The elements of Euclid, explained and demonstrated in a new and most easie method with the uses of each proposition in all the parts of the mathematicks / by Claude Francois Milliet D'Chales, a Jesuit ; done out of French, corrected and augmented, and illustrated with nine copper plates, and the effigies of Euclid, by Reeve Williams ...

THE TWELFTH BOOK OF Euclid's Elements.

EUclid, after having given in the preceding Books, the general Principles of Solids, and ex∣plained the way of measuring the most Regular, that is to say those which are terminated by plain Superficies; treateth in this, of Bodies inclosed under curved Sur∣faces, as are the Cylinder, the Cone, and the Sphere, and comparing them with each other, he giveth the Rules of their Solidity, and the way of measuring of them. This Book is very useful, seeing we find herein those Principles whereon the most skilful Geometricians have established so many curious Demonstrations on the Cylinder, the Cone, and of the Sphere.

PROPOSITION I. THEOREM.

LIke Polygons inscribed in a Circle, are in the samo Ratio as are the Squares of the Diameters of those Circles.

* 1.1 If the Polygons ABCDE, GFHKL, inscribed in Circles, are like; they shall be in the same Ratio as are the Squares of the Diameters AM, FN. Draw the Lines BM, GN, FH, AC.

Demonstration. It is supposed that the Polygons are like, that is to say that the Angles B, and G, are equal; and that there is the same Ratio of AB to BC, as of FG to GH, whence I conclude (by the 6th. of the 6th.) that the Triangles ABC, FGH, are equi∣angular, and that the Angles ACB, FHG, are equal; so then (by the 21st. of the 3d.) the Angles AMB, FNG, are equal. Now the Angles ABM, FGN, being in a semi-circle, are Right (by the 31st. of the 3d.) and consequently the Triangles ABM, FGN, are equi∣angular, Therefore (by the 5th. of the 6th.) there shall be the same Ratio of

AB to FG, as of AM to FN; and (by the 22d. of the 11th.) if one describe Two similar Polygons on AB, FG, which are those proposed; and also Two other like Polygons AM, and FN, which shall be their Squares, there shall be the same Ratio of the Polygon ABCDE to the Polygon FGNKL, as of the Square of AM to the Square of FN.

This Proposition is necessary to Demon∣strate the following Proposition.

LEMMA.

IF a Quantity be less than a Circle, there may be inscribed in the Circle a Regular Polygon, greater than that quantity.

* 1.2 Let the Figure A be less than the Circle B; there may be inscribed in the Circle a Regular Polygon greater than the figure A. Let the figure G, be the difference of the figure A, and the Circle in such sort that the figures A and G taken together be equal to the Circle B. Inscribe in the Circle B, the Squares CDEF (by the 6th. of the

4th.) if that Square were greater than the figure A; we should have what we pretend to. If it be less, divide the Qua∣drants of the Circle CD, DE, EF, FC, into Two equally in the Points H, I, K, L, in such sort as you may have an Octogon. And if that Octogon be less than the figure A; subdivide the Arks, and you shall have a Polygon of Sixteen Sides, then Thirty Two of Sixty Four, &c. I say in fine you shall have a Polygon greater than the figure A; that is to say, a Polygon having less difference from the Circle than the figure A, in such sort that the difference shall be less than the figure G.

Demonstration. The Square inscribed is more than the one half of the Circle, it being the half of the Square described about the Circle, and in describing the Octagon, you take away more than the half of the Remainder, that is to say, the four Segments CHD, DIE, EKF, CLF. For the Triangle CHD, is the half of the Rectangle CO, (by the 34th. of the 1st.) it is then more than the half of the Segment CHD; it is the same of the other Arks. In like manner, in describing a Polygon of Sixteen Sides, you take away more than one half of that which remains in the Circle; and so of the rest. You will then at last leave

a lesser quantity than G. For it is evident that having proposed Two unequal Quanta∣ties, if you take away more than the half of the greater; and again more than the half of the remainder, and again more than the half of that remainder, and so forward; that which remains shall be less than the second quantity. Let us suppose that the second is contained One Hundred times in the first; it is evident that by Dividing the first into One Hundred parts, in such sort that the first may have a greater Reason to the second, than of Two to One; the second shall be less than the Hundreth part. So then you shall at length meet with a Polygon which shall be less surpassed by the Circle, than the Circle doth the figure A; that is to say, that which remains of the Circle, having taken away the Polygon, shall be less than G. The Polygon shall be greater than the figure A.

