 n. 2.

77 n. 1.

77 n. 2.

2. By reason of the Equiangular ▵▵ FNO and HAI (for FNO is Equiangular to the ▵ FKQ, and that to the ▵ HQM, by reason of the Vertical Angles at Q; and that also to the ▵ HAI by reason of the Common Angle at H) and you have also,

As HA,

Or DH to AI, so FN to NO.

〈 math 〉〈 math 〉.

Wherefore now, 3. you'l have evidently, the □ DH to FM into AI as DP to NO.

〈 math 〉〈 math 〉. DH NO DP.

And Inversly as oea to a so oeb to b.

SCHOLIUM IX.

SInce therefore the Radius DH or a is known, and also NO the Difference of the versed Sines BL and BK, it is evident, that DP the versed Sine of the angle A will be known also; supposing that the first Quantity oea is likewise known. But this may be had by another Antecedent Inference, if you make, as AH to FM so AI to a fourth oea.

a ea oa

Hence therefore arises, 1. the Rule: Having given the 3 Sides of an Oblique-angled Triangle, to find any one of the Angles, viz. by inferring,

1. As the Sine of T to the Sine of R, one of the sides com∣prehending AC; so the sine of the other side AB to a fourth, 〈 math 〉〈 math 〉

2. As this fourth to the sine of T, so the difference of the versed sines of the third side, BC, and the differences of the o∣thers to the versed sine of the Angle sought, viz.

〈 math 〉〈 math 〉.

But since the sides of a Spherical Triangle may be changed

into Angles, and contrariwise the sides being continued [as 〈◊〉〈◊〉 the side AB of the given Triangle ABC (Fig. 47.) be con••¦nued a Circle, the rest into Semicircles from the Poles b and 〈◊〉〈◊〉 and likewise the Semicircle HI from the Pole A, and the Sem••¦circle FG from the Pole B, and the Semicircle EA from th•• Pole C, you'l have a new Triangle a, b, c, the 3 angles o•• which will be equal to the 3 sides of the former ABC; as th•• angle a or its measure IG, is equal to the side AB, by reaso•• each makes a Quadrant joined with the third arch AG; b•• the measure of the angle b, is the side AC (viz. in this cas•• wherein the side AC is a Quadrant, in the other wherein 〈◊〉〈◊〉 would be greater or less than a Quadrant, it would be th•• measure of the angle of the Compl. for then the Semicircle H•••• described from the Pole A, would not pass thro' C but beyon•• or on one side of C. See Pitisc. lib. 1. Prop. 61. p. m. 25.) — the angle c or its measure KL, is equal to the 〈◊〉〈◊〉 BC, because with the third KC they make the Quadrants BK and CL] Therefore, 2. Having given the three Angles 〈◊〉〈◊〉 the Oblique-angled Triangle abc, you may find any side, e. g. ac, if there be sought the Angle ABC, or rather its Comple∣ment KBF, or its measure FK=ac, — from th•• 3 sides given of the ▵ ABC, by the preceding Rule, by infer∣ring, viz. 1. As S. T. to the sine R, of one side comprehending the angle of one side AB (i. e. of one angle a adjacent to the side sought) so the sine of the other side BC (i. e. of the other an∣gle C) to a fourth.

2. As the fourth to the S. T. so the difference of the ver∣sed Sines of the third side AC, and the differences of the others (i. e. of the 3d angle b, and the differences of the rest) to the versed Sine of the comprehended angle, or Complement to a Semicircle (i. e. of the side sought ac.)

Proposition XXXV.

SImilar Plane Figures (α) are to one another in Duplicate Proportion of their Homologous Sides.

Demonstration.

For, 1. the Bases of 2 similar Triangles or Parallelograms

for which any 2 Homologous Sides, e. g. AB and AB (Fig. 75.) may be taken) and Perpendiculars let fall thereon DE and DE, will be by Consect. 1. Prop. 34, as a to ea, b to eb. Therefore the Parallelograms and Triangles themselves, will be as ba to eeba, by Consect. 7 and 8. Def. 12. i. e. by Def. 34. in duplicate Reason of their Perpendiculars or assumed Sides, which is most conspicuous in Squares, which putting a for the ••ide of one, and ea for the other, are to one another as aa to ••eaa.

2. Like Polygons are resolved into like Tri∣••ngles, when the Triangles ABC and ABC,(α) 1.47 and ••lso AED and AED are Equiangular, by Con∣sect. 3. Prop. 34. but CAD and CAD, are also Equiangular, because each of their angles are the remainder of ••qual ones, after equal ones are taken from them. Wherefore ••he first Triangles are in duplicate Proportion of the sides BC ••nd BC; the second likewise of the sides CD and CD the ••hird are also in the same Proportion of the sides DE and DE, &c. i. e. (since by the Hypoth. BC has the same reason to BC ••s CD to CD, and DE to DE) each to each is in duplicate Pro∣portion of the sides BC to BC, or CD to CD, by the first of ••his. Therefore by a Syllepsis, the whole Polygons are in du∣plicate Proportion of the same Sides: Which is the second thing ••o be demonstrated.

3. Circles and their like Sectors, are as the Squares of their Diameters, by Prop. 32. therefore in duplicate Proportion of them, by the first of this: Which is the third thing: Therefore simi∣••ar Plane Figures, &c. Q. E. D.

CONSECTARYS.

THerefore 2 similar Plane Figures are one to another, as the first Homologous Side, to a third Proportional, by vertue of Definition 34.

II. Any two Figures described on 4 Proportional Lines(α) 1.48 and similar to 2 others, are likewise Proportional, and contrariwise; for if the simple Reasons or Proportions of Lines be the same, their duplicate Proportions will be the same also, and reciprocally.

SCHOLIUM.

BUT as this second Consectary confirms Prop. 22 and i•• Scholium, so the first teaches us a twofold Geometric•• Praxis. 1. A Way to express the Proportion of similar F••¦gures by two Right Lines, viz. by finding a third Proportio¦nal to their Homologous Sides. For as the side of the first 〈◊〉〈◊〉 this third, so will be the first Figure to the second. 2. A wa•• to augment or diminish any given Figure in a given Reaso•• or Proportion, viz. by finding a mean Proportional between a•••• side of the given Figure, and another Line which shall be 〈◊〉〈◊〉 that in a given Proportion, and then by describing thereon 〈◊〉〈◊〉 similar or like Figure.

Proposition XXXVI.

SImilar or like Solid Figures, are to one another in triplicate Pr••¦portion of their Homologous Sides.

Demonstration.

For, 1. The similar Bases of two similar Parallelepipedo•• (and consequently also of Prisms and Cylinders, by Consect. •••• and 5. Definit. 16. and also of Pyramids and Cones, by Consect 3 and 4, of Definit. 17.) are, as ab to eeab, by (α) the pr••¦ceding Proposition, and their Altitudes as c to ec, by Consect. •• Prop. 34.

Therefore Parallelepipeds, Cylinders and Prisms (and so the thi•••• part of these, Cones and Pyramids) will be as abc to e{powerof3} abc, by Co••¦sect. 3, 4, 5. Definit. 16. i. e. they will be, by Definit. 34. •••• Consect. 1 and 2. Prop. 34. in Triplicate Proportion of the•• Perpendiculars or Homologous Sides. Which is especially Co••••¦spicuous in Cubes; which, putting a for the Side of one, an•• ea for the side, are to one another as a{powerof3} to e{powerof3} a{powerof3}.

2. Polyedrous or many sided Figures, may be resolved in•• Pyramids of similar Bases and Altitudes; which is evident 〈◊〉〈◊〉 Regular ones, from the Consect. of Definit. 21. and cannot b•• difficult to understand also of Irregular ones; because the lik••

••nclination of their Planes every where similar and equal in number, necessarily require that the whole Altitudes of similar Polyedrous Solids, as well as similar Parallelepipedons, by Con∣sect. 2. Prop. 34. should be in subduplicate Proportion of their Bases, and so these being likewise divided in C and C (Fig. 76. N^o 1.) the parts of their heigths GC and GC, will be in the same Proportion: Whence, e. g. 2 Pyramids standing on simi∣••ar Bases ABDEF and ABDEF, and having like Altitudes GC and GC, will necessarily be like or similar; and the same ••hing may be likewise judged of others.

Or yet, to shew it more evidently, the Polyedrous Solids may be resolved into like Triangular Prisms; for, e. g. each of the Trian∣gles of their similar Bases ABDEF and ABDEF (N^o 2.) are similar, viz. abf and abf, ABF and ABF, by the pre∣ceding Prop. N^o 2. The Planes ABba and ABba, also AafF and A af F, are similar by the Hypoth. and consequently also the Planes BbfF and BbfF (fb is to ba as fb to ba, and also in the one ba to bB, as ba to bB in the other; therefore ex equo as fb to bB so f b to bB, &c.) and so the whole Triangular Prisms will be similar, by Definit. 35. and so of others. There∣fore similar Polyedrous Solids will be in the same Proportion as similar Pyramids, or Triangular Prisms, i. e. by the first of this in Triplicate Proportion of their Sides.(α) 1.49

3. Spheres are as the Cubes of their Diame∣ters(β) 1.50 by Consect. 3. Prop. 32. Therefore they are by the first of this, as a{powerof3} to e{powerof3} a{powerof3}. Therefore similar or like Solids are in Triplicate Proportion of their Homologous Sides. Q. E. D.

CHAP. VI. Of the Proportions of Magnitudes of divers sorts com∣pared together.

Proposition XXXVII.

THE Parallelogram ABCD (Fig. 77. N. 1.) is to the Triangle BCD upon the same base DC, and of the same heighth as 2 〈◊〉〈◊〉 1. This has been already Demonstrated in Consect. 3. De••••∣nit. 12. Here we shall give you another

Demonstration.

Suppose, 1. the whole Base CD divided into four equ•••• parts by the transverse Parallel Lines EG, HK, LN, then w•••• (by reason of the similitude of the ▵▵ DGF, DKI, DN•• DCB) GF be 1, KI 2, NM 3, CB 4; and having further∣more continually Bisected the Parts of the Base, the Indivisible or the Portions of the Lines drawn transversly thro' the Trian∣gle will be 1, 2, 3, 4, 5, 6, 7, 8, &c. ad infinitum, all a∣long in an Arithmetical Progression, beginning from the Poi•••• D, as 0; to which the like number of Indivisibles always an∣swer in the Parallelogram equal to the greatest, viz. the Li•••• BC. Wherefore by the 4th Consect. of Prop. 16. all the In••••¦visibles of the Triangle, to all those of the Parallelogram take•••• together, i. e. the Triangle it self to the Parallelogram, is as 〈◊〉〈◊〉 to 2. Q. E. D.

