BOOK XII.
HISTORICAL NOTE.
The predominant feature of Book XII. is the use of the
method of exhaustion, which is applied in Propositions
2,
3-5,
10, 11, 12, and (in a slightly different form) in
Propositions 16-18. We conclude therefore that for the content of this Book Euclid was greatly indebted to Eudoxus, to whom the discovery of the method of exhaustion is attributed. The evidence for this attribution comes mainly from Archimedes. (1) In the preface to
On the Sphere and Cylinder I., after stating the main results obtained by himself regarding the surface of a sphere or a segment thereof, and the volume and surface of a right cylinder with height equal to its diameter as compared with those of a sphere with the same diameter, Archimedes adds: “Having now discovered that the properties mentioned are true of these figures, I cannot feel any hesitation in setting them side by side both with my former investigations and
with those of the theorems of Eudoxus on solids which are held to be most irrefragably established, namely that
any pyramid is one third part of the prism which has the same base with the pyramid and equal height [i.e. Eucl.
XII. 7], and that
any cone is one third part of the cylinder which has the same base with the cone and equal height [i.e. Eucl.
XII. 10]. For, though these properties also were naturally inherent in the figures all along, yet they were in fact unknown to all the many able geometers who lived before Eudoxus and had not been observed by any one.”
(2) In the preface to the treatise known as the
Quadrature of the Parabola Archimedes states the “lemma”
assumed by him and known as the “Axiom of Archimedes”
(see note on
X. 1 above) and proceeds: “Earlier geometers (
οἱ πρότερον γεωμέτραι) have also used this lemma; for it is by the use of this same lemma that they have shown that
circles are to one another in the duplicate ratio of their diameters [Eucl.
XII. 2], and that
spheres are to one another in the triplicate ratio of their diameters [Eucl.
XII. 18], and further that
every pyramid is one third part of the prism which has the same base with the pyramid and equal height [Eucl.
XII. 7]; also, that
every cone is one third part of the cylinder which has the same base with the cone and equal height [Eucl.
XII. 10] they proved by assuming
a certain lemma similar to that aforesaid.”
Thus in the first passage two theorems of Eucl. XII. are definitely attributed to Eudoxus; and, when Archimedes says, in the second passage, that “earlier geometers”
proved these two theorems by means of the lemma known as the “Axiom of Archimedes”
and of a lemma similar to it respectively, we can hardly suppose him to be alluding to any other proof than that given by Eudoxus. As a matter of fact, the lemma used by Euclid to prove both propositions (
XII. 3-5 and
7, and
XII. 10) is the theorem of Eucl.
X. 1. As regards the connexion between the two “lemmas”
see note on
X. 1.
We are not, however, to suppose that none of the
results obtained by the method of exhaustion had been discovered before the time of Eudoxus (fl. about 368-5 B.C.). Two at least are of earlier date, those of
Eucl. XII. 2 and
XII. 7.
(
a) Simplicius (
Comment. in Aristot. Phys. p. 61, ed. Diels) quotes Eudemus as saying, in his
History of Geometry, that Hippocrates of Chios (fl. say 430 B.C.) first laid it down (
ἔθετο) that similar segments of circles are in the ratio of the squares on their bases and that he proved this (
ἐδείκνυεν) by proving (
ἐκ τοῦ δεῖξαι) that the squares on the diameters have the same ratio as the (whole) circles. We know nothing of the method by which Hippocrates proved this proposition; but, having regard to the evidence from Archimedes quoted above, it is not permissible to suppose that the method was the fully developed
method of exhaustion as we know it.
(
b) As regards the two theorems about the volume of a pyramid and of a cone respectively, which Eudoxus was the first to prove, we now have authentic evidence in the short treatise by Archimedes discovered by Heiberg in a MS. at Constantinople in 1906 and published in
Hermes the following year (see now
Archimedis opera omnia, ed. Heiberg, 2. ed., Vol. II., 1913, pp. 425-507; T. L. Heath,
The Method of Archimedes, Cambridge, 1912). The said treatise, complete in all essentials, bears the title
Ἀρχιμήδους περὶ τῶν μηχανικῶν θεωρημάτων πρὸς Ἐρατοσθένην ἔφοδος. This “Method”
(or “Plan of attack”
), addressed to Eratosthenes, is none other than the
ἐφόδιον on which, according to Suidas, Theodosius wrote a commentary, and which is several times cited by Heron in his
Metrica; its discovery adds a new and important chapter to the history of the integral calculus. In the preface to this work Archimedes alludes to the theorems which he first discovered by means of mechanical considerations, but proved afterwards by geometry, because the investigation by means of mechanics did not constitute a rigid proof; he observes, however, that the mechanical method is of great use for the
discovery of theorems, and it is much easier to provide the rigid proof when the fact to be proved has once been discovered than it would be if nothing were known to begin with. He goes on: “Hence too, in the case of those theorems the proof of which was first discovered by Eudoxus, namely those relating to the cone and the pyramid, that the cone is one third part of the cylinder, and the pyramid one third part of the prism, having the same base and equal height, no small part of the credit will naturally be assigned to Democritus, who was the first to make the statement (of the fact) regarding the said figure [i.e. property], though without proving it.”