PROPOSITION II. THEOREM.

CIrcles are in the same Ratio as are the Squares of their Diameters.

I Demonstrate that the Circles A and B are in the same Ratio as are the

Squares of CD, EF. For if they were not in the same Ratio, the Circle A would have a greater Ratio to the Circle B, than the Square of CD to the Square of EF. Let the figure G have the same Ratio to the Circle B, as hath the Square of CD to the Square of EF: the figure G shall be lesser than the Circle A; (by the preceding Lemma) there may be inscribed a Regular Poly∣gon greater than G in the Circle A. Let there also be inscribed in the Circle B, a like regular Polygon.

Demonstration. The Polygon A to the Polygon B, hath the same Ratio, as the Square of CD to the Square of EF; that is to say, the same as hath G to the Circle B: Now the quantity G is lesser than the Polygon inscribed in A: so then (by the 14th. of the 5th.) the Circle should be less than the Polygon which is inscribed, which is evidently false. It must then be said, that the figure G, less than the Circle A, cannot have the same Ratio to the Circle B, as the Square of CD to the Square of EF; and by consequence, that Circle A hath not a greater Ratio to the Circle B, than the Square of CD to the Square of EF. Neither hath it less, because

that the Circle B to the Circle A, would have a greater Ratio, and there would be applyed the same Demonstration.

Coroll. 1. Circles are in duplicate Ratio of that of their Diameters; be∣cause that Squares being like or similar, are in duplicate Ratio of that of their sides.

Coroll. 2. Circles are in the same Ratio, as are the similar Polygons in∣scribed in them.

Coroll. 3. This general Rule must be well taken notice of, when like figures inscribed in other like figures, in such sort that they become nearer and nearer, and degenerate, in fine, into those figures, they are always in the same Ratio. I would say, that if like regular Polygons be described in several Circles, they are in the same Ratio, as are the Squares of the Diameters; and that the greater number of sides they are made to have, they become so much the nearer the Circles; the Circle shall have the same Ratio as the Squares of their Diame∣ters. This way of measuring Bodies by inscription is very necessary.

USE.

THis Proposition is very universal, and is the way of our reasoning on Circles, after the same manner as on Squares. For example, we say (in the 47th. of the first,) that in a Rectangular Triangle, that the Square of the Base is equal to the Square of the other sides taken together. We may say the same of Circles; that is to say, that the Circle described on the Base of a Rectangular Triangle is equal to the Circles which hath the sides for Diameters, and after this we may augment or diminish a Circle into what Proportion we list. We prove also in Opticks, that the Light decreaseth in duplicate Ratio of that of the distance of the Luminous Body.

PROPOSITION III. THEOREM.

EVery Pyramid whose Base is Trian∣gular, may be divided into two equal Prisms, which are greater than half of the Pyramid, and into Two equal Pyramids.

There may be found in the Pyramid

ABCD, two equal Prisms EBFI, EHKC, which shall be greater than half the Pyramid. Divide the six Sides of the Pyramid into equal parts in G, F, E, I, H, K; and draw the Lines EG, GF, FE, HI, IK, EK.

Demonstration. In the Triangle ABD, there is the same Ratio of AG to GB, as of AF to FD; seeing they are equal: thence (by the 2d. of the 6th.) GF, BD, are parallels; and GF shall be the half of BD, that is to say, equal to BH. In like manner GE, BI; FE, HI, shall be parallel and equal; and (by the 15th. of the 11th.) the Planes GFE, BHI, shall be parallel; and by consequence EBFI, shall be a Prism. I say the same of the figure HEKF, which shall also be a Prism equal to the former; and (by the 40th. of the 11th.) seeing the Parallelogram Base HIKD, is double to the Triangle BHI, (by the 41th. of the first.)