SCHOLIUM.

NOW if any one should doubt whether the Triangle 〈◊〉〈◊〉 Parallelogram may be rightly said to consist of an in••••¦nite number of Indivisible Lines, he may, with Dr. Wa••••••

[illustration]

Pag. 115.

78

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81

82

83

84

85

86

87

88

instead of Lines, conceive infinitely little Parallelograms of the same infinitely little Heighth, and it will do as well. For ha∣ving cut the Base (N^o 2) into 4 equal parts by transverse Pa∣rallels, there will be circumscribed about the Triangle so many Parallelograms of equal heighth, being in the same Proportion as their Bases, by Prop. 28. i. e. increasing in Arithmetical Progression. In the following Bisection, there will arise 8 such Parallelograms approaching nearer to the Triangle, in the next 16, &c. so that at length infinite such Parallelograms of in∣finitely less heighth, and ending in the Triangle itself, will constitute or make an infinite Series of Arithmetical Propor∣tionals, beginning not from 0 but 1; to which there will an∣swer in the Parallelogram infinite little Parallelograms of the same heighth, equal to the greatest. Whence it again follows, by Consect. 9. Prop. 21. that the one Series is to the other, i. e. the Triangle to the Parallelogram as 1 to 2; which being here thus once explained, may be the more easily applied to Cases of the like nature hereafter.

CONSECTARYS.

I. SInce in like manner in the Circle (Fig. 79.) the Periphe∣rys at equal intervals from one another, as so many Ele∣ments of the Circle, increase in Arithmetical Progression; the Sum of these Elements, i. e. the Circle it self will be to the Sum of as many Terms equal to the greatest Periphery, i. e. to a Cylindrical Surface, whose Base is the greatest Periphery, and its Altitude the Semidiameter, as 1 to 2.

II. Hence the Curve Surface of a Cylinder circumscribed a∣bout a Sphere, i. e. whose Altitude is equal to the Diameter, is Quadruple to its Base.

III. Also the Sector of the Circle bac, to a Cylindrical Sur∣face, whose Base is the Arch bc, but its Altitude the Semidia∣meter ac, is as 1 to 2.

IV. And because the Surface of the Cone BCD is to its cir∣cular Base, as BC to CA, i. e. as the √2 to 1. by Schol. Prop. 17. the Cylindrical Surface, the Conical Surface and the Cir∣cular

one we have hitherto made use of will be as 2, √2 and 1, and consequently continually Proportional.

SCHOLIUM

ALL which may also abundantly appear this way, viz. by putting for the Diameter of the Circle a, for the Semidiameter ½ a and for the Circumference ea, you'l have the Area of the Circle ¼ eaa by Consect. 1 Definit. 31. and Multi¦plying ••he heighth of the Cylinder AB i. e. ½ a by the Periph•• ea yo••'l have the Cylindrical Surface ½ eaa by Consect. 6. Definit 1•• as now is evident also by Consect. 1, 2 and 3. Now if y•••• would also have the Surface of the Cone, since it's side by the Pythagorick Theorem is 〈 math 〉〈 math 〉 and the half of that 〈 math 〉〈 math 〉 i. e. (b•• N^o 2 of Schol. Prop. 22.) 〈 math 〉〈 math 〉 and this half being multi¦plied by the Periphery of the Base ea, you'l have (by virtue o•• Consect. 4. Definit. 1.8.) the Su••f••ce of the Cone 〈 math 〉〈 math 〉 i. e. (b•• the Schol. just now cited) 〈 math 〉〈 math 〉: So that now appears also th•• 4th Consect. of this; because the Rectangle of those Extreme•••• eaa and ¼ eaa is ⅛ eaa{powerof4} as well as the square of the mean.

Proposition XXXVIII.

A Parallelepiped(α) 1.51 BF (Fig. 77. N^o 3.) is to a Pyra•••••• ABCDE upon the same Base BD and of the same heighth, 〈◊〉〈◊〉 3 to 1. This was D••monstrated in Consect. 3. Definit. 17. bu•• here we shall give you another.

Demonstration.

Suppose 1 the whole Al••itude BE divided into 3 equal Pa•••• by transverse P••ains Parallel to the Base, then will (by reason 〈◊〉〈◊〉 the Similitude of the Pyramids abcd ACBD•• and AECDE) the Bases abcd ABCD and ABC•• be by Consect. 2. Prop. 34 and Consect. 3. Defi•••••• 17 in duplicate Proportion of the Altitudes i. e. in duplicate Arithmetical Progression 1, •• 9, mo••eover 2, bisecting the parts of the Altitude, the qu••¦d••angular Sections now double in Number (as the Indivisibles o•• Elements of the proposed Pyramid) will be as 1, 4, 9, 16, 2••

36, &c. ad Infinitum, all along in a duplicate Arithmetical Pro∣portion; while in the mean time there answer to them as many Elements in the Parallelepiped equal to the greatest ABCD, wherefore by Consect. 10. Prop. 21. all the Indivisibles of the Pyramid taken together will be to all the Indivisibles of the Parallelepiped also taken together, i. e. the Pyramid it self to the Parallelepiped, as 1 to 3. Q. E. D.

CONSECTARY.

THis Demonstration may be easily accommodated to all other Pyramids and Prisms, and also Cones and Cylinders,(α) 1.52 since here also (Fig. 78.) the circular Planes ba, BA, and BA are as the squares of the Diameters, and so as 1, 4, 9. and so likewise all the other Elements of the Cone by continual bisection are in duplicate Arithmetical Progression; when in the mean time there answer to them in the Cylinder as many Ele∣ments equal to the greatest BA &c.

Proposition XXXIX.

A Cylinder is to a Sphere inscribed in it i. e. of the same Base and Altitude as 3 to 2.

Demonstration.

Suppose 1 (Fig. 80.) the half Altitude GH (for the same pro∣portion which will hold when demonstrated of the half Cylinder AK and Hemisphere AGB, will also hold the same of the whole Cylinder to the whole Sphere) to be divided into 3 equal parts, then will AH, C1, E2, be mean proportionals between the Segments of the Diameter by Prop. 34 Schol. 2 N^o 3, and so by Prop. 17. the Rectangles LHG, L1 G, L2 G equal to the Squares AH, C1, E2, being in order as 9, 8 and 5. and also 2dly, having bisected the former parts of the heighth, the six Squares cutting the Sphere Cross ways will be found to be as 36, 35, 32, 27, 20, 11. &c. in the progression we have shewn at large in Consect. 12. Prop 21. Wherefore since all the Indivisibles of the Hemisphere, viz. the circular Planes answering to the Squares of the said ••aniverse

Diameters have the same proportion of Progression, by Pr•••• 32. and there answer to them the like number of Elements i•• the Cylinder equal to the greatest AH: All thes•• these taken together will be to all the other take•• together i. e, the whole Cylinder AK to the whol•• Hemisphere AGB by vertue of the aforesaid Co••¦sect. 12. as 3 to 2(α) 1.53 Q. E. D.

SCHOLIUM. I.

HOnora••us Fabri elegantly deduces this Prop. a priori, in a g••¦netick Method in his Synopsis Geom. p. 318. (which a•• Carotus Renaldinus performs from the same common Foundationi•• lib. 1. de Compos. and Resol. p. 301, and the following, but aft•••• a more obscure way and from a demonstration further fetch'd Fabri's is after this Method: The whole Figure (81) AL being turned round about BZ, the Quadrant ADLBA will describe a Hemisphere, the Square AZ a Cylinder and the triangle BM•• a Cone all of the same Base and Altitude. Since therefore Circl•• are as the squares of their Diameters by Prop. 32. and the Squa•• of GE=to the Squares of GD and GF taken together (f•• the Square of GF i. e. GB×□ GD is = □ BD or BA or G•• by the Pythag. Theor.) and so the Circle described by GE •••• be = to 2 Circles described by GD and GF taken together then taking away the Common Circle described by GF there w•••• remain the circle described by GF within the Cone equal t•• the Annulus or Ring described by DE about the Sphere. A•• since this may be demonstrated after the same way in any othe•• case, viz. that a circle described by g f, will be equal to an A••¦nulus described by de; it will follow, that all Rings or An•••• described by the Lines DE or de (i. e. all that Solid that conceived to be described by the trilinear Figure ADLM turne round) will be equal to all the Circles described by GF or gf (i•• to the Cone generated by the Triangle BLM;) and so as th•• Cone i•• ⅓ part of the Cylinder generated by AL, by the C••••sect. of Prop. 38. so also the Solid made by the Triline•• ADLM (viz. the Excess of the Cylinder above the Sphere will be ⅓ of the Cylinder, and consequently the Hemisphere Q. E. D.

CONSECTAYS.

HEnce you have a further Confirmation of Consect. 2. Prop. 32. and Prop. 36. N. 3.

II. Hence also naturally flows a Confirmation of Consect. 2. Definit. 20. and consequently the Dimension of the Sphere both as to its solidity and Surface. For putting a for the Dia∣meter of the Sphere and circumscribed Cylinder, and Ea for the Circumference, the Basis of the greatest Circle will be ¼ eaa, and ••hat multiplied by the Altitude, gives ¼ ea{powerof3} for the Cylinder. Therefore by the present Proposition, ⅙ ea{powerof3} gives the Solidity of the Sphere (by making as 3 to 2 so ¼ to ⅙) This divided by ⅙ a, will give, by vertue of Consect. 1. of the aforesaid Def. 20. and Consect. 3 Definit. 17. the Surface of the Sphere eaa.

III. Therefore the(α) 1.54 Surface of the Sphere eaa, is manifest∣ly Quadruple of the greatest Circle ¼ eaa.

IV. The Surface of the Cylinder, without the Bases, made by multiplying the Altitude a by the Circular Periphery of the Base ea, will be eea, equal to the Surface of the Sphere.

V. Adding therefore the 2 Bases, each whereof is ¼ eaa, the whole Surface of the Cylinder 1 ½ eaa, will be to the Sur∣face of the Sphere eaa as 3 to 2.

VI. The Square of the Diameter aa to the Area of the Cir∣cle ¼ eaa, is as a to ¼ ea, i. e. as the Diameter to the 4th part of the circumference.

VII. A Cone of the same Base and Altitude with the Sphere and Cylinder, will be by Consect. 2. of this, Prop. and the Con∣sect. of Prop. 38. 〈 math 〉〈 math 〉 ea{powerof3}, and of the Cylinders ¼ or 〈 math 〉〈 math 〉 ea{powerof3}. There∣fore a Cone, Sphere, and Cylinder, of the same heighth and dia∣meter, are as 1, 2, 3. The Cone therefore is equal to the Excess of the Cylinder above the Sphere; as is otherwise evident in Scholium 1. of this.