Hence the
discovery of the two theorems must now be attributed to Democritus (fl. towards the end of 5th cent. B.C.). The words “without proving it”
(
χωρὶς ἀποδείξεως) do not mean that Democritus gave no sort of proof, but only that he did not give a proof on the rigorous lines required later; for the same words are used by Archimedes of his own investigations by means of mechanics, which, however, do constitute a reasoned argument. The character of Archimedes' mechanical arguments combined with a passage of Plutarch about a particular question in infinitesimals said to have been raised by Democritus may perhaps give a clue to the line of Democritus' argument as regards the pyramid. The essential feature of Archimedes' mechanical arguments in this tract is that he regards an area as the sum of an infinite number of
siraight lines parallel to one another and terminated by the boundary or boundaries of the closed figure the area of which is to be found, and a volume as the sum of an infinite number of
plane sections parallel to one another: which is of course the same thing as taking (as we do in the integral calculus) the sum of an infinite number of strips of breadth
dx (say), when
dx becomes indefinitely small, or the sum of an infinite number of parallel laminae of depth
dz (say), when
dz becomes indefinitely small. To give only one instance, we may take the case of the area of a segment of a parabola cut off by a chord.
Let
CBA be the parabolic segment,
CE the tangent at
C meeting the diameter
EBD through the middle point of the chord
CA in
E, so that
.
Draw
AF parallel to
ED meeting
CE produced in
F. Produce
CB to
H so that , where
K is the point in which
CH meets
AF; and suppose
CH to be a lever.
Let any diameter
MNPO be drawn meeting the curve in
P and
CF,
CK,
CA in
M,
N,
O respectively.
Archimedes then observes that
(“for this is proved in a lemma”
), whence
, so that, if a straight line
TG equal to
PO be placed with its middle point at
H, the straight line
MO with centre of gravity at
N, and the straight line
TG with centre of gravity at
H, will balance about
K.
Taking all other parts of diameters like
PO intercepted between the curve and
CA, and placing equal straight lines with their centres of gravity at
H, these straight lines collected at
H will balance (about
K) all the lines like
MO parallel to
FA intercepted within the triangle
CFA in the positions in which they severally lie in the figure.
Hence Archimedes infers that an area equal to that of the parabolic segment hung at
H will balance (about
K) the triangle
CFA hung at its centre of gravity, the point
X (a point on
CK such that ), and therefore that
,
from which it follows that
.
The same sort of argument is used for solids,
plane sections taking the place of
straight lines.
Archimedes is careful to state once more that this method of argument does not constitute a
proof. Thus, at the end of the above proposition about the parabolic segment, he adds: “This property is of course not proved by what has just been said; but it has furnished a sort of
indication (
ἔμφασίν τινα) that the conclusion is true.”
Let us now turn to the passage of Plutarch (
De Comm. Not. adv. Stoicos XXXIX 3) about Democritus above referred to. Plutarch speaks of Democritus as having raised the question in natural philosophy (
φυσικῶς): “if a cone were cut by a plane parallel to the base [by which is clearly meant a plane indefinitely near to the base], what must we think of the surfaces of the sections, that they are equal or unequal? For, if they are unequal, they will make the cone irregular, as having many indentations, like steps, and unevennesses; but, if they are equal, the sections will be equal, and the cone will appear to have the property of the cylinder and to be made up of equal, not unequal circles, which is very absurd.”
The phrase “
made up of equal...circles”
(
ἐξ ἴσων συγκείμενος...κύκλων) shows that Democritus already had the idea of a solid being the sum of an infinite number of parallel planes, or indefinitely thin laminae, indefinitely near together: a most important anticipation of the same thought which led to such fruitful results in Archimedes. If then one may hazard a conjecture as to Democritus' argument with regard to a pyramid, it seems probable that he would notice that, if two pyramids of the same height and equal triangular bases are respectively cut by planes parallel to the base and dividing the heights in the same ratio, the corresponding sections of the two pyramids are equal, whence he would infer that the pyramids are equal as being the sum of the same infinite number of equal plane sections or indefinitely thin laminae. (This would be a particular anticipation of Cavalieri's proposition that the areal or solid contents of two figures are equal if two sections of them taken at the same height, whatever the height may be, always give equal straight lines or equal surfaces respectively.) And Democritus would of course see that the three pyramids into which a prism on the same base and of equal height with the original pyramid is divided (as in Eucl.
XII. 7) satisfy this test of equality, so that the pyramid would be one third part of the prism. The extension to a pyramid with a polygonal base would be easy. And Democritus may have stated the proposition for the cone (of course without an absolute proof) as a natural inference from the result of increasing indefinitely the number of sides in a regular polygon forming the base of a pyramid.
BOOK XII. PROPOSITIONS.
PROPOSITION 1.
Similar polygons inscribed in circles are to one another as the squares on the diameters.
Let
ABC,
FGH be circles, let
ABCDE,
FGHKL be similar polygons inscribed in them, and let
BM,
GN be diameters of the circles; I say that, as the square on
BM is to the square on
GN, so is the polygon
ABCDE to the polygon
FGHKL.
For let
BE,
AM,
GL,
FN be joined.
Now, since the polygon
ABCDE is similar to the polygon
FGHKL, the angle
BAE is equal to the angle
GFL, and, as
BA is to
AE, so is
GF to
FL. [VI. Def. I]
Thus
BAE,
GFL are two triangles which have one angle equal to one angle, namely the angle
BAE to the angle
GFL, and the sides about the equal angles proportional; therefore the triangle
ABE is equiangular with the triangle
FGL. [
VI. 6]
Therefore the angle
AEB is equal to the angle
FLG.
But the angle
AEB is equal to the angle
AMB, for they stand on the same circumference; [
III. 27] and the angle
FLG to the angle
FNG; therefore the angle
AMB is also equal to the angle
FNG.