Secondly, the Pyramids AEFG, ECKI, are like and equal.

Demonstration. The Triangles AFG, FDH, are equal (by the 3d. of the first,) as also FDH, EIK. In like manner the Triangles AGE, EIC, and so of the other Triangles of the Pyramid;

they are then equal (by the 10th. and 11th. Def.) they are also similar to the great Pyramid ABDC; for the Tri∣angles, ABC, AGE; are like (by the 2d. of the 6th.) the Lines GE being pa∣rallel; which I could Demonstrate in all the Triangles of the lesser Pyramids.

In fine, I conclude that the Prisms are more than the one half of the first Pyramid. For if each were equal to one of the lesser Pyramids, the Two Prisms would be the half of the great Pyramid: Now they are greater than one of the Pyramids, as the Prism GHE, contains a Pyramid AGFE, which may easily be proved from the pa∣rallelism of their Sides; whence I con∣clude that the Two Prisms taken toge∣ther, are greater than the Two Pyra∣mids, and consequently greater than half the greater Pyramid.

PROPOSITION. IV. THEOREM.

IF Two Triangular Pyramids of equal height, be divided into Two Prisms and Two Pyramids, and that the last Pyramids be divided after the same manner, all the Prisms of the one Pyramid shall have the same Ratio to all those of the other, as the Base of the one Pyramid hath to the Base of the other.

If one divide the Two Pyramids ABCD, DEFG, of equal height, and of Triangular Bases, into Two Prisms, and into Two Pyramids, ac∣cording to the method of the Third Proposition; and if one should subdivide after the same manner the Two little Pyramids, and so consecutively, in such sort that there be as many divisions in the one as in the other, there then being the same number of Prisms in both. I say that all the Prisms of the one, to all the Prisms of the other, shall be in the same Ratio as are the Bases.

Demonstration. Seeing the Pyramids are of the same height; the Prisms pro∣duced by the first division, shall also

have the same height, seeing each hath the half of that of their Pyramids: Now Prisms of equal height, are in the same Ratio as are their Bases, (by the Coroll. of the 32d. of the 11th.) the Bases BTV, EPX, are like unto the Bases BDC, EGF, and having for Sides the half of those of the great Bases, they shall be but the one fourth of the great Bases; so then they shall be in the same Ratio as are the great Bases. There∣fore the first Prisms shall have the same Ratio as the greater Bases. I prove after the same manner, that the Prisms produced by the second Division, that is to say of the little Pyramids, shall be in the same Ratio, as the greater Bases. Therefore all the Prisms of the one, have the same Ratio to all the Prisms of the other, as the Base to the Base.

USE.

THese Two Propositions were necessary to compare Pyramids the one with the other, and to measure them.

PROPOSITION V. THEOREM.

TRiangular Pyramids of the same height, are in the same Ratio as their Bases.

The Pyramids ABCD, EFGH, are in the same Ratio as are their Bases. For if they were not, one of the Pyramids, for example ABCD, would have a greater Ratio to the Pyramid EFGH, than the Base BCD to the Base FGH. So that a quantity L, less than ABCD, would have the same Ratio to the Pyra∣mid EFGH, as the Base BCD to the Base FGH. Divide the Pyramid ABCD after the manner of the Third Proposition; divide also the Pyramid which will re∣sult of the first division into Two Prisms, and into two Pyramids; and those again into Two other Prisms; continue divi∣ding after this manner as long as need requires: Seeing the Prisms of the first division are more than the half of the Pyramid ABCD (by the 3d.) and that the Prisms of the second Division is greater than half the Remainder; that is

to say than the little Pyramid, and that those of the Third Division is more than the half Remainder; it is evident, that by making many sub-divisions, there will at length remain a quantity less than the excess of the Pyramid ABCD above the quantity L; that is to say, that all the Prisms being put together, shall be greater than the quantity L. Make as many Divisions in the Pyramid EFGH, untill you have as many Prisms as there are in ABCD.