SCHOLIUM II.

AND thus we have briefly and directly demonstrated the chief Propositions of Archimedes, in his 1. Book de Sphaer•• & Cylind. which he has deduced by a tedious Apparatus, and 〈◊〉〈◊〉 ••••directly. And now if you have a mind to Survey the 〈◊〉〈◊〉 and more perplext way of Archimedes, and compare it with this shorter cut we have given you; take it thus: Archimedes thought it necessary first of all to premise this Lemma; Th•••• all the Conical Surfaces of the Conical Body made by Circum∣volution of the Polygon, or many angled Figure A, B, C, D, E, &c. (Fig. 82.) inscribed in a Circle, according to Definit. 19. I say, those Conical Surfaces taken all together, will be equal to a Circle, whose Radius is a mean Proportional between the Diameter AE and a transverse Line BE, drawn from one ex∣tremity of the Diameter E to the end of the side AB next to the other extremity. This we will thus demonstrate by the help of specious Arithmetick: Since BN, HN are the Right Sines of equal Arches, CK and CK whole Sines, &c. and the Lines BH, GC, &c. parallel; having drawn obliquely the transverse Lines HC, GD, all the angles at H, C, G, D, &c. will be equal by Consect. 1. Definit. 11. and consequently all the Tri∣angles BNA, HNI, ICK, &c. equiangular, both among them∣selves, and to the ▵ ABE; since the angle at B is a Right one, by Consect. 1 Prop. 33. and the angle at A common with the ▵ BNA. Wherefore as

• BN to NA
• or HN to NI

so

• CK to CI.
• & GK to KL.

and so

• DM to ML
• FM to ME

so EB to BA; and so by making BN, HN, DM, FM=a CK and GK=b, EB=c, for NA, NI, ML and ME, you may rightly put ea for KI and KL eb for AB, ec. Which being done you may easily obtain the Conical Surfaces of the inscrib'd Solid, and the Area of a Circle whose Radius shall be a mean Proportional between AE and FB, and it will be evidently manifest, that these two are equal. For, 1. (for Conical Surfaces) the Diameter of the Base BH=2a, and the side of the Cone AB=ec: Therefore (making here o the name of the Reason between the Diameter and Circumference) the circumference will be 2 oa

which multiplied by half the side ½ ec, gives the Conical Surface oaec, by Consect. 4. Definit. 18. And since the Circumference BH is as before 2 oa, and the circumference CG = 2ob, half of the sum 2oa+2ob, viz. oa+ob is the equated Circum∣ference: which multiplied by the Side BC=ec gives the Sur∣face of the truncated Cone BHGC=oaec−obec, by Con∣sect 5. of the aforesaid Def. since, lastly, the Surface of the truncated Cone DFGC is equal to one, and likewise the conical Surface EDF to the other, by adding you'l have the Sum of all 40oaec+2obec.

2. (for the Area of the Circle) the Diameter AE is =4ea+2eb, and BE=c: the Rectangle of these is =4eac+2 ebc = (which also is evidently equal to the Rectangle of all the transverse Lines BH, CG, DF into the side AB, as Ar∣chimedes proposes in the matter) = to the Square of the Radius in the Circle sought, because the Radius is a mean Proportio∣nal between AE and BE, and so equal to 〈 math 〉〈 math 〉, so that the whole Diameter is 〈 math 〉〈 math 〉. There∣fore, 2. the circumference of this Circle will be 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉: which multiplyed by half the Semidiameter, i. e. by 〈 math 〉〈 math 〉 gives the Area of the Circle sought 〈 math 〉〈 math 〉. But this Root extracted is 〈 math 〉〈 math 〉, equal to the superiour Sum of the conical Surfaces. Q. E. D.

Having thus demonstrated the Lemma, we will easily demon∣strate with Archimedes (tho not after his way) That the Sur∣face of any Sphere, is Quadruple of the greatest Circle in it, which is already evident from the 3d Consect. For since all the conical Surfaces of the inscribed Solid taken together, by the preceding Lemma, are equal to the Area of a Circle whose Ra∣dius is a mean Proportional between the Diameter AE and the Transverse EB; and this mean Proportional approaches always so much nearer to the Diameter AE, and those Surfaces so much nearer to the Surface of the Sphere, by how many the more sides the inscribed Figure is conceived to have, by Con∣sect. 1 and 2. Def. 18. if you conceive in your mind the Bi∣section of the Arches AB, BC, &c. to be continued in Infinitum, it will necessarily follow, that all those conical Surfaces will at length end in the Surface of the Sphere it self, and that mean

Pro∣portional in the diameter AE, and so the Surface of the Sphere will be equal to a Circle, whose Radius is the Diameter AE, But that Circle would be Quadruple of the greatest Circle in the present Sphere, by Prop. 35. Therefore the Surface of the Sphere is Quadruple of that Circle also. Q. E. D.

Hence also it would be very easie to deduce with Archime∣des (tho again after another way) that celebrated Proposition, which we have already demonstrated from another Principle in the Prop. of this Schol. viz. That a Cylinder is to a Sphere of the same Diameter and Altitude, as 3 to 2. For by putting a for the Diameter and Altitude, and ea for the Circumference, the Area of the Circle, will be ¼ eaa; and this Area being multiplied by the Altitude a, gives ¼ ea{powerof3} for the Cylinder, by Consect. 5. Definit. 16. and the same Quadruple, i. e. eaa mul∣tiplied by ⅙ a gives ⅙ ea{powerof3} for the Sphere, by Consect. 1. Definit. 20. and Consect. 3. Definit. 17. Wherefore the Cylinder will be to the Sphere as ¼ to ⅙, i. e. in the same Denominator as 〈 math 〉〈 math 〉 to 〈 math 〉〈 math 〉, i. e. as 6 to 4, or 3 to 2. Q.E.D.

Whence it is evident, that the Dimension of the Sphere would be every ways absolute if the Proportion of the Diame∣ter to the Circumference were known; which now with Ar∣chimedes, we will endeavour to Investigate.

Proposition. XL.

THE Proportion of the Periphery of a Circle(α) 1.55 to the Diame∣ter, is less than 3 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 to 1. and greater than 3 〈 math 〉〈 math 〉 to 1.

Demonstration.

The whole force of this Proposition consists in these, that, 1. Any Figure circumscribed about a Circle, has a greater Peri∣phery than the Circle, but any inscrib'd one a less. 2. The Peri∣phery of a circumscribed Figure of 96 sides, has a less Pro∣portion to the Diameter, than 3 〈 math 〉〈 math 〉 to 1.

To demonstrate this second, we will enquire 3. the Proportion of one side of such a Figure whether circumscribed or inscribed after the fol∣lowing way.

For the first part of the Proposition.

Suppose the Arch BC (Fig. 84. N. 1.) of 30 degrees, and its Tangent BC making with the Radius AC a Right Angle, to make the Triangle ABC half of an Equilateral one, so that AB shall be to BC in double Reason, viz. as 1000 to 500; which being supposed, AC will be the Root of the Difference of the Squares BC and AB, i. e. a little greater than 866, but not quite 〈 math 〉〈 math 〉.

Then continually Bisecting the Angles BAC by AG, GAC by AH, HAC by AK, KAC by AL, BC is half the side of a circumscribed Hexagon, GC the half side of a Dodecagon (or 12 sided Figure) HC of a Polygon of 24 sides, KC of one of 48; lastly, LC of one of 96 sides; and by N. 3. Schol. 3. Prop. 34. GC will be to AC as BC to BA+AC, and also HC to AC as GC to GA+AC, &c. Wherefore

In the first Bisection, of what parts GC is 500, of the same will AC be 1866 and a little more, and AG (which is the Root of the Sum of the □ □ GC and AC) 1931 〈 math 〉〈 math 〉+.

In the second Bisection, of what parts HC is 500, of the same will AC be found to be 3797 〈 math 〉〈 math 〉+and AH 3830 〈 math 〉〈 math 〉+.

In the third Bisection, of what parts KL is 500 of the same will AC be 7628 〈 math 〉〈 math 〉+ and AK 7644 〈 math 〉〈 math 〉+.

In the 4th Bisection, of what parts LC is 500 of the same will AC be 15272 〈 math 〉〈 math 〉+.

Now therefore LC taken 96 times, will give 48000 the Semi-periphery of the Polygon, which has the same Proportion to the Semi-diameter AC 15272 〈 math 〉〈 math 〉 as the whole Periphery to the whole Diameter. But 48000 contains 15272 〈 math 〉〈 math 〉, 3 times, and moreover 2181 〈 math 〉〈 math 〉 remaining parts, which are less than 〈 math 〉〈 math 〉 part of the division, for multiplied by 7 they give only 1526 〈 math 〉〈 math 〉+.

Therefore it is evident, that the Periphery of this Polygon (and much more the Periphery of a less Circle than that) will have a less Proportion to the Diameter, than 3 〈 math 〉〈 math 〉 to 1. Which is one thing we were to demonstrate.

For the 2d Part of the Prop.

SUppose the Arch BC to be (N^o 2.) of 60 Degr, that is the Angle BAC at the Periphery of 30 Since the Angle at B is a right one by Consect. 1. Prop. 33. the Triangle ABC will be again half an Equilateral one, and BC the whole side of an Hexagon, and GC of a Dodecagon, &c. So that putting for BC 1000 (as before we put 500 for the side of the Hexagon) let AC be 2000 and AB the Root of the difference of the Squares BC and AC i. e. less than 1732 〈 math 〉〈 math 〉 viz. 1732 and not quite 〈 math 〉〈 math 〉.

Then bisecting continually the Angles BAC, GAC, &c. since the Angles at the Periphery BAG, GAC, GCB, standing on equal Arches BG and GC, are equal by Prop. 33. and the Angle at C, (common to the Triangles GCF and GCA) and the o∣thers at H, K, L are all right ones by Consect. 1. of the afore∣said Prop. these 2 Triangles CGF and CGA are equiangular and consequently by Prop. 34. the Perpendicular GC in the one will be to the Perpendicular GA in the other as the Hypothenuse CF in the one to the Hypoth. AC in the other i. e. (by the foun∣dation we have laid in the former part of the Demonstration of N^o 3, Schol. 3. Prop. 34.) as BC to AB+AC; and in like man∣ner in the following HC will be to HA as GC to AG+AC, &c. Wherefore.

