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

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

Proposition XLIV.

IN Right-angled Triangles(α) 1.1 (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) 1.2 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) 1.3 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) 1.4 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) 1.5 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) 1.6 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) 1.7 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) 1.8 (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) 1.9 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) 1.10 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) 1.11 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) 1.12 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) 1.13 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) 1.14 〈◊〉〈◊〉 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) 1.15 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) 1.16 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) 1.17 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

  Euc. 47. Lib. 1.
• (α) 1.2

  Eucl. prop. 12. lib. 2.
• (α) 1.3

  Eucl. Prop. 13. l. 2.
• (α) 1.4

  Eucl. 36. Lib. 3.
• (α) 1.5

  Ptol. lib. I. Almagesh.
• (α) 1.6

  Ptol. lib. I. Almagesh.
• (α) 1.7

  Eucl. 12 lib. 12.
• (α) 1.8

  Eucl. Prop. 13. lib. 13.
• (α) 1.9

  Eucl. Prop. 10. lib. 13.
• (α) 1.10

  Eucl. Prop. 1. lib. 14.
• (α) 1.11

  Eucl. 9. lib. 13. Corol.
• (α) 1.12

  Eucl. 13. lib. 13.
• (α) 1.13

  Eucl. 14. lib. 13.
• (α) 1.14

  Eucl. 15. lib. 13.
• (α) 1.15

  Eucl. Consect. 1. Prop. 17. lib. 13.
• (α) 1.16

  Eucl. Prop. 16. Co∣roll. lib. 13.
• (α) 1.17

  Eucl. Prop. 18. and los••, lib. 13.