Demonstration. The Prisms of ABCD hath the same Ratio to the Prisms of EFGH, as hath the Base BCD to the Base FGH: Now the Ratio of the Base BCD to the Base FGH, is the same with that of the quantity L to the Pyramid EFGH. There is therefore the same Ratio of the Prisms of ABCD to the Prisms of EFGH, as of the quantity L to the Pyramid EFGH. Again, the Prisms of ABCD are grea∣ter than the quantity L; thence (by the 14th. of the 5th) the Prisms compre∣hended in the Pyramid EFGH, would be greater than the same Pyramid EFGH; which is evidently false, the part not being capable of being greater than the whole. We must then acknowledge

that a Magnitude less than one of the Pyramids, cannot have the same Ratio to the other, as hath the Base to the Base; and consequently, neither of the Pyra∣mids hath a greater Ratio to the other, than the Base to the Base.

PROPOSITION VI. THEOREM.

ALL sorts of Pyramids, of the same height, have the same Ratio as their Bases.

The Pyramids ABC, DEFG, of the same height, are in the same Ratio as are their Bases BC, EFG. Divide the Bases into Triangles.

Demonstration. The Triangular Py∣ramids AB, DE, of the same height, are in the same Ratio as are their Bases (by the 5th.) In like manner, the Tri∣angular Pyramids AC, DF, are in the same Ratio as their Bases. There shall be then the same Ratio of the Pyramid ABC to the Pyramid DEF, as of the Base BC to the Base EF (by the 12th. of the 5th.) Again, seeing there is the same Ratio of the Pyramid DEF to the Pyramid ABC, as of the Base EF to

the Base BC; and that there is also the same Ratio of the Pyramid DG to the Pyramid ABC, as of the Base G to the Base BC; there will be also the same Ratio of the Pyramid DEFG to the Pyramid ABC, as of the Base EFG to Base BC.

PROPOSITION VII. THEOREM.

EVery Pyramid is the Third part of a Prism of the same Base, and of the same height.

Let there be proposed in the first place a Triangular Prism AB; I say that a Pyramid which hath for its Base one of the Triangles ACE, BDF, and which shall be of the same height, as the Pyra∣mid ACEF, shall be the Third part of the Prism. Draw the Three Diagonals AF, DC, FC, of the same parallelo∣gram.

Demonstration. The Prism is divi∣ded into Three equal Pyramids ACFE, ACFD, CFBD; thence each of them shall be the Third part of the Prism. The Two first, having for Base the Tri∣angles AFE, AFD, which are equal

(by the 34th. of the first) and for height the perpendicular drawn from their top C to the Plane of their Base AF, they shall be equal (by the preceding.) The Pyramids ACFD, CFBD, which have for Bases equal Triangles ADC, DCB, and the same Vertex F, shall also be equal (by the preceding.) Thence one of these Pyramids; for example AFEC, which hath the same Base ACE, as the Prism; and the same height, which would be the perpendi∣cular drawn from the point F, to the Plane of the Base ACE; is the third part of the same. If the Prism were a Polygon, it must be then divided into several Triangular Prisms; and the Pyra∣mid which would have the same Base, and the same height, would be also di∣vided into so many Triangular Pyramids, each of which would be the Third part of its Prism. Therefore (by the 32d. of the 5th.) the Polygonal Pyramid would be the Third part of the Polygonal Prism.

PROPOSITION VIII. THEOREM.

LIke Pyramids are in triple Ratio of their homologous Sides.

If the Pyramids are Triangular, they may be converted into Prisms, which shall be also like, seeing they would have some Planes, which would be the same as those of the Pyramids. Now like Prisms are in Triple Ratio of their homologous sides, (by the 39th. of the 11th.) thence the Pyramids which are the Third of them (by the preceding) shall be in Triple Ratio of their homologous sides.

If the Pyramids are Polygons, they may be reduced into Triangular Py∣ramids.

PROPOSITION IX. THEOREM.

EQual Pyramids have their height and Bases Reciprocal, and those Pyra∣mids whose height and Bases are reciprocal, are equal.