But the right angle
BAM is also equal to the right angle
GFN; [
III. 31] therefore the remaining angle is equal to the remaining angle. [
I. 32]
Therefore the triangle
ABM is equiangular with the triangle
FGN.
Therefore, proportionally, as
BM is to
GN, so is
BA to
GF. [
VI. 4]
But the ratio of the square on
BM to the square on
GN is duplicate of the ratio of
BM to
GN, and the ratio of the polygon
ABCDE to the polygon
FGHKL is duplicate of the ratio of
BA to
GF; [
VI. 20] therefore also, as the square on
BM is to the square on
GN, so is the polygon
ABCDE to the polygon
FGHKL.
Therefore etc. Q. E. D.
PROPOSITION 2.
Circles are to one another as the squares on the diameters.
Let
ABCD,
EFGH be circles, and
BD,
FH their diameters; I say that, as the circle
ABCD is to the circle
EFGH, so is the square on
BD to the square on
FH.
For, if the square on
BD is not to the square on
FH as the circle
ABCD is to the circle
EFGH, then, as the square on
BD is to the square on
FH, so will the circle
ABCD be either to some less area than the circle
EFGH, or to a greater.
First, let it be in that ratio to a less area
S.
Let the square
EFGH be inscribed in the circle
EFGH; then the inscribed square is greater than the half of the circle
EFGH, inasmuch as, if through the points
E,
F,
G,
H we draw tangents to the circle, the square
EFGH is half the square circumscribed about the circle, and the circle is less than the circumscribed square; hence the inscribed square
EFGH is greater than the half of the circle
EFGH.
Let the circumferences
EF,
FG,
GH,
HE be bisected at the points
K,
L,
M,
N, and let
EK,
KF,
FL,
LG,
GM,
MH,
HN,
NE be joined; therefore each of the triangles
EKF,
FLG,
GMH,
HNE is also greater than the half of the segment of the circle about it, inasmuch as, if through the points
K,
L,
M,
N we draw tangents to the circle and complete the parallelograms on the straight lines
EF,
FG,
GH,
HE, each of the triangles
EKF,
FLG,
GMH,
HNE will be half of the parallelogram about it, while the segment about it is less than the parallelogram; hence each of the triangles
EKF,
FLG,
GMH,
HNE is greater than the half of the segment of the circle about it.
Thus, by bisecting the remaining circumferences and joining straight lines, and by doing this continually, we shall leave some segments of the circle which will be less than the excess by which the circle
EFGH exceeds the area
S.
For it was proved in the first theorem of the tenth book that, if two unequal magnitudes be set out, and if from the greater there be subtracted a magnitude greater than the half, and from that which is left a greater than the half, and if this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.
Let segments be left such as described, and let the segments of the circle
EFGH on
EK,
KF,
FL,
LG,
GM,
MH,
HN,
NE be less than the excess by which the circle
EFGH exceeds the area
S.
Therefore the remainder, the polygon
EKFLGMHN, is greater than the area
S.
Let there be inscribed, also, in the circle
ABCD the polygon
AOBPCQDR similar to the polygon
EKFLGMHN; therefore, as the square on
BD is to the square on
FH, so is the polygon
AOBPCQDR to the polygon
EKFLGMHN. [
XII. 1]
But, as the square on
BD is to the square on
FH, so also is the circle
ABCD to the area
S; therefore also, as the circle
ABCD is to the area
S, so is the polygon
AOBPCQDR to the polygon
EKFLGMHN; [
V. 11] therefore, alternately, as the circle
ABCD is to the polygon inscribed in it, so is the area
S to the polygon
EKFLGMHN. [
V. 16]
But the circle
ABCD is greater than the polygon inscribed in it; therefore the area
S is also greater than the polygon
EKFLGMHN.
But it is also less: which is impossible.
Therefore, as the square on
BD is to the square on
FH, so is not the circle
ABCD to any area less than the circle
EFGH.
Similarly we can prove that neither is the circle
EFGH to any area less than the circle
ABCD as the square on
FH is to the square on
BD.
I say next that neither is the circle
ABCD to any area greater than the circle
EFGH as the square on
BD is to the square on
FH.
For, if possible, let it be in that ratio to a greater area
S.
Therefore, inversely, as the square on
FH is to the square on
DB, so is the area
S to the circle
ABCD.
But, as the area
S is to the circle
ABCD, so is the circle
EFGH to some area less than the circle
ABCD; therefore also, as the square on
FH is to the square on
BD, so is the circle
EFGH to some area less than the circle
ABCD: [
V. 11] which was proved impossible.
Therefore, as the square on
BD is to the square on
FH, so is not the circle
ABCD to any area greater than the circle
EFGH.
And it was proved that neither is it in that ratio to any area less than the circle
EFGH; therefore, as the square on
BD is to the square on
FH, so is the circle
ABCD to the circle
EFGH.
Therefore etc. Q. E. D.
LEMMA.
I say that, the area
S being greater than the circle
EFGH, as the area
S is to the circle
ABCD, so is the circle
EFGH to some area less than the circle
ABCD.
For let it be contrived that, as the area
S is to the circle
ABCD, so is the circle
EFGH to the area
T.
I say that the area
T is less than the circle
ABCD.
For since, as the area
S is to the circle
ABCD, so is the circle
EFGH to the area
T, therefore, alternately, as the area
S is to the circle
EFGH, so is the circle
ABCD to the area
T. [
V. 16]
But the area
S is greater than the circle
EFGH; therefore the circle
ABCD is also greater than the area
T.