In the first Bisection, of what parts GC is 1000 of the same AG will be a little less then 3732 〈 math 〉〈 math 〉; and AC (which is the Root of the Sum of the □ □ AG and GC) will be a little less then 3863 〈 math 〉〈 math 〉.

In the second Bifection, of what parts GC is 1000 of the same will AH be a little less then 7595 〈 math 〉〈 math 〉 and AC a little less then 766 〈 math 〉〈 math 〉.

In the third Bisection, of what parts KC is 1000 of the same will AK be a little less then 15257 〈 math 〉〈 math 〉 and AC a little less 15290 〈 math 〉〈 math 〉.

In the fourth Bisection, of what parts LC is 1000 of the same will AL be a little less then 30547 〈 math 〉〈 math 〉; and AC a little less then 30564, and consequently if LC be put 500, AC will be less then 15282.

Now therefore LC taken 96 times will give 48000 for the Periphery of the inscrib'd Polygon, and 15282 and a little less for the Diameter AC. But 48000 contains 15282 thrice, and moreover a remainder of 2154 parts, which are more then 〈 math 〉〈 math 〉 of the Divisor; for 〈 math 〉〈 math 〉 of this number makes 215 〈 math 〉〈 math 〉 and so 〈 math 〉〈 math 〉 makes 2150 〈 math 〉〈 math 〉 i. e. 2152 〈 math 〉〈 math 〉. Therefore it is evident that the Peri∣phery of this Inscribed Polygon (and much more the Periphery of a Circle greater then that) will be to it's Diameter in a greater proportion then 3 〈 math 〉〈 math 〉 to 1, which is the 2d thing.

SCHOLIUM

IF any one had rather make use of the small numbers of Archimedes, which he chose for this purpose, by putting in the first part of the Demonstration, for AB 306 and for BC 153, in the second for AC 1560 and for BC 780, by the like process of Demonstration, he may infer the same with Archimedes. We like our Numbers best, tho' somewhat large, because they may be remember'd, and are more proportionate to things, and make also the latter part of our Demonstration like the former. The Proportion in the mean while of the Dia∣meter to the Periphery of the Circle by the Archimedean way is included within such narrow Limits, that they only differ from one another 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 parts; for 〈 math 〉〈 math 〉 Subtracted from 〈 math 〉〈 math 〉 leave 〈 math 〉〈 math 〉, as 3 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 and 3 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 if they are reduced to the same Deno∣mination make on the one hand 〈 math 〉〈 math 〉 on the other 〈 math 〉〈 math 〉. Hence it would be easy, having divided the difference 〈 math 〉〈 math 〉 into 2 Parts, to express a middle proportion of the Periphery to the Diameter be∣tween the 2 Archimedean and Extreme ones as in these Numbers, 1561 to 4970, (or by dividing both sides by 5) as 3123 to 994, or (dividing both sides by 7) as 446 〈 math 〉〈 math 〉 to 142, or (by dividing again by 2) as 223 〈 math 〉〈 math 〉 to 71, &c.

While these Numbers become as fit for use as those of Archi∣medes, which we therefore use before any other, particularly in Dimensions that dont require an exact Niceness; where they do those may be made use of exhibited by Ptolomey Vieta, Ludolphus a Ceulea, Metius, Snellius, Lansbergius, Hugeus, &c. as if

The Diameter be

The Circumference will be

10,000,000

31, 416, 666. Ptolomey.

10000,000,000

31, 415, 926, 535. Vieta.

100,000,000,000,000,000000

314, 159, 265, 358, 979, 323, 846 ½, &c. Ludolph. a Ceulen.

Proposition XLI.

THe Area of a Circle has the same proportion to the Square of it Diameter, as the 4th part of the Circumference to the Dia∣meter.

Demonstration.

Though we have before Demonstrated this Truth in Consect. 6. Prop. 39. yet here we will give it you again after another way. Since therefore the Circumference is a little less then 3 〈 math 〉〈 math 〉, and •• little more than 3 〈 math 〉〈 math 〉 Diameters, for this excess putting z if the Diameter be 1 we will call the Circumference 3+z therefore the 4th Part of it will be 〈 math 〉〈 math 〉: And the Area of the Circle (by Multiplying the half Semidiameter) i. e. ¼ by the Circumfe∣rence, you'l have both 〈 math 〉〈 math 〉 and the square of the Diameter =•• Q. E. D.

CONSECTARY.

THerefore if the(α) 1.56 Proportion of Archimedes be near e∣nough truth to be made use of, viz. 22 to 7; the Ar•••• of the Circle will be to the Square of the Diameter as 11 to 14 because the quarter part of 22, i. e. 5 ½ or 〈 math 〉〈 math 〉 to the Diam. 7 〈 math 〉〈 math 〉 is in the same Proportion.

Proposition XLII.

THE Diameter(β) 1.57 of a Square AC (Fig. 83.) is incom∣mensurable to the side AB (and consequently also to th•• whole Periphery) i. e. it bears a Proportion to 〈◊〉〈◊〉 that cannot be exactly expressed by Numbers.

Demonstration.

For, if for AB you put 1, BC will be als••

1, and by the Pythagorick Theorem AC will be = √2. There∣fore by Consect. 4. Definit. 30. AC is incommensurable to the side AB, &c. Q.E.D.

CONSECTARY I.

IT is notwithstanding Commensurable in Power; for its Square is to the Square of the Side as 2 to 1.

CONSECTARY II.

NOW if the Proportion of the side or whole Periphery ABCDA, to the Diam. AC is to be expressed by Num∣bers somewhat near, as we have done in the Diameter and Cir∣cumference of a Circle; then making the side AB=100, the Diameter is greater than 141 〈 math 〉〈 math 〉, less than 141 〈 math 〉〈 math 〉.

Proposition XLIII.

THE Area of a Circle is Incommensurable to the Square of the Dia∣meter.

Demonstration.

For dividing the Semidiameter CD (Fig. 85.) into two equal Parts (and consequently the Diameter DF into 4) AC will be 2. viz. √4 and AC √3. by Schol. 2. Prop. 34. N. 3. the sum √4+√3, and the sum of as many equal to the greatest AC 4. Having moreover Bisected the Parts of the Semidiameter, AC will =4 or the √16, ac=√15, AC=√12, AC=√7; the sum 〈 math 〉〈 math 〉; and the sum of as many e∣qual to the greatest AC is =16, &c. And thus the last sums will be the Square Numbers increasing in Quadruple Pro∣portion; but the former Sums will be always composed of the Ra∣tional Root of every such Square, and of several other irrational Roots of Numbers unevenly decreasing; so that it will be im∣possible to express those former Sums by any Rational Number, by what we have said in Schol. 2. Definit. 30. Wherefore all the Indivisibles of the Quadrant ADC are to as many of the Square ACDE equal to the greatest, i. e. the Quadrant it self

ADC to the Square ACDE (and consequently the whole Area of the Circle to this circumscrib'd Square) will be as a Surd Quantity to a true and truly Square Number, i. e. the Area of the Circle will be Incommensurable to the Square of the Diameter, by Consect. 4. of the said Definit. Q. E. D.

CONSECTARY.

AND because the fourth part of the Circumference ha•••• the same Proportion to the Diameter, as the Area of the Circle to the Square of the Diameter, by Prop. 41. There∣fore also that will be Incommensurable to this, and consequently the whole Circumference will be so to the Diameter.

SCHOLIUM.

WHerefore it is somewhat Wonderful, which G. G. Leib∣nitius(α) 1.58 tells us, that the Square of the Diameter be∣ing 1, the Area of the Circle will be 〈 math 〉〈 math 〉, &c. ad Infinitum. i. e. by adding 1 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉, &c.) to 〈 math 〉〈 math 〉, &c. i. e. to the Sum of infinite Fractions whose common Numerator is 2. But their Denominators Squares les∣sen'd by Unity, and taken out of the Series of the Squares of Natural Numbers by every fourth, omitting the Intermediate ones: Which Sum might seem expressible in Numbers, since all its parts are Fractions redu∣cible to a common Denominaton; while notwith∣standing Leibnitius himself confesses, that the Cir∣cle is not Commensurable to the Square, nor ex∣pressible by any Number.

CHAP. VII. Of the Powers of the Sides of Triangles, and other Regular Figures, &c.

Proposition XLIV.

IN Right-angled Triangles(α) 1.59 (ABC, Fig. 86.) the Square of the Side (BC) that subtends the Right-angle, is equal to the Squares of the other Sides (AB and AC) taken together.

Demonstration.

Though we have demonstrated this Truth more than once in the foregoing Proposition; yet here we will confirm it again as follows. Having described on each side of the Square BE a Se∣micircle, which will all necessarily touch one another in one point, and be equal to the Semicircle, BAC, if you conceive as many Triangles inscribed also equal to BAC; it will be evident that the Square BE will contain the said 4 Triangles; and besides the little Square FGHI, whose side FI, v. g. is the difference be∣tween the greater side of the Triangle CI, and the less CF, (for because the less side CF=BA, lying in the first Semicircle, if it be continued to I in the se∣cond Semicircle makes CI=CA the greater side of the other Triangle, and so in the others. From thence it is evident, That as the Angles ABC, and ACB together make one right one; so likewise BCF (= CBA) and ECF make also one right one; and consequently ECF is =ACB, and the Arch and the Line EI= to the Arch and the Line AB, &c.) Where∣fore, if the greatest side of the given Triangle BC or BD, &c. be called a, and AC, b and the least AC, or CF, &c. be called c; the 〈◊〉〈◊〉 of the side BC, will be =aa, and the Area of each Triangle ½ bc: and so the 4 Triangles together 2 bc: but the side of the middle little Square will be b−c, and its Square bb+cc=2bc: Wherefore if you add to this the 4 Trian∣gles▪

The Sum of all, i. e. the whole Square BE will be bb+cc=aa. Q. E. D.

CONSECTARYS.

I. HEnce having the sides that comprehend the Right-ang•••• given, AC=b and AB=c, the Hypothenuse or Ba•• that subtends the Right-angle BC will be = √bb+cc.

II. But if BC be given = a and AC=b, and you are 〈◊〉〈◊〉 find AB=x; because xx+bb=aa; you'l have (taking awa•• from both sides bb) xx=aa−bb: therefore x, i. e. AB=√aa−bb.

III. If 2 Right-angled Triangles have their Hypothenus•• and one Leg equal, the other will also be equal.

Proposition XLV.

IN Obtuse-angled Triangles (Fig. 87. N. 1. the Square of the Ba•• or greatest Side BC that subtends the Obtuse-angle BAC, is eq•••• to the Squares of(α) 1.60 the other 2 Sides (AB and AC) taken togethe•• and also to 2 Rectangles (CAD) made by one of 〈◊〉〈◊〉 Sides which contain the Obtuse-angle (AC) and 〈◊〉〈◊〉 continuation AD to the Perpendicular BD let fall fr•••• the other side.