Let there be proposed Two Triangu∣lar Pyramids, they shall have their Bases and heights reciprocal. Let there be made Prisms of equal heights and Bases; seeing those Prisms are Triple each to its Pyramid (by the 7th.) they shall be also equal. Now equal Prisms, have their Bases and heights reciprocal (by the 4th. Coroll. of the 39th.) thence the Bases and heights of the Pyramids, which are the same with those of the Prisms shall be reciprocal to them.

Secondly, if the Bases and heights of Pyramids be reciprocal, the Prisms shall be equal, as also the Pyramids, which are the Third parts of them.

If the Pyramids were Polygons, they ought to be reduced into Triangular Py∣ramids.

Coroll. There might be made some other Propositions; for example, that Pyramids of equal height, are in the same Ratio as their Bases, and those which have the same Base, are in the same Ratio as their heights.

USE

FRom these Propositions is gathered a way of measuring of Pyramids; which is, to multiply the Base by the one third of its height. Then there might be made some other Propositions, as that if a Prism be equal to a Pyramid, the Bases and height of the Prism, with the Third part of the height of the Pyramid shall be reciprocal; that is to say, if there be the same Ratio of the Pyramid to the Base of the Prism, as of the height of the Prism to the third part of the height of the Pyramid; the Pyra∣mid and Prism shall then be equal.

LEMMA.

IF there be proposed a quantity less than a Cylinder, there may be in∣scribed in that Cylinder a Polygon Prism greater than that quantity.

If the Quantity A be less than the Cy∣linder which hath the Circle B for its Base; there may be inscribed in the Cylinder a Polygon Prism greater than the quantity

A. The Square CDEF is inscribed, GHIK is circumscribed, CLDMENFO is an Octogon. Draw the Tangent PLQ, and continue the Sides ED, FC, to P and Q; and imagine as many Prisms of the same height as the Cylinder, which have for their Bases those Polygons. That which hath for its Base the Circumscribed Square incompasseth the Cylinder, and that on the inscribed Square is also inscribed in the Cylinder.

Demonstration. The Prisms which have the same height, are in the same Ratio as are their Bases (by the Coroll. of the 39th. of the 11th.) and the inscribed Square being the half of the circumscribed Square, its Prism shall be the half of the other; it shall then be more than the half of the Cylinder. Making a Prism which hath the Octagon for Base, there will be taken away more than the half of that which remained for the Cylinder, having taken away the Prism of the inscribed Square; because the Triangle CLD is half the rectangle CQ; and because that Prisms of equal or the same height, are in Ratio as are their Bases, the Prism which hath for its Base the Triangle CLD, shall be the half of the Prism of the Rect∣angle DCPQ; it shall then be more than

the half of the part of the Cylinder, which hath for Base the Segment DLC. It is the same of the other Segments. I demonstrate af∣ter the same manner, that in making a Polygon Prism of Sixteen Sides, I take away more than half of what remained of the Cylinder, by taking away the Octogonal Prism; there will remain a part of the Cylinder, less than the excess of the Cylinder above the quan∣tity A. We shall then have a Prism inscri∣bed in the Cylinder, which will be less ex∣ceeded by the Cylinder, than the quantity A. One might argue after the same method about inscribed Pyramids in a Cone.

PROPOSITION X. THEOREM.

A Cone is the Third part of a Cylinder of the same Base, and the same height.

If a Cone, and a Cylinder have the Circle A for Base, and the same height; the Cylinder shall be triple to the Cone. For if the Cylinder had a greater Ratio to the Cone, than the Triple; a quan∣tity B less than the Cylinder, would have the same Ratio to the Cone, as

Three to One; and (by the preceding Lemma) there might be inscribed in the Cylinder, a Polygonal Prism greater than the quantity B. Let us suppose that that quantity is that which hath for Base the Polygon CDEFGH. Make also on the same Base, a Pyramid inscri∣bed in the Cone.

Demonstration. The Cylinder, the Cone, the Prism, and the Pyramid, are all of the same height; thence the Prism is triple to the Pyramid (by the 7th.) Now the quantity B is also the triple of the Cone; there is then therefore the same Ratio of the Prism to the Pyramid, as of the quantity B to the Cone; and (by the 14th. of the 5th.) seeing the Prism is greater than the quantity B, the Py∣ramid should be greater than the Cone in which it is inscribed, which cannot be.