Hence, as the area
S is to the circle
ABCD, so is the circle
EFGH to some area less than the circle
ABCD. Q. E. D.
PROPOSITION 3.
Any pyramid which has a triangular base is divided into two pyramids equal and similar to one another,
similar to the whole and having triangular bases,
and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.
Let there be a pyramid of which the triangle
ABC is the base and the point
D the vertex; I say that the pyramid
ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.
For let
AB,
BC,
CA,
AD,
DB,
DC be bisected at the points
E,
F,
G,
H,
K,
L, and let
HE,
EG,
GH,
HK,
KL,
LH,
KF,
FG be joined.
Since
AE is equal to
EB, and
AH to
DH, therefore
EH is parallel to
DB. [
VI. 2]
For the same reason
HK is also parallel to
AB.
Therefore
HEBK is a parallelogram; therefore
HK is equal to
EB. [
I. 34]
But
EB is equal to
EA; therefore
AE is also equal to
HK.
But
AH is also equal to
HD; therefore the two sides
EA,
AH are equal to the two sides
KH,
HD respectively, and the angle
EAH is equal to the angle
KHD; therefore the base
EH is equal to the base
KD. [
I. 4]
Therefore the triangle
AEH is equal and similar to the triangle
HKD.
For the same reason the triangle
AHG is also equal and similar to the triangle
HLD.
Now, since two straight lines
EH,
HG meeting one another are parallel to two straight lines
KD,
DL meeting one another, and are not in the same plane, they will contain equal angles. [
XI. 10]
Therefore the angle
EHG is equal to the angle
KDL.
And, since the two straight lines
EH,
HG are equal to the two
KD,
DL respectively, and the angle
EHG is equal to the angle
KDL, therefore the base
EG is equal to the base
KL; [
I. 4] therefore the triangle
EHG is equal and similar to the triangle
KDL.
For the same reason the triangle
AEG is also equal and similar to the triangle
HKL.
Therefore the pyramid of which the triangle
AEG is the base and the point
H the vertex is equal and similar to the pyramid of which the triangle
HKL is the base and the point
D the vertex. [
XI. Def. 10]
And, since
HK has been drawn parallel to
AB, one of the sides of the triangle
ADB, the triangle
ADB is equiangular to the triangle
DHK, [
I. 29] and they have their sides proportional; therefore the triangle
ADB is similar to the triangle
DHK. [
VI. Def. 1]
For the same reason the triangle
DBC is also similar to the triangle
DKL, and the triangle
ADC to the triangle
DLH.
Now, since the two straight lines
BA,
AC meeting one another are parallel to the two straight lines
KH,
HL meeting one another, not in the same plane, they will contain equal angles. [
XI. 10]
Therefore the angle
BAC is equal to the angle
KHL.
And, as
BA is to
AC, so is
KH to
HL; therefore the triangle
ABC is similar to the triangle
HKL.
Therefore also the pyramid of which the triangle
ABC is the base and the point
D the vertex is similar to the pyramid of which the triangle
HKL is the base and the point
D the vertex.
But the pyramid of which the triangle
HKL is the base and the point
D the vertex was proved similar to the pyramid of which the triangle
AEG is the base and the point
H the vertex.
Therefore each of the pyramids
AEGH,
HKLD is similar to the whole pyramid
ABCD.
Next, since
BF is equal to
FC, the parallelogram
EBFG is double of the triangle
GFC.
And since, if there be two prisms of equal height, and one have a parallelogram as base, and the other a triangle, and if the parallelogram be double of the triangle, the prisms are equal, [
XI. 39] therefore the prism contained by the two triangles
BKF,
EHG, and the three parallelograms
EBFG,
EBKH,
HKFG is equal to the prism contained by the two triangles
GFC,
HKL and the three parallelograms
KFCL,
LCGH,
HKFG.
And it is manifest that each of the prisms, namely that in which the parallelogram
EBFG is the base and the straight line
HK is its opposite, and that in which the triangle
GFC is the base and the triangle
HKL its opposite, is greater than each of the pyramids of which the triangles
AEG,
HKL are the bases and the points
H,
D the vertices, inasmuch as, if we join the straight lines
EF,
EK, the prism in which the parallelogram
EBFG is the base and the straight line
HK its opposite is greater than the pyramid of which the triangle
EBF is the base and the point
K the vertex.
But the pyramid of which the triangle
EBF is the base and the point
K the vertex is equal to the pyramid of which the triangle
AEG is the base and the point
H the vertex; for they are contained by equal and similar planes.
Hence also the prism in which the parallelogram
EBFG is the base and the straight line
HK its opposite is greater than the pyramid of which the triangle
AEG is the base and the point
H the vertex.
But the prism in which the parallelogram
EBFG is the base and the straight line
HK its opposite is equal to the prism in which the triangle
GFC is the base and the triangle
HKL its opposite, and the pyramid of which the triangle
AEG is the base and the point
H the vertex is equal to the pyramid of which the triangle
HKL is the base and the point
D the vertex.
Therefore the said two prisms are greater than the said two pyramids of which the triangles
AEG,
HKL are the bases and the points
H,
D the vertices.
Therefore the whole pyramid, of which the triangle
ABC is the base and the point
D the vertex, has been divided into two pyramids equal to one another and into two equal prisms, and the two prisms are greater than the half of the whole pyramid. Q. E. D.
PROPOSITION 4.