Demonstration.

If BC be called a, AB=c, AC=b, AD=x, CD will b•• =b+x. Therefore □ BD=cc−xx, by Consect. 2. of the pr••¦ceding Prop. In like manner if □ CD= 〈 math 〉〈 math 〉 be sub¦tracted from the □ BC=aa, you'l have 〈 math 〉〈 math 〉 to the same □ BD. Therefore 〈 math 〉〈 math 〉, i. e. (adding on both sides xx) 〈 math 〉〈 math 〉. i. e. (adding on both sides bb and 2bx) 〈 math 〉〈 math 〉. Q. E. D.

CONSECTARY.

IF in this last Equation you subtract from both Sides cc+bb then will 〈 math 〉〈 math 〉 and (if you moreover divide both Sides by 2b) you'l have 〈 math 〉〈 math 〉: Which is the Rule, when you have the Sides of an Obtuse-angled Triangle given, to find the Segment AD, and consequently the Perpen∣dicular BD.

Proposition XLVI.

IN Acute-angled Triangles(α) 1.61 the Square of any side (e. g. B. C, Fig. 87. N. 2.) subtending any of the Angles, as A is equal to the Squares of the other 2 sides (AB and AC) taken together, less 2 Rect∣angles (CAD) made by one side, containing the Acute-angle (CA) and its Segment AD reaching from the Acute-angle (A) to the Perpen∣dicular (BE) let fall from the other side.

Demonstration.

Make again BC=a, AC=b, AB=c, AD=x; then will CD=b−x. Therefore cc−xx= □ BD, and 〈 math 〉〈 math 〉 (i. e. □ BC−□CD) will also be = □ BD.

Therefore 〈 math 〉〈 math 〉.

i. e. (adding to both sides xx)

〈 math 〉〈 math 〉,

i. e. (adding on both sides bb, and subtracting 2bx)

〈 math 〉〈 math 〉. Q. E. D.

CONSECTARYS.

I. IF in the last Equation, except one, you add on both sides bb, and subtract aa, you'l have 〈 math 〉〈 math 〉, and, if moreover you divide hoth sides by 2b, you'l have 〈 math 〉〈 math 〉: Which is the Rule, having 3 sides given in an Acute-angled Triangle, to find the Segment AD, and consequently the Perpendicular BD.

Knowing therefore the Segments AD and CD, and also the Perpendicular BD in Oblique-angled Triangles, whether Ob∣tuse-angled or Acute-angled, when moreover the sides BC an•• AB are likewise given, the Angles of either Right-angled Tri∣angles or Oblique-angled ones, will be known; so that the la•••• Case of Plane Trigonometry, which we deferr'd from Prop. 34 to this place, may hence receive its solution.

Proposition XLVII.

THE Square of the Tangent of a(α) 1.62 Circle, is equal to a Rectangle contain'd under the whole Secant DA, and that part of it whi•••• is without the Circle DE, whether the Secant pass thro' the Centre o•• not.

Demonstration.

For in the first Case, if CB and CE are = b, DE=x, then will CD=b+x, & AD=2b+x: therefore

□ ADE=2 〈 math 〉〈 math 〉. there∣fore, if from the □ CD you substract □ CB=bb the remainde•• will be 〈 math 〉〈 math 〉 =□ BD=□ ADE. Q.E.D.

In the second Case, the lines remaining as before, make D••=y, FE or FA=Z: therefore the □ ADE will be = 〈 math 〉〈 math 〉, but the □ FC equal to the □ EC−□FE= 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉. But the same Square FD is = 〈 math 〉〈 math 〉. Wherefore taking away from these equal Square•• the common one ZZ you'l have 2 〈 math 〉〈 math 〉 i. e. pr. 1st Case =□ BD. Q. E. D.

CONSECTARYS.

I. THerefore the Rectangles of diverse secants (as of AD•• & ade in Fig. 88, n. 1.) which are equal to the sam•• Square. BD, are equal also to one another; which the la•••• Equation in our Demonstration (〈 math 〉〈 math 〉 is an ocular proof of.

[illustration]

Pag. 133.

89

90

91

92

93

94

95

96

II. Therefore by Prop. 19. as a D to AD so is reciprocally DE to D e in the Fig. of the 2 Case.

III. Tangents to the same Circle from the same Point, as DB & D b (n. 3.) are equal; because the Square of each is equal to the same Rectangle.

IV. Nor can there be more Tangents drawn from the same point then two: For if besides DB & D b, D c could also touch the Circle, then it would be equal to them. By Cons. 2. but that is absurd by Cons. 3. Def. 7.

SCHOLIUM I.

HEnce is evident, the Original of the Geometrical Constru∣ctions, which Cartes makes use of, p. 6, and 7. in resol∣ving these 3 Equations, 〈 math 〉〈 math 〉, & 〈 math 〉〈 math 〉. For in the First case, since he makes NO or NL (Fig. 89. 11. 1.) or NP ½ a, and the Tangent ZMb, make MNO through the Center xx will =Z the quantity sought; which thus appears: Making MO=Z, NM will =Z½ a, and its □ 〈 math 〉〈 math 〉. But the □ OMP (which is by the present Prop. = □ LM, i. e. bb) together with the □ NL or NP i. e. ¼aa is = □ NM by the Pathagor. Theor. Therefore 〈 math 〉〈 math 〉. i e. (taking away from both sides ¼ aa) ZZ−a Z=bb i. e. (by adding on both sides aZ) ZZ=aZ+bb: Which is the Equation proposed. In the Second Case, if you make PM=Z (as Cartes makes it) you'll have 〈 math 〉〈 math 〉, and to this again as before you'l have = bb+¼aa. Therefore ZZ+aZ=bb: Therefore ZZ=−aZ+bb: Which is the very Equation of the second Case. In the third Case, whether you make the w ole Secant RM (N. 2) or that part of it without the Circle QM=Z, the Root sought, there will come out on both sides the same Equa∣tion of the third Case; and so it is manifest, that this Equation has those 2 Roots. For if RM be =Z (adding + to the Fig. of Cartes the Line NO which shall bisect QR, and makes OM=LN) OR or OQ will be =Z−½a, and so the □ OQ=ZZ−aZ+ ¼aa, and this together with the □ RMQ (which is by vertue of the pres. Prop. = □ LM) =□ NQ. i. e. OM, i. e. ZZ−

〈 math 〉〈 math 〉, i. e. (by adding aZ and taking away ½ aa) 〈 math 〉〈 math 〉; i. e. (taking away bb) ZZ=aZ−bb: Which is the the very Equation of the third Case. But if QM be made =Z, OQ or OR will = ½−Z, and its □ = 〈 math 〉〈 math 〉, as well as the former, and so all the rest. Q. E. D.

SCHOLIUM. II.

NOW if you would immediately deduce these Rules by the present Proposition, without the Pythagorick Theorem. It may (e. g. in the first Case) be done much shorter thus: If MO=Z and NO or NP=½ a, then will PM=Za: There∣fore □ OMP= 〈 math 〉〈 math 〉, or the □ LM, by the pres. Pro∣position; by adding therefore to both sides aZ, you'l have ZZ=aZ+bb, which is the very Equation of the first Case. In the second Case, if MR be Z, QM or PR will =a−Z: Therefore □ RMP 〈 math 〉〈 math 〉 =bb, i. e. aZ=bb+ZZ, i. e. aZ−bb=ZZ; but if QM=Z, RM will = a−Z. There∣fore □ RMP aZ−ZZ=bb, as before, &c.

SCHOLIUM III.

FRom the 2 Consect. of the present Proposition, flows ano∣ther Rule for solving the last Case of Plain Trigonometry, which we solved in the Consect. of the foregoing Prop. viz. If you have all the three sides of the Oblique-angled Triangle BCD (Fig. 90) given, if from the Center C, at the distance of the lesser side CB you describe a Circle, then will, by Consect. 2. of the present Proposition, BD the Base of the Triangle (here we call the greatest side of the Triangle the Base, or in an Equicrural Triangle, one of the greatest) will be to AD, (the sum of the Sides DC+CB) as DE the difference of the Sides, to DF the Segment of the Base without the Ciecle; which being found, if the remainder of the Base within the Circle be divided into two equal parts, you'l have both FG and GB, as also DG; which being given, by help of the Right-angled ▵▵ GBC and GDC all the Angles required may be found.

Proposition XLVIII.

IN any Quadrilateral Figure(α) 1.63 ABCD (Fig. 91. N. 1.) in∣scribed in a Circle, the ▭ of the Diagonals AC and BD is equal to the two Rectangles of the opposite sides AB into CD, and AD into BC.

Demonstration.

Having drawn AE so that the Angle BAE shall be equal to the Angle CAD, the Triangles thereby formed (for they have the other Angles EBA and ACD in the same Segment equal, by ver∣tue of Consect. 1. Prop. 33.) will be Equiangular one to another and consequently (by Prop. 34) as AC to AD so AB to BE. Wherefore by making AC=a and CD=ea, and AB=b, BE will be =eb. In like manner when in the ▵ ▵ BAC and EAD, the respective Angles are equal (viz. adding the common part EAF to BAE and CAD, equal by Constr.) and besides the an∣gles BCA and EDA in the same Segment are also equal; those Triangles will also be Equiangular, and AD will be to DE as AC to CB; wherefore by putting, as before, a for AC, and oa for CB, and c for AD, DE will =oc. Therefore the whole BD=eb+oc. The Rectangle therefore of AC into BD will = eba+oca=; the Rectangle of AB into CD=eba+□ of AD into BC=oac. Q. E. D.

SCHOLIUM

IN Squares and Rectangles (N. 2.) the thing is self-evident. For in Squares if the side be a, the Diagonals AC and BD will be √2aa, and so their Rectangle =2aa will be manifestly equal to the two Rectangles of the opposite sides. In Oblongs, if the two opposite sides are a and the others b. the Diago∣nals will be 〈 math 〉〈 math 〉, and their Rectangle aa+bb manifestly equal to the two Rectangles of the opposite Sides.

Proposition XLIX.

THe side (AB) of an Equilateral Triangle (ABC, Fig. 92. N. 1.) inscribed in a(α) 1.64 Circle, is in Power triple of the Radius (AD) i. e. of the □ of AD.

Demonstration.