But if it were said that the Cone hath a greater Ratio to the Cylinder, than one to three; there might be taken the same method to Demonstrate the con∣trary.

PROPOSITION XI. THEOREM.

CYlinders and Cones of the same height, are in the same Ratio as are their Bases.

There is proposed two Cylinders or two Cones of the same height, which have the Circles A and B for their Bases: I say they are in the same Ratio as are their Bases. For if they are not in the same Ratio; one of them, for example A, shall have a greater Ratio to the Cylinder B, than the Base A to the Base B; so that a quantity L, less than the Cylinder A, would have the same Ratio to the Cylinder B, as hath the Base A to the Base B. There may then be in∣scribed a Polygon Prism in the Cylinder A, greater than the quantity L. Let it be that which hath for Base the Polygon CDEF; and let there be inscribed a like Polygon GHIK, in the Base B, which serveth for Base to the Cylinder of the same height.

Demonstration. The Prisms A and B are in the same Ratio as are their Polygon Bases (by the Coroll. 4. of the 39th. of the 11th.) and the Polygons are in the same Ratio as are the Circles (by the Coroll. of the 2d.) so then the Prism A shall be in the same Ratio to the Prism B, as the Circle A to the Circle B. Now as the Circle A is to the Cirle B, so is the quan∣tity L to the Cylinder B; therefore as the Prism A is to the Prism B, so is the quantity L to the Cylinder B. The Prism A is greater than the quantity L; by consequence (according to the 14th. of the 5th.) the Prism B inscribed in the Cylinder B, would be greater than it, which cannot be. Therefore neither Cylinder hath greater Ratio to the other, than that of its Base to the Base of the other.

Coroll. Cylinders are triple to Cones of the same height, therefore Cones of the same height, are in the same Ratio as are their Bases.

PROPOSITION XII. THEOREM.

LIke Cylinders and Cones are in Triple Ratio to that of the Diameters of their Bases.

Let there be proposed two like Cy∣linders, or two Cones, which have for their Bases the Circles A and B. I say that the Ratio of the Cylinder A to the Cylinder B, is in Triple Ratio of the Diameter DC to the Diameter EF. For if it hath not a Triple Ratio, let the quantity G, less than the Cylinder A, have to the Cylinder B a Triple Ratio of that of the Diameter DC to the Dia∣meter EF; and let there be inscribed in the Cylinder A a Prism, which let be greater than G; and another like thereto in the Cylinder B: they shall have the same Altitude or height as had the Cy∣linders, for like Cylinders have their heights and the Diameters of their Bases proportional, as well as the Prisms (by the 11th. Def. of the 11th.)

Demonstration. The Diameter DC hath the same Ratio to the Diameter

EF, as hath the Side DI to the Side DL, or DC, to EF, (as I Demonstra∣ted in the first.) Now the Prisms are in Triple Ratio of their homologous Sides (by the Coroll. 4. of the 39 of 11.) therefore the Prism A to the Prism B, is in Triple Ratio of DC to EF. We have sup∣posed that the quantity G, in respect of the Cylinder B, was in Triple Ratio of the Diameter DC to the Diameter EF. So then there will be the same Ratio of the Prism A to the Prism B, as of the quantity G to the Cylinder B; and (by the 14th. of the fifth) seeing that the Prism A, is greater than the quantity G; the Prism B, that is to say, descri∣bed in the Cylinder B, would be greater than the same Cylinder, which is im∣possible. Therefore like Cylinders are in Triple Ratio of the Diameters of their Bases.

Cones are the Third parts of Cy∣linders (by the Tenth;) therefore like Cones are in Triple Ratio of that of the Diameters of their Bases.

PROPOSITION XIII. THEOREM.

IF a Cylinder be cut by a Plane parallel to its Base, the parts of the Axis shall be in the same Ratio, as are the parts of the Cylinder.