If there be two pyramids of the same height which have triangular bases,
and cach of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms,
then,
as the base of the one pyramid is to the base of the other pyramid,
so will all the prisms in the one pyramid be to all the prisms,
being equal in multitude,
in the other pyramid.
Let there be two pyramids of the same height which have the triangular bases
ABC,
DEF, and vertices the points
G,
H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [
XII. 3] I say that, as the base
ABC is to the base
DEF, so are all the prisms in the pyramid
ABCG to all the prisms, being equal in multitude, in the pyramid
DEFH,
For, since
BO is equal to
OC, and
AL to
LC, therefore
LO is parallel to
AB, and the triangle
ABC is similar to the triangle
LOC.
For the same reason the triangle
DEF is also similar to the triangle
RVF.
And, since
BC is double of
CO, and
EF of
FV, therefore, as
BC is to
CO, so is
EF to
FV.
And on
BC,
CO are described the similar and similarly situated rectilineal figures
ABC,
LOC, and on
EF,
FV the similar and similarly situated figures
DEF,
RVF; therefore, as the triangle
ABC is to the triangle
LOC, so is the triangle
DEF to the triangle
RVF; [
VI. 22] therefore, alternately, as the triangle
ABC is to the triangle
DEF, so is the triangle
LOC to the triangle
RVF. [
V. 16]
But, as the triangle
LOC is to the triangle
RVF, so is the prism in which the triangle
LOC is the base and
PMN its opposite to the prism in which the triangle
RVF is the base and
STU its opposite; [Lemma following] therefore also, as the triangle
ABC is to the triangle
DEF, so is the prism in which the triangle
LOC is the base and
PMN its opposite to the prism in which the triangle
RVF is the base and
STU its opposite.
But, as the said prisms are to one another, so is the prism in which the parallelogram
KBOL is the base and the straight line
PM its opposite to the prism in which the parallelogram
QEVR is the base and the straight line
ST its opposite. [
XI. 39; cf.
XII. 3]
Therefore also the two prisms, that in which the parallelogram
KBOL is the base and
PM its opposite, and that in which the triangle
LOC is the base and
PMN its opposite, are to the prisms in which
QEVR is the base and the straight line
ST its opposite and in which the triangle
RVF is the base and
STU its opposite in the same ratio [
V. 12]
Therefore also, as the base
ABC is to the base
DEF, so are the said two prisms to the said two prisms.
And similarly, if the pyramids
PMNG,
STUH be divided into two prisms and two pyramids, as the base
PMN is to the base
STU, so will the two prisms in the pyramid
PMNG be to the two prisms in the pyramid
STUH.
But, as the base
PMN is to the base
STU, so is the base
ABC to the base
DEF; for the triangles
PMN,
STU are equal to the triangles
LOC,
RVF respectively.
Therefore also, as the base
ABC is to the base
DEF, so are the four prisms to the four prisms.
And similarly also, if we divide the remaining pyramids into two pyramids and into two prisms, then, as the base
ABC is to the base
DEF, so will all the prisms in the pyramid
ABCG be to all the prisms, being equal in multitude, in the pyramid
DEFH. Q. E. D.
LEMMA.
But that, as the triangle
LOC is to the triangle
RVF, so is the prism in which the triangle
LOC is the base and
PMN its opposite to the prism in which the triangle
RVF is the base and
STU its opposite, we must prove as follows.
For in the same figure let perpendiculars be conceived drawn from
G,
H to the planes
ABC,
DEF; these are of course equal because, by hypothesis, the pyramids are of equal height.
Now, since the two straight lines
GC and the perpendicular from
G are cut by the parallel planes
ABC,
PMN, they will be cut in the same ratios. [
XI. 17]
And
GC is bisected by the plane
PMN at
N; therefore the perpendicular from
G to the plane
ABC will also be bisected by the plane
PMN.
For the same reason the perpendicular from
H to the plane
DEF will also be bisected by the plane
STU.
And the perpendiculars from
G,
H to the planes
ABC,
DEF are equal; therefore the perpendiculars from the triangles
PMN,
STU to the planes
ABC,
DEF are also equal.
Therefore the prisms in which the triangles
LOC,
RVF are bases, and
PMN,
STU their opposites, are of equal height.
Hence also the parallelepipedal solids described from the said prisms are of equal height and are to one another as their bases; [
XI. 32] therefore their halves, namely the said prisms, are to one another as the base
LOC is to the base
RVF. Q. E. D.
PROPOSITION 5.
Pyramids which are of the same height and have triangular bases are to one another as the bases.
Let there be pyramids of the same height, of which the triangles
ABC,
DEF are the bases and the points
G,
H the vertices; I say that, as the base
ABC is to the base
DEF, so is the pyramid
ABCG to the pyramid
DEFH.
For, if the pyramid
ABCG is not to the pyramid
DEFH as the base
ABC is to the base
DEF, then, as the base
ABC is to the base
DEF, so will the pyramid
ABCG be either to some solid less than the pyramid
DEFH or to a greater.
Let it, first, be in that ratio to a less solid
W, and let the pyramid
DEFH be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; then the two prisms are greater than the half of the whole pyramid. [
XII. 3]
Again, let the pyramids arising from the division be similarly divided, and let this be done continually until there are left over from the pyramid
DEFH some pyramids which are less than the excess by which the pyramid
DEFH exceeds the solid
W. [X. I]
Let such be left, and let them be, for the sake of argument,
DQRS,
STUH; therefore the remainders, the prisms in the pyramid
DEFH, are greater than the solid
W.