MAke AD or FD=a, and so its Square aa. Since There∣fore, having drawn DF thro' the middle of AB, or the middle of the Arch AFB let DE be = ½a; for the angles at E are right ones, by Consect. 5. Definit. 8. and the Hypothe∣nuses AD, AF, are equal, by Schol. of Definition 15. but the side AE is common. Therrefore the other Sides FE and ED are equal by, by Consect. 3. Prop. 43.) and the □ of the latter is ¼ aa, which subtracted from aa leaves ¾ aa for the □ of AE. Therefore the line AE is 〈 math 〉〈 math 〉, and conse∣quently AB 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉: therefore □ AB=3aa. Q E. D.

CONSECTARYS.

I. IF the Radius of a Circle be = a, the side of an Inscribed Regular Triangle will be 〈 math 〉〈 math 〉, e. g. if AD be 10, AB will be 〈 math 〉〈 math 〉; and if AD be 10,000,000, AB will be 〈 math 〉〈 math 〉, i. e. 17320508, and the Perpen∣dicular DE 5000,000.

II. Hence it is evident, that in the genesis of a Tetraedrum pro∣posed in Def. 22, that the elevation CE (Fig. 44, N. 1.) is to the remaining part of the Diameter of the Sphere CF as 2 to 1; for making the Radius CB=a and its □ aa, the □ of AB or BE will =3aa, by the present Proposition. Therefore the □ of CB being subtracted from the □ of BD or BE, there remains the □ of CE=2aa. But since CE, CB, CF, are continual Proportionals, by N. 3. Schol. 2. Prop. 34. CE will be to CF as the □ of CE to the Square of CB, by vertue of Prop. 35. i. e. as 2 to 1.

SCHOLIUM.

HEnce you have the Euclidean way of generating(α) 1.65 a Te∣traedrum, and inscribing it in a given Sphere, when he bids you divide the Diameter EF of a given Sphere so that EC shall be 2 and CF 1, and then at EF to erect the Perpendicular CA, and by means thereof to describe the Circle ABD, and to inscribe therein an Equilateral Triangle, &c.

Proposition L.

THE Side (AB) of a Regular Tetragon (or Square) (ABCD, Fig. 92. N. 2.) is double in Power of the Radius (AD).

Demonstration.

For having drawn the Diameter AC and BD, the Triangle AOB is Right-angled, and consequently, by the Pythagorick Theo∣rem, if the □ of AO and BO be made equal to aa, then will the □ of AB=2aa, Q. E. D.

CONSECTARY.

THerefore when the Radius of the Circle AO is made = a, the side of the □ AB will =√2aa, e. g. if AO be 10, AB will be √200; and if AO be 10,000,000, AB will be √200,000,000,000,000, i. e. 14142136.

Proposition LI.

THE side AB of a Regular Pentagon(α) 1.66 (ABCDE) (Fig. 93. N. 1.) is equal in Power to the side of an Hexagon and Deca∣gon inscribed in the same Circle, i. e. the □ of AB is equal to the Squares AF and AO taken together.

Demonstration.

Make AO=a and AF=b, AB=x: We are to demonstrate that xx=aa+bb: which may be done by finding the side AB by the parts BH and HA,(α) 1.67 after the following way: First of all the angle AOB is 72°, and the o∣thers in that Triangle at A and B 54°. But BGG is also 54°, as subtending the Decagonal Arch BF of 36°, and also one half of it FG of 18°. Therefore the ▵▵ ABO and HBO are Equiangular, and you'l have

As AB to BO so BO to BH 〈 math 〉〈 math 〉

Secondly, in the Triangle BFA the Angles at B and A are e∣qual by N. 3. Consect. 5. Def. 8. and by vertue of the same also the Angles at F and A in the ▵ FHA are so too. Wherefore the ▵▵ BFA and FFA are Equiangular, and you'l have

As BA to AF so AF to AH 〈 math 〉〈 math 〉

Therefore the whole line AB (because the part AH is found =〈 math 〉〈 math 〉 and BH=〈 math 〉〈 math 〉) will be 〈 math 〉〈 math 〉, which was first made = x; so that now 〈 math 〉〈 math 〉 is = x, and multiplying both sides by x, aa+bb=xx. Q. E. D.

CONSECTARY I.

THerefore if the Radius of a Circle be (a) the side of a Pen∣tagon AB will be 〈 math 〉〈 math 〉.

CONSECTARY II.

THerefore the □ AI= 〈 math 〉〈 math 〉 and □ OI= 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉. Therefore OI= 〈 math 〉〈 math 〉. Which yet may be expressed otherwise, viz. OI= 〈 math 〉〈 math 〉.

Demonstration.

Make(α) 1.68 OA or OF (N. 2.) as before = a, AF=b, and FI now = x; then will OI=a−x and having drawn the Arch FK at the Interval AF, so that AK may be equal to this, and FI=

IK=x; then will the angle IKA=F72°. Therefore the an∣gle AKO=108; and since KOA is 36°, KAO will be also 36°, and so KA=KO=AF=b. Wherefore OI is = b+x, which was above a−x. Therefore 2OI= 〈 math 〉〈 math 〉, i. e. a+b. Therefore OI= 〈 math 〉〈 math 〉.

CONSECTARY III.

THerefore the difference between the Perpendicular of the Triangle DE (Fig. 92 and 93. N. 1.) and the Perpen∣dicular of the Pentagon OI is =½ b, by vertue of Consect. 2. of this and of the Demonstrat. of Prop. 49.

CONSECTARY IV.

HEnce is also evident, by the present Proposition, that which in Fabri's Genesis of an Icosaedr. Def. 22. we said, viz. that (See Fig. 46.) B a is equal to the side of a Pentagon BA, because, viz. Fa is =to the Semidiameter OB, and BF is the side of a Decagon.

Proposition LII.

THE side of an Hexagon is in Power equal to the Radius, as being it self equal to it by N. 1. Schol. Def. 15.

Proposition LIII.

THE side of a Regular Octagon (ABCD, &c. Fig. 94.) is equal in Power to half the side of the Square, and the difference (PB) of that half side from the Radius, taken together.

Demonstration.

For that the □ of AB is = to the □ of AP+□ BP, is evident from the Pythag. Theor. But that PO is =PA half the side of the Square, is evident from the equality of the Angles PAO and POA, since each is a half right one or 45°. Where∣fore

the side of the Octagon is equal in Power to half the side of the Square, &c. Q. E. D.

CONSECTARY.

THerefore, if the Radius be =a, AP will be, by vertue of the Pythag. Theor. =〈 math 〉〈 math 〉 and PB= 〈 math 〉〈 math 〉 of that □ ½ aa; and of this □ 〈 math 〉〈 math 〉. Therefore the Sum of □ AB= 〈 math 〉〈 math 〉. Therefore the side of the Octagon = 〈 math 〉〈 math 〉. e. g. If AO be 10, AB will be 〈 math 〉〈 math 〉, and if the Radius AO 10000000, AB will be 〈 math 〉〈 math 〉, by vertue of the present Cons. or from the Prop. it self, if AO be 10000000, AP will be vertue of the Cons. of Prop. 50.=7071068, and consequently BP=2928932 The Squares of these added together give the □ AB, and the Root thence extracted AB=7653 668.

Proposition LIV.

THE side of a Regular Decagon(α) 1.69 is equal in Power to the greatest part of the side of an Hexagon cut in mean and extreme Reason.

Demonstration.

Suppose BD (Fig. 95.) divided in mean and extreme Reason in E, and BA to be joined to it long ways = to the side of a Decagon inscribed in the same Circle, whose Radius is BC or AC=BD. Now we are to demonstrate that DE the greatest part of the de of the Hexagon BD divided in mean and extreme Rea∣son, is equal to the side of a Decagon BA, and the Power of the one equal to the Power of the other. Because the Angle ACB is 36°, ABC and A72°, and consequently CBD 108°, BCD and D will be each 36°, and so the whole ACD 72°. (that so CD may pass precisely thro' the other end of the side of the Pen∣tagon

AF.) Wherefore the ▵ ▵ ABC and ADC are Equi∣angular, and

• AD to AC i. e. BD
  • as AC i. e. BD

to AB. Therefore the whole line AD is divided in mean and extreme Reason. But BD is al∣so divided in the same Reason by Hyp. Wherefore

• As AD to DB and DB to BA,
• So DB to DE and DE to EB.

Therefore DB is in the same Proportion to DE as DB to BA. Therefore DE is =BA, and the Power of the one to the Power of the other. Q. E. D.

CONSECTARYS.

THerefore, if the Radius of the side of the Hexagon is a the side of the Decagon will be 〈 math 〉〈 math 〉, by Schol. 2. Prop. 27. e. g. if the Radius be 10, the side of the Decagon will be 〈 math 〉〈 math 〉, and if the Radius be put 10000000, the side of the Decagon will be = 〈 math 〉〈 math 〉, viz. by adding the Square of the Radius and the Square of half the Radius into one Sum; whence you'l have the side of the Decagon =6180340; the half whereof 3090170 gives the difference between the Perpendiculars of the Triangle and the Pentagon, by Cons. 3. Prop. 51.

II. The side therefore of the Pentagon is by Prop. 〈 math 〉〈 math 〉; for the Square of the Hexagon is aa or 〈 math 〉〈 math 〉 aa, the □ of the Decagon 〈 math 〉〈 math 〉: the Root extracted out of the Sum of these is the side of the Pentagon, viz. 〈 math 〉〈 math 〉 e. g. if the Radius be 10, the side of the Pentagon will be 〈 math 〉〈 math 〉, and if the Radius be put 10000000, since the side of the Hexagon is equal to it, and the side of the Decagon 6180340, their Squares being added into one Sum, the Root extracted out of that Sum will give the side of the Pentagon, 1755704 nearly; and the sides being collected into

one sum, the half of it 8090170 will give the Perpendicular in the Pentagon OI, by Consect. 2. Prop. 51.

SCHOLIUM I.

TO illustrate what we have deduced in the Consectarys of Prop. 51, you may take the following Notes. If a be put =10 or 〈 math 〉〈 math 〉, the side of the Decagon will be = 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉 nearly = b; therefore the □ aa=〈 math 〉〈 math 〉 and the □ bb=〈 math 〉〈 math 〉: therefore aa+bb or the □ AB=〈 math 〉〈 math 〉, the Perpen∣dicular OI=〈 math 〉〈 math 〉 divided by 2, that is, 〈 math 〉〈 math 〉. Now the □ of AI is ¼ of the □ AB=〈 math 〉〈 math 〉 the □ OI=〈 math 〉〈 math 〉 =〈 math 〉〈 math 〉. Now if you add the □ AI and the □ OI, the sum will be = □ AO= 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉, or near 100. Thus likewise, since the Perpendicular above found OI, in Consect. 1. may be also determined by 〈 math 〉〈 math 〉, since aa is = 〈 math 〉〈 math 〉, and 3 aa 〈 math 〉〈 math 〉, subtracting from it bb=〈 math 〉〈 math 〉 the Remainder will be 〈 math 〉〈 math 〉, and this be∣ing divided by 4, you'l have the □ OI=〈 math 〉〈 math 〉. and the Root of it extracted 〈 math 〉〈 math 〉 nearly; so that those two different quantities in Consect. 1. will rightly express the same Perpendicular OI.