Let the Cylinder AB, be cut by the Plane DC, parallel to its Base. I say that there shall be the same Ratio of the Cylinder AF to the Cylinder FB, as of the Line AF to the Line FB. Draw the Line BG perpendicular to the Plane of its Base A, draw also in the Planes of the Circles DC, and A, the Lines FE, AG.

Demonstration. The Plane of the Triangle BAG, cutteth the parallel Planes A and DC; thence the Sections EF, and AG, are parallel (by the 16th. of the 11th.) So then there is the same Ratio of AF to FB, as of the height GE to EB. Let there be taken an Aliquot part of EB, and having divi∣ded GE, EB, into parts equal to that Aliquot part, let there be drawn Planes parallel to the Base AG: You shall have so many C linders of the same height, which

having their Bases and heights equal, they shall be equal (by the 11th.) Again, the Lines AF and FB, shall be divi∣ded after the same manner as EG, EB, (by the 16th. of the first) So then the Line AF contains as many times the Aliquot part of the Line FB, as the Cylinder AF contains a like Aliquot parts of FB. There is thence the same Ratio of the parts of the Cylinder, as of the parts of the Axis.

Coroll. The parts of the perpendicular are in the same Ratio as are the parts of the Cylinder.

PROPOSITION XIV. THEOREM.

CYlinders and Cones having the same Base, are in the same Ratio as are their height.

Two Cylinders AB; CD, of equal Bases, being proposed; cut out of the greater a Cylinder of the same height, with the lesser, by drawing a Plane EF parallel to its Base. It is evident that the Cylinders EF, AB, are equal (by the 11th.) and that CF to CD, hath

the same Ratio as GI to GH, or (by the Coroll. of the preceding) as the height of the Cylinder CF to the height of CD, there is thence the same Ratio of AB to CD, as of the height of EF or AB, to the height of CD.

As to Cones, seeing they are the one thirds of Cylinders, if they have equal Bases, they shall be also in the same Ratio as are their height.

PROPOSITION XV. THEOREM.

EQual Cylinders and Cones, have their Bases and height reciprocal, and those which have their Bases and heights reciprocal, are equal.

If the Cylinders AB, CD, are equal; there shall be the same Ratio of the Base B to the Base D, as of the height CD to the height AB. Let the height DE be equal to the height AB.

Demonstration. There is the same Ratio of the Cylinder AB to the Cylin∣der DE, of the same height, as of the Base B to the Base D (by the 14th.) Now as the Cylinder AB is to the Cy∣linder DE; so is the Cylinder CD

equal to AB, to the Cylinder DE; that is to say, so is the height CD to the height AB or DE. Therefore as the Base B is to the Base D, so is the height CD to the height AB.

Secondly, If there be the same Ratio of the Base B to the Base D, as of the height CD to the height AB; the Cylinders AB, CD, shall be equal. For the Cylinder AB is to the Cylin∣der DE, as is the Base B to the Base D, and the Cylinder CD shall have the same Ratio to the Cylinder DE, as CD to DE; there shall thence be the same Ratio of AB to DE, as of CD to DE; and (by the 8th. of the 5th.) the Cylinders AB, and CD, shall be equal.

The Propositions 16 and 17 are very difficult, and are only of use in the 18th. which I will Demonstrate after a more easie method by the Two following Lemma's.

LEMMA I.

IF there be proposed a quantity less than a Sphere; there may be inscri∣bed in the same Sphere a number of Cy∣linders, which taken together shall be of the same height, and greater than that quantity.

* 1.3 Let ABC be half of a great Circle of the Sphere, and let the quantity D be less than the same hemi sphere; I say that there may be inscribed therein several Cylinders of the same height, which taken together, shall be greater than the quantity D. For if the hemi-sphere surpasseth in bigness the quantity D, it will surpass it by a certain quantity; let that quantity be the Cylinder MP. In such sort that the quantity D, and MP taken together, be equal to the hemi-sphere. Make or conceive it so made, that there be the same Ratio of a great Circle of the Sphere to the Base MO, as of the height MN to the height R. Divide the Line EB into as many equal parts as you please, each of them less than R; and drawing parallels to the Line AG, describe inscribed and circumscribed parallelograms. The Number of the circumscribed shall surpass by a unit those of the inscribed.