Let the pyramid
ABCG also be divided similarly, and a similar number of times, with the pyramid
DEFH; therefore, as the base
ABC is to the base
DEF, so are the prisms in the pyramid
ABCG to the prisms in the pyramid
DEFH. [
XII. 4]
But, as the base
ABC is to the base
DEF, so also is the pyramid
ABCG to the solid
W; therefore also, as the pyramid
ABCG is to the solid
W, so are the prisms in the pyramid
ABCG to the prisms in the pyramid
DEFH; [V. II] therefore, alternately, as the pyramid
ABCG is to the prisms in it, so is the solid
W to the prisms in the pyramid
DEFH. [
V. 16]
But the pyramid
ABCG is greater than the prisms in it; therefore the solid
W is also greater than the prisms in the pyramid
DEFH.
But it is also less: which is impossible.
Therefore the prism
ABCG is not to any solid less than the pyramid
DEFH as the base
ABC is to the base
DEF.
Similarly it can be proved that neither is the pyramid
DEFH to any solid less than the pyramid
ABCG as the base
DEF is to the base
ABC.
I say next that neither is the pyramid
ABCG to any solid greater than the pyramid
DEFH as the base
ABC is to the base
DEF.
For, if possible, let it be in that ratio to a greater solid
W; therefore, inversely, as the base
DEF is to the base
ABC, so is the solid
W to the pyramid
ABCG.
But, as the solid
W is to the solid
ABCG, so is the pyramid
DEFH to some solid less than the pyramid
ABCG, as was before proved; [
XII. 2, Lemma] therefore also, as the base
DEF is to the base
ABC, so is the pyramid
DEFH to some solid less than the pyramid
ABCG: [V. II] which was proved absurd.
Therefore the pyramid
ABCG is not to any solid greater than the pyramid
DEFH as the base
ABC is to the base
DEF.
But it was proved that neither is it in that ratio to a less solid.
Therefore, as the base
ABC is to the base
DEF, so is the pyramid
ABCG to the pyramid
DEFH. Q. E. D.
PROPOSITION 6.
Pyramids which are of the same height and have polygonal bases are to one another as the bases.
Let there be pyramids of the same height of which the polygons
ABCDE,
FGHKL are the bases and the points
M,
N the vertices; I say that, as the base
ABCDE is to the base
FGHKL, so is the pyramid
ABCDEM to the pyramid
FGHKLN.
For let
AC,
AD,
FH,
FK be joined.
Since then
ABCM,
ACDM are two pyramids which have triangular bases and equal height, they are to one another as the bases; [
XII. 5] therefore, as the base
ABC is to the base
ACD, so is the pyramid
ABCM to the pyramid
ACDM.
And,
componendo, as the base
ABCD is to the base
ACD, so is the pyramid
ABCDM to the pyramid
ACDM. [
V. 18]
But also, as the base
ACD is to the base
ADE, so is the pyramid
ACDM to the pyramid
ADEM. [
XII. 5]
Therefore,
ex aequali, as the base
ABCD is to the base
ADE, so is the pyramid
ABCDM to the pyramid
ADEM. [
V. 22]
And again
componendo, as the base
ABCDE is to the base
ADE, so is the pyramid
ABCDEM to the pyramid
ADEM. [
V. 18]
Similarly also it can be proved that, as the base
FGHKL is to the base
FGH, so is the pyramid
FGHKLN to the pyramid
FGHN.
And, since
ADEM,
FGHN are two pyramids which have triangular bases and equal height, therefore, as the base
ADE is to the base
FGH, so is the pyramid
ADEM to the pyramid
FGHN. [
XII. 5]
But, as the base
ADE is to the base
ABCDE, so was the pyramid
ADEM to the pyramid
ABCDEM.
Therefore also,
ex aequali, as the base
ABCDE is to the base
FGH, so is the pyramid
ABCDEM to the pyramid
FGHN. [
V. 22]
But further, as the base
FGH is to the base
FGHKL, so also was the pyramid
FGHN to the pyramid
FGHKLN.
Therefore also,
ex aequali, as the base
ABCDE is to the base
FGHKL, so is the pyramid
ABCDEM to the pyramid
FGHKLN. [
V. 22] Q. E. D.
PROPOSITION 7.
Any prism which has a triangular base is divided into three pyramids equal to one another which have triangular bases.
Let there be a prism in which the triangle
ABC is the base and
DEF its opposite; I say that the prism
ABCDEF is divided into three pyramids equal to one another, which have triangular bases.
For let
BD,
EC,
CD be joined.
Since
ABED is a parallelogram, and
BD is its diameter, therefore the triangle
ABD is equal to the triangle
EBD; [
I. 34] therefore also the pyramid of which the triangle
ABD is the base and the point
C the vertex is equal to the pyramid of which the triangle
DEB is the base and the point
C the vertex. [
XII. 5]
But the pyramid of which the triangle
DEB is the base and the point
C the vertex is the same with the pyramid of which the triangle
EBC is the base and the point
D the vertex; for they are contained by the same planes.
Therefore the pyramid of which the triangle
ABD is the base and the point
C the vertex is also equal to the pyramid of which the triangle
EBC is the base and the point
D the vertex.
Again, since
FCBE is a parallelogram, and
CE is its diameter, the triangle
CEF is equal to the triangle
CBE. [
I. 34]
Therefore also the pyramid of which the triangle
BCE is the base and the point
D the vertex is equal to the pyramid of which the triangle
ECF is the base and the point
D the vertex. [
XII. 5]
But the pyramid of which the triangle
BCE is the base and the point
D the vertex was proved equal to the pyramid of which the triangle
ABD is the base and the point
C the vertex; therefore also the pyramid of which the triangle
CEF is the base and the point
D the vertex is equal to the pyramid of which the triangle
ABD is the base and the point
C the vertex; therefore the prism
ABCDEF has been divided into three pyramids equal to one another which have triangular bases.