SCHOLIƲM II.

NOW therefore as we have Practical Rules to determine A∣rithmetically the sides of the Pentagon and Decagon, so also they may be found Geometrically by what we have demon∣strated. For if the Semidiameter CB (Fig. 96. N. 1.) be divided into 2 parts, EC will =½ a; and erecting perpendicularly the Radius CD=aDE will = 〈 math 〉〈 math 〉. Moreover if you cut of EF equal to it, FC will be = 〈 math 〉〈 math 〉= to the side of the Decagon, by Consect. 1. Having therefore drawn DF, which is equal in Power to the Radius or Side of the Hexagon DC, and the side of the Decagon FC together, by the Pythag. Theorem

it will be the side of the Pentagon sought. Much to the same purpose is also this other new Construction of the same Problem, wherein BG (Numb. 2.) is the side of the Hexagon BD the side of the Square, to which GF is made equal, so that FC is that side of the Decagon, and DF of the Pentagon; which we thus demonstrate after our way: Having bisected GH the side of an Equil. Triangle, the Square of GE will be 〈 math 〉〈 math 〉, by Prop. 48. which being subtracted from the Square of GF=2 aa, viz. 〈 math 〉〈 math 〉 aa, by Prop. 49. there will remain for the Square of EF 〈 math 〉〈 math 〉 aa, and for the line EF 〈 math 〉〈 math 〉, and for FC 〈 math 〉〈 math 〉, which is the side of the Decagon, as DF of the Pentagon, after the same way as before.

Proposition LV.

THE side of a Quindecagon (or 15 sided Figure) is equal in Power to the half Difference between the side of the Equil. Triangle and the side of the Pentagon, & moreover to the Difference of the Perpendi∣culars let fall on both sides taken together.

Demonstration.

For if AB (Fig. 97.) be the side of an Inscribed Triangle, and De the side of a Pentagon parallel to it; AD will be the side of the Quindecagon to be inscribed, by Consect. 4. Def. 15. But this side AD in the little Right-angled Triangle, is equal in Power to the side AH (which is the half Difference between AB and DE) and the side HD (which is the difference be∣tween the Perpendiculars CF and CG) taken together by the Pythag. Theor. Q. E. D.

CONSECTARY.

HEnce if we call the Side AB of the Equil. ▵ c, and mak•• the side of the Pentagon DE=d, AH will = 〈 math 〉〈 math 〉 HD is =½b, by Consect. 3. of Prop. 51. Since therefore the ▵ AH is = 〈 math 〉〈 math 〉 and the ▵ HD=〈 math 〉〈 math 〉 the ▵ AD w•••••• 〈 math 〉〈 math 〉

Therefore the side of the Quindecagon = 〈 math 〉〈 math 〉that is, Collecting the Square of the half difference of the side of the Triangle and Pentagon, and the Square of the difference of the Perpendiculars into one Sum, and then Extracting the Square Root of that Sum, you'l have the side of the Quindeca∣gon sought. E. 9. if rhe Radius CI be made 10000000 the difference of the sides of the Triangle 17320508, and of the side of the Pentagon 11757704 will be 5564804, and the half of this 2782402; but the difference of the Perpendicular CF from the Perpendicular CG, is 3090170.

The Squares therefore of these Two last Numbers being Collected into one Sum, nnd the Root Extracted will give the side of the Quindecagon 4158234 nearly.

SCHOLIUM.

HEre we will shew the Excellent use of these last Proposition•• in making the Tables of Signs. For having found above, supposing the Radius of 10000000 parts, the sides of the chief Regular Figures, if they are Bisected, you will have so many Primary Sines; viz. from the side of the Triangle the side of 60 Degrees 8660754, from the Side of the Square, the Sine of 45° 7071068; from the Side of the Pentagon, the Sine of 36° 5877853; from the Side of the Hexa∣gon, the Sine of 30° 5000000; from the Side of the Octagon, the Sine of 22° 30 3826843; from the Side

of the Decagon the sine of 18°=3090170; from the side of the quindecagon lastly the sine of 12°=2079117. From these seven primary sines you may find afterwards the rest, and consequently all the Tangents and Secants according to the Rule we have deduc'd, n. 3. Schol. 5. Prop. 34. and which Ph. Lansbergius illustrates in a prolix Example in his Geom. of Triangles Lib. 2. p. 7. and the following. But af∣ter what way, having found these greater numbers of sines, Tangents, &c. Logarithms have been of late accommodated to them, remains now to be shewn, which in brief is thus; viz. the Logarithms of sines, &c. might immediately be had from the Logarithms of vulgar numbers, if the tables of vul∣gar numbers were extended so far, as to contain such large numbers; and thus the sine e. g. of o gr. 34. which is 98900 the Logarithm in the Chiliads of Vlacquus is 49951962916. But because the other sines which are greater than this are not to be found among vulgar numbers (for they ascend not be∣yond 100000, others only reaching to 10000 or 20000) there is a way found of finding the Logarithms of greater num∣bers, than what are contained in the Tables. E g. If the Logarithm of the sine of 45° which is 7071068 is to be found, now this whole number is not to be found in any vulgar Ta∣bles, yet its first four notes 7071 are to be found in the vul∣gar Tables of Strauchius with the correspondent Logarithm 3. 8494808, and the five first 70710 in the Tables of Vlac∣quus with the Log. 4. 8494808372. One of these Loga∣rithms, e g. the latter, is taken out, only by augmenting the Characteristick with so many units, as there remain notes out of the number proposed, which are not found in the Tables, so that the Log. taken thus out will be 6. 8494808372. Then multiply the remaining notes of the proposed number by the difference of this Logarithm from the next following, (which for that purpose is every where added in the Vlacquian Chiliads, and is in this case 61419) and from the Product 4176492 cast away as many notes as adhere to the propo∣sed number beyond the tabular ones, in this case 2; for of the remainder 41764, if they are added to the Logarithm before taken out, there will come the Logarithm requi∣red 6. 8494850136, viz. according to the Tables of Vlac∣q••us, wherein for the Log of 10 you have 10,000,000,

000; but according to those of Strauchius which have for the Logarithm of 10 only 10,000,000, you must cut off the three last notes, that the Logarithm of the given sine may be 6.8494850; as is found in the Strauchian and other tables of sines, except that instead of the Characteristick 6 there precedes the Characteristick 9, whereof we will add this reason: If the Characteristicks had been kept, as they were found by the rule just now given, the Logarithm of the whole sine (which is in the Strauchian Tables 10,000,000) would have come out 70,000,000, incongruous enough in Trigo∣nometrical Operations. Wherefore that Log. of the whole sine might begin from 1, for the easiness of Multiplication and Division they have assumed 100,000,000; the Cha∣racteristick being augmented by three, wherewith it was con∣sequently necessary to augment also all the antecedent ones; and hence e. g. the Logarithm of the least sine 2909 begins from the Characteristick 6, which otherwise according to the Tables of vulgar numbers would have been 3.

Having found after this way the Logarithms of all the sines (altho' here also if you have found the Logarithms of the signs of 45° and moreover the Logarithm of 30, the Loga∣rithms of all the rest may be compendiously found by addition and substraction from a new principle which now we shal omit) the Logarithms of the Tangents and Secants may easi∣ly be found also, only by working, but now Logarithmically, according to the Rules of Schol. 5. Prop. 34. n. 5. and 6.

Proposition LVI.

THE side of a Tetraëdrum(α) 1.70 or equ•• Pyramid is in power to the Diameter of a circumscribed Sphere, as 2 to 3.

Demonstration.

For because by the genesis of the Tetraëdrum Def. 22 (see its Fig. 44 n. 1.) and Schol. Prop. 49. OC is ⅓ of the se∣midiameter OB, which we will call a the □ of CB will be =〈 math 〉〈 math 〉aa by the Pythag. Theor. and so the power (or Sq.) of the side of the Tetraëdrum = 〈 math 〉〈 math 〉 aa by Prop. 49. but the pow∣er

of the Diam. 2a or 〈 math 〉〈 math 〉 a is 〈 math 〉〈 math 〉 aa. Therefore the power of the side of the Tetraëdrum is to the power of the Diam. as 24 to 36. i. e. (dividing each side by 12) as 2 to 3. Q. E. D.

Or more short.

The □ of CB is =2 by Schol of Prop. 49. and the □ of EC=4. Therefore the □ EB=6. But the □ EF is =9. Therefore the □ of EB is to the □ of EF as 6 to 9, i. e. as 2 to 3. Q. E. D.

CONSECTARY.

THerefore if the Diam. EF be made =a, the side EB will be =〈 math 〉〈 math 〉.

Proposition LVII.

THE side of the Octaëdrum(α) 1.71 is in power one half of the Diameter of the circum∣scribed Sphere.

Demonstration.

For since by the genesis of the Octaëdrum Def 22. (see Fig. 44. n. 2.) CA, CB, CF, &c. are so many radii of great circles, if for Radius you put a, the square of AF will be =2aa by the Pythag. Theor. But the square of the Diam. FG=2a is 4aa. Therefore the power of the side is to the power of the Diam as 2 to 4, i. e. as 1 to 2. Q. E. D.

More short.

Because AF is also the side of a square inscribed in the grea∣test circle by the gen. of the Octaëdrum; the □ of AF will be by Prop. 50. to the □ of FC as 2 to 1: Therefore to the square of FG as 2 to 4, by Prop. 35. Q. E D.

CONSECTARY.

THerefore if the Diam. of the sphere be made (a) the side of the Octaëdrum AF will be 〈 math 〉〈 math 〉.

Proposition LVIII.

THE side of the Hexaëdrum or Cube(α) 1.72 〈◊〉〈◊〉 in Power subtriple of the diameter of the circumscribed sphere.

Demonstration.

Making a the side of the inscribed cube GF or FE (Fig. 98.) the square of the diagonal GE of the base of the cube will be 2aa by the Pythag. Theor. and by the same Reason the square of the diam. of the cube and the circumscribed sphere GD will be = to the square of GE+□ DE=3aa. Q. E. D.

CONSECTARYS.