Now all the circumscribed rectangles sur∣pass the inscribed, by the little Rectangles through which the Circumference of the Circle passeth; and all those Rectangles taken together are equal to the Rectangle AL. Imagine that the Semi-Circle is made to roul about the Diameter EB; the Semi-Circle shall describe a hemi-sphere, and the inscribed Rectangles will describe inscribed Cylinders in the semi-sphere, and the Circumscribed will describe other Cylinders.

Demonstration. The Circumscribed Cy∣linders surpass more the inscribed, than doth the hemi-sphere surpass the same in∣scribed Cylinders; seeing that they are comprehended within the Circumscribed Cy∣linders. Now the Circumscribed surpass the inscribed by so much as is the Cylinder AL; therefore the hemi-sphere shall sur∣pass by less the inscribed Cylinders, than doth the Cylinder made by the Rectangle AL. The Cylinder AL, is less than the Cylinder MP; for there is the same Ratio of a great Circle of the Sphere which serveth for Base to the Cylinder AL, as of MP to R; so then by the foregoing, a Cylinder which hath for Base a great Circle of the Sphere, and the Altitude R, would be equal to the Cylinder MP.

Consequently the hemi-sphere which sur∣passeth the quantity D, by the Cylinder MP; and the inscribed Cylinders by a quantity less than AL, surpasseth the in∣scribed Cylinders by less than the quantity D. Therefore the quantity D, is less than the inscribed Cylinders.

What I have said of a hemi sphere, may be said of a whole Sphere.

LEMMA II.

LIke Cylinders inscribed in Two Spheres, are in Triple Ratio of the Diameters of the Spheres.

* 1.4 If the two like Cylinders CD, EF, be inscribed in the Spheres A, B; they shall be in Triple Ratio of the Diameters LM, NO. Draw the Lines GD, IF.

Demonstration. The like Right Cylin∣ders CD, EF, are like; so then there is the same Ratio of HD to DR, as of QF to FS; as also the same Ratio of KD to KG, as of PF to PI. And consequently the Triangles GDK, IFP, are like (by the 6th. of the 6th.) so there shall be the same Ratio of KD to PF, as of GD to IF, or of LM to ON. Now the like Cylinders CD, EF, are in Triple

Ratio of KD and PF, the Diameters of their Bases (by the 12th.) therefore the like Cylinders CD, EF, inscribed in the Spheres A and B, are in triple Ratio of the Diameters of the Spheres.

PROPOSITION XVIII. THEOREM.

SPheres are in triple Ratio of their Dia∣meters.

The Spheres A and B are in Triple Ratio of their Diameters CD, EF. For if they be not in Triple Ratio; one of the Spheres, as A, shall be in a greater Ratio than Triple, of that of CD to EF; therefore a quantity G less than the Sphere A, shall be in Triple Ratio of that of CD to EF; and so one might (according to the first Lemma) inscribe in the Sphere A, Cylinders of the same height, greater than the quantity G.

Let there be inscribed in the Sphere B, as many like Cylinders as those of the Cylinder A.

Demonstration. The Cylinders of the Sphere A, to those of the Sphere B, shall be in Triple Ratio of that of CD to EF (by the preceding:) Now the quantity G

in respect of the Sphere B, is in Triple Ratio of CD to EF; there is then the same Ratio of the Cylinders of the Sphere A, to the like Cylinders of the Sphere B. So then the Cylinders of A being greater than the quantity G; the Cylinder B, that is to say, inscribed in the Sphere B, would be greater than the Sphere B, which is impossible. There∣fore the Spheres A and B are in Triple Ratio of that of their Diameters.

Coroll. Spheres are in the same Ratio as are the Cubes of their Diameters; seeing that the Cubes being like solids, are in Triple Ratio of that of their Sides.

Notes

• * 1.1

  Plate VIII.
• * 1.2

  Lemma. Fig. II.
• * 1.3

  Lemma. Fig. I.
• * 1.4

  Lemma. Fig. II.