And, since the pyramid of which the triangle
ABD is the base and the point
C the vertex is the same with the pyramid of which the triangle
CAB is the base and the point
D the vertex, for they are contained by the same planes, while the pyramid of which the triangle
ABD is the base and the point
C the vertex was proved to be a third of the prism in which the triangle
ABC is the base and
DEF its opposite, therefore also the pyramid of which the triangle
ABC is the base and the point
D the vertex is a third of the prism which has the same base, the triangle
ABC, and
DEF as its opposite.
PORISM.
From this it is manifest that any pyramid is a third part of the prism which has the same base with it and equal height. Q. E. D.
PROPOSITION 8.
Similar pyramids which have triangular bases are in the triplicate ratio of their corresponding sides.
Let there be similar and similarly situated pyramids of which the triangles
ABC,
DEF, are the bases and the points
G,
H the vertices; I say that the pyramid
ABCG has to the pyramid
DEFH the ratio triplicate of that which
BC has to
EF.
For let the parallelepipedal solids
BGML,
EHQP be completed.
Now, since the pyramid
ABCG is similar to the pyramid
DEFH, therefore the angle
ABC is equal to the angle
DEF, the angle
GBC to the angle
HEF, and the angle
ABG to the angle
DEH; and, as
AB is to
DE, so is
BC to
EF, and
BG to
EH.
And since, as
AB is to
DE, so is
BC to
EF, and the sides are proportional about equal angles, therefore the parallelogram
BM is similar to the parallelogram
EQ.
For the same reason
BN is also similar to
ER, and
BK to
EO; therefore the three parallelograms
MB,
BK,
BN are similar to the three
EQ,
EO,
ER.
But the three parallelograms
MB,
BK,
BN are equal and similar to their three opposites, and the three
EQ,
EO,
ER are equal and similar to their three opposites. [
XI. 24]
Therefore the solids
BGML,
EHQP are contained by similar planes equal in multitude.
Therefore the solid
BGML is similar to the solid
EHQP.
But similar parallelepipedal solids are in the triplicate ratio of their corresponding sides. [
XI. 33]
Therefore the solid
BGML has to the solid
EHQP the ratio triplicate of that which the corresponding side
BC has to the corresponding side
EF.
But, as the solid
BGML is to the solid
EHQP, so is the pyramid
ABCG to the pyramid
DEFH, inasmuch as the pyramid is a sixth part of the solid, because the prism which is half of the parallelepipedal solid [
XI. 28] is also triple of the pyramid. [
XII. 7]
Therefore the pyramid
ABCG also has to the pyramid
DEFH the ratio triplicate of that which
BC has to
EF. Q. E. D.
PORISM.
From this it is manifest that similar pyramids which have polygonal bases are also to one another in the triplicate ratio of their corresponding sides.
For, if they are divided into the pyramids contained in them which have triangular bases, by virtue of the fact that the similar polygons forming their bases are also divided into similar triangles equal in multitude and corresponding to the wholes [
VI. 20], then, as the one pyramid which has a triangular base in the one complete pyramid is to the one pyramid which has a triangular base in the other complete pyramid, so also will all the pyramids which have triangular bases contained in the one pyramid be to all the pyramids which have triangular bases contained in the other pyramid [
V. 12], that is, the pyramid itself which has a polygonal base to the pyramid which has a polygonal base.
But the pyramid which has a triangular base is to the pyramid which has a triangular base in the triplicate ratio of the corresponding sides; therefore also the pyramid which has a polygonal base has to the pyramid which has a similar base the ratio triplicate of that which the side has to the side.
PROPOSITION 9.
In equal pyramids which have triangular bases the bases are reciprocally proportional to the heights; and those pyramids in which the bases are reciprocally proportional to the heights are equal.
For let there be equal pyramids which have the triangular bases
ABC,
DEF and vertices the points
G,
H; I say that in the pyramids
ABCG,
DEFH the bases are reciprocally proportional to the heights, that is, as the base
ABC is to the base
DEF, so is the height of the pyramid
DEFH to the height of the pyramid
ABCG.
For let the parallelepipedal solids
BGML,
EHQP be completed.
Now, since the pyramid
ABCG is equal to the pyramid
DEFH, and the solid
BGML is six times the pyramid
ABCG, and the solid
EHQP six times the pyramid
DEFH, therefore the solid
BGML is equal to the solid
EHQP.
But in equal parallelepipedal solids the bases are reciprocally proportional to the heights; [
XI. 34] therefore, as the base
BM is to the base
EQ, so is the height of the solid
EHQP to the height of the solid
BGML.
But, as the base
BM is to
EQ, so is the triangle
ABC to the triangle
DEF. [
I. 34]
Therefore also, as the triangle
ABC is to the triangle
DEF, so is the height of the solid
EHQP to the height of the solid
BGML. [
V. 11]
But the height of the solid
EHQP is the same with the height of the pyramid
DEFH, and the height of the solid
BGML is the same with the height of the pyramid
ABCG, therefore, as the base
ABC is to the base
DEF, so is the height of the pyramid
DEFH to the height of the pyramid
ABCG.
Therefore in the pyramids
ABCG,
DEFH the bases are reciprocally proportional to the heights.