I. THerefore if the diameter of the sphere be made = a, the side of the cube AB will be 〈 math 〉〈 math 〉.

II. The diameter of the sphere is equal in power to the side of the Tetraëdrum and cube taken together. For if the power of the diam. of the sphere be made aa the power of the side of the Tetraëdrum will be ⅔ aa by Consect. Prop. 56. and the power of the side of the cube ⅓aa by the pres. Consect. 1. Wherefore these two powers jointly make aa. Q. E. D.

Proposition LIX.

THE side of the Dodecaëdrum(α) 1.73 is equal in power to the greater part of the side of the cube divided in mean and extreme reason.

[illustration]

Pag. 150.

97

98

99

100

101

102

103

Demonstration.

For if to the side of the cube AB (vid. Fig. 45. n. 1.) and to its upper base ABCD you conceive to be accommodated or fitted a regular Pentagon according to the genesis of a Dodeca∣ëdrum laid down in Def. 22, and at the interval Be you make the arch ef, the ▵ ▵ ABe and Aef will be equiangular; (for the angles at A and B being 36°, and AeB 108, having drawn ef, the angles Bef and Bfe are each 72°; therefore Aef the remaining angle will be 36°) wherefore as AB to Be (i. e. Bf) so Ae (i. e. Be or Bf) to Af. Therefore the side of the cube AB is divided in mean and extream reason in f, and Be the side of the Dodecaëdrum is = to the greater part Bf. Q. E. D.

SCHOLIƲM.

HEnce would arise a new method of dividing a given line in mean and extream reason, viz. if you apply to the given line a part of the equilateral Pentagon by means of the angles A and B 36°, and at the interval Be you cut off Bf. This angle may be had geometrically, if another regular Pen∣tagon be inscribed in a circle, and having drawn also a like subtense, if the angles at the subtense are made at A and B equal, by Eucl. 23. lib. 1.

Proposition LX.

THE side of an Icosaëdrum(α) 1.74 is equal in power to the side of a Pentagon in a circle containing only five sides of the Icosaëdrum; and the semi diameter of this circle is in power sub∣quintuple of the Diam. of the sphere of the circumscribed Icosa∣ëdrum.

Demonstration.

Both these are evident from the genesis of the Icosaëdrum in Def. 22. The first immediately hence, because all the other

sides of the triangles (Fig. 99.) Aa, Ba, &c. are made equal to the side of the Pentagon AB by Consect. 4. Prop. 51. The latter from this inference; if for OA the radius of the circle you put a (since the side of the Pentagon, which here is also the side of the Icosaëdrum, it will be equal in power to the radius and side of the Decagon taken together by the aforesaid Prop.) the altitude OG will be the side of the Decagon = 〈 math 〉〈 math 〉 by Consect. 1. Prop. 54. to which the equal in∣ferior part oH being added, and the intermediate altitude Oo =a, you'l have the whole diameter of the circumscribed sphere GH= 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 i e. 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 and so the square of the diameter of the sphere will be 5aa: Therefore the square of the diameter of the sphere is to the square of the semi-diam. of the circle containing the five sides of the Icosaëdrum as 5 to 1. Q. E. D.

SCHOLIƲM.

IT is also evident that a sphere described on the diameter GH will pass thro' the other angles of this Icosaëdrum; for assuming the center between O and o the radius FG will be = 〈 math 〉〈 math 〉. But FA is also = 〈 math 〉〈 math 〉; for the □ of FO is =¼aa, and the □ AO=aa: Therefore the sum is =〈 math 〉〈 math 〉aa = □ FA. Q. E. D.

CONSECTARY I.

THerefore, if the radius of the circle ABCDE remain a, you'l have the altitude OG 〈 math 〉〈 math 〉, and the side of the Icosaëdrum 〈 math 〉〈 math 〉, by Cons. 1. and 2. Prop. 54. and the diam of the circumscribed Sphere 2 〈 math 〉〈 math 〉, as is evident from the Demonstration.

CONSECTARY II. Being a general one of the five last Propositions.

IF AB (Fig. 100) be the diameter of a sphere(α) 1.75 divided in D so that AD shall be ⅓ AB, then (having erected the perpendicular DF) BF will be the side of the Tetraëdrum by Prop. 56. and AF the side of the Hexaëdrum by Prop. 58. Cons. 2. and BE or AE (erecting from the center the perpendicular CE) will be the side of the Octaëdrum by Prop. 57. Now if AF be cut in mean and extreme reason in O, you'l have AO the side of the Dodecaëdrum by Prop. 59. Lastly, if you erect BG double of CB, HI will be double of CI, and the □ of HI=4 □ of CI; consequently the □ CH or CB=5 □ CI. Therefore the □ of AB (double of CH) is also=to 5 □ of HI (which is double of CI) therefore HI is the radius of the circle circumscribing the Pentagon of the Icosaëdrum, and IB the side of the Decagon inscribed in the same circle, and HB the side of the Pentagon, and also the side of the Ico∣saëdrum, by Prop. 60.

Notes

• (α) 1.1

  Eucl. lib. 6. Definit. 1.
• (β) 1.2

  lib. II. Def. 9.
• (α) 1.3

  Eucl. 13. l. 1.
• (α) 1.4

  Eucl. lib. 2. prop. 3. A. C. also the third.
• (β) 1.5

  Eucl. Prop. 4. lib. 2.
• (α) 1.6

  Eucl. & Clav. 5.
• (α) 1.7

  Eucl. & Clav. 6.
• (b) 1.8

  Eucl. & Clav. prop. 7. l. 2
• (α) 1.9

  Eucl. & Clav. prop. 8.
• (β) 1.10

  Eucl & Clav. prop. 9.
• (α) 1.11

  Eucl. & Clav. x.
• (α) 1.12

  Eucl. 17 l. 6 & 20 l. 7.
• (β) 1.13

  lib. I. de Sphaer. & Cyl. Prop. 14.
• (a) 1.14

  lib. I. de Sphaer. & Cyl. prop. 15.
• (b) 1.15

  E••cl. 16 l. 6 & 19. l. 7.
• (α) 1.16

  Eucl. lib. 6. prop. 22.
• (α) 1.17

  Eucl. 15, 16. v. 9 10, 13, vi••.
• (β) 1.18

  18, v.
• (γ) 1.19

  17, v.
• (α) 1.20

  Eucl. 15, 16. v. 9 10, 13, vi••.
• (β) 1.21

  Eucl. 3, 20, 22. lib. v. 14. vii.
• (γ) 1.22

  Eucl. 21, 23. lib. v.
• (α) 1.23

  Eucl. 1. 12. v. 5. 6. 12. vii.
• (α) 1.24

  Eucl. 5 & 19. lib. v. 7 & 11. l. 7.
• (β) 1.25

  Besides se∣veral other Prop. see also the 17 & 18 lib. vii.
• (α) 1.26

  Eucl. 11. lib. II. & 30. lib. vi.
• (α) 1.27

  Eucl. Prop. 1. lib. vi.
• (β) 1.28

  Prop. 5.6. lib. xii. 25, 32. xi. & Cons. 30 & 31 of the same
• (γ) 1.29

  Prop. 11. lib. xii.
• (δ) 1.30

  Prop. 35, 36, 37, 38, 39, 40, lib. lib. 1 & 29, 30, 31. lib. xi.
• (*) 1.31

  Schol prop. 1. l. 6, 12, 13, 14
• ‖ 1.32

  lib. 6. prop. 14. 15. l. 11.34. l. 12.11, and its Coroll. also Prop. 15.
• (α) 1.33

  Prop. 23. lib. 6.
• (β) 1.34

  Eucl. Prop. 2. l. 12.
• (α) 1.35

  Prop. 18 lib. 12.
• (β) 1.36

  Eucl. Prop. 20. l. 3.
• (α) 1.37

  Eucl. prop. 21. 27. 31. lib. 3.
• (β) 1.38

  prop. 32. l. 1
• (α) 1.39

  Eucl. prop. 21. 27. 31. lib. 3.
• (α) 1.40

  Eucl prop. 6. lib. 6.
• (β) 1.41

  Eucl. 2. lib. 6.
• (γ) 1.42

  Eucl. l. 9 & 10. l. 6.
• (α) 1.43

  Eucl. 11 & 12, l. 6.
• (β) 1.44

  Eucl. 13. lib. 6 & and Eucl. 8. lib. 6.
• (α) 1.45

  Eucl. 3. lib. 6.
• (α) 1.46

  Lansberg. Geom. Triang. lib. 4. Prop. 12.
• (α) 1.47

  Eucl. 19 & 20 lib. 6.
• (α) 1.48

  Eucl. prop. 22. lib. 6.
• (α) 1.49

  Eucl. prop. 12. lib. 12. of Cones and Cylinders.
• (β) 1.50

  Eucl. 18. lib. 12.
• (α) 1.51

  Euclid. Prop 7 Coroll. lib. 12.
• (α) 1.52

  Euclid. Prop. 10. lib. 12.
• (α) 1.53

  A••chim. 32 (al. 31.) lib. I de Sphaer.
• (α) 1.54

  Archim. lib. I. de Sph. & Cylind. Prop. 31. (al. 30).
• (α) 1.55

  Archim. Cyclom. prop. 2
• (α) 1.56

  Archim. Prop. 3. Cy∣clom.
• (β) 1.57

  Eucl. last Prop. lib. 10.
• (α) 1.58

  Acta E∣rudit. Ann. 82. p. 44 & the following.
• (α) 1.59

  Euc. 47. Lib. 1.
• (α) 1.60

  Eucl. prop. 12. lib. 2.
• (α) 1.61

  Eucl. Prop. 13. l. 2.
• (α) 1.62

  Eucl. 36. Lib. 3.
• (α) 1.63

  Ptol. lib. I. Almagesh.
• (α) 1.64

  Ptol. lib. I. Almagesh.
• (α) 1.65

  Eucl. 12 lib. 12.
• (α) 1.66

  Eucl. Prop. 13. lib. 13.
• (α) 1.67

  Eucl. Prop. 10. lib. 13.
• (α) 1.68

  Eucl. Prop. 1. lib. 14.
• (α) 1.69

  Eucl. 9. lib. 13. Corol.
• (α) 1.70

  Eucl. 13. lib. 13.
• (α) 1.71

  Eucl. 14. lib. 13.
• (α) 1.72

  Eucl. 15. lib. 13.
• (α) 1.73

  Eucl. Consect. 1. Prop. 17. lib. 13.
• (α) 1.74

  Eucl. Prop. 16. Co∣roll. lib. 13.
• (α) 1.75

  Eucl. Prop. 18. and los••, lib. 13.