Next, in the pyramids
ABCG,
DEFH let the bases be reciprocally proportional to the heights; that is, as the base
ABC is to the base
DEF, so let the height of the pyramid
DEFH be to the height of the pyramid
ABCG; I say that the pyramid
ABCG is equal to the pyramid
DEFH.
For, with the same construction, since, as the base
ABC is to the base
DEF, so is the height of the pyramid
DEFH to the height of the pyramid
ABCG, while, as the base
ABC is to the base
DEF, so is the parallelogram
BM to the parallelogram
EQ, therefore also, as the parallelogram
BM is to the parallelogram
EQ, so is the height of the pyramid
DEFH to the height of the pyramid
ABCG. [
V. 11]
But the height of the pyramid
DEFH is the same with the height of the parallelepiped
EHQP, and the height of the pyramid
ABCG is the same with the height of the parallelepiped
BGML; therefore, as the base
BM is to the base
EQ, so is the height of the parallelepiped
EHQP to the height of the parallelepiped
BGML.
But those parallelepipedal solids in which the bases are reciprocally proportional to the heights are equal; [
XI. 34] therefore the parallelepipedal solid
BGML is equal to the parallelepipedal solid
EHQP.
And the pyramid
ABCG is a sixth part of
BGML, and the pyramid
DEFH a sixth part of the parallelepiped
EHQP; therefore the pyramid
ABCG is equal to the pyramid
DEFH.
Therefore etc Q. E. D.
PROPOSITION 10.
Any cone is a third part of the cylinder which has the same base with it and equal height.
For let a cone have the same base, namely the circle
ABCD, with a cylinder and equal height; I say that the cone is a third part of the cylinder, that is, that the cylinder is triple of the cone.
For if the cylinder is not triple of the cone, the cylinder will be either greater than triple or less than triple of the cone.
First let it be greater than triple, and let the square
ABCD be inscribed in the circle
ABCD; [
IV. 6] then the square
ABCD is greater than the half of the circle
ABCD.
From the square
ABCD let there be set up a prism of equal height with the cylinder.
Then the prism so set up is greater than the half of the cylinder, inasmuch as, if we also circumscribe a square about the circle
ABCD [
IV. 7], the square inscribed in the circle
ABCD is half of that circumscribed about it, and the solids set up from them are parallelepipedal prisms of equal height, while parallelepipedal solids which are of the same height are to one another as their bases; [
XI. 32] therefore also the prism set up on the square
ABCD is half of the prism set up from the square circumscribed about the circle
ABCD; [cf.
XI. 28, or
XII. 6 and 7, Por.] and the cylinder is less than the prism set up from the square circumscribed about the circle
ABCD; therefore the prism set up from the square
ABCD and of equal height with the cylinder is greater than the half of the cylinder.
Let the circumferences
AB,
BC,
CD,
DA be bisected at the points
E,
F,
G,
H, and let
AE,
EB,
BF,
FC,
CG,
GD,
DH,
HA be joined; then each of the triangles
AEB,
BFC,
CGD,
DHA is greater than the half of that segment of the circle
ABCD which is about it, as we proved before. [
XII. 2]
On each of the triangles
AEB,
BFC,
CGD,
DHA let prisms be set up of equal height with the cylinder; then each of the prisms so set up is greater than the half part of that segment of the cylinder which is about it, inasmuch as, if we draw through the points
E,
F,
G,
H parallels to
AB,
BC,
CD,
DA, complete the parallelograms on
AB,
BC,
CD,
DA, and set up from them parallelepipedal solids of equal height with the cylinder, the prisms on the triangles
AEB,
BFC,
CGD,
DHA are halves of the several solids set up; and the segments of the cylinder are less than the parallelepipedal solids set up; hence also the prisms on the triangles
AEB,
BFC,
CGD,
DHA are greater than the half of the segments of the cylinder about them.
Thus, bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles prisms of equal height with the cylinder, and doing this continually, we shall leave some segments of the cylinder which will be less than the excess by which the cylinder exceeds the triple of the cone. [
X. 1]
Let such segments be left, and let them be
AE,
EB,
BF,
FC,
CG,
GD,
DH,
HA; therefore the remainder, the prism of which the polygon
AEBFCGDH is the base and the height is the same as that of the cylinder, is greater than triple of the cone.
But the prism of which the polygon
AEBFCGDH is the base and the height the same as that of the cylinder is triple of the pyramid of which the polygon
AEBFCGDH is the base and the vertex is the same as that of the cone; [
XII. 7, Por.] therefore also the pyramid of which the polygon
AEBFCGDH is the base and the vertex is the same as that of the cone is greater than the cone which has the circle
ABCD as base.
But it is also less, for it is enclosed by it: which is impossible.
Therefore the cylinder is not greater than triple of the cone.
I say next that neither is the cylinder less than triple of the cone,
For, if possible, let the cylinder be less than triple of the cone, therefore, inversely, the cone is greater than a third part of the cylinder.
Let the square
ABCD be inscribed in the circle
ABCD; therefore the square
ABCD is greater than the half of the circle
ABCD.
Now let there be set up from the square
ABCD a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone, seeing that, as we proved before, if we circumscribe a square about the circle, the square
ABCD will be half of the square circumscribed about the circle, and if we set up from the squares parallelepipedal solids of equal height with the cone, which are also called prisms, the solid set up from the square
ABCD will be half of that set up from the square circumscribed about the circle; for they are to one another as their bases. [
XI. 32]
Hence also the thirds of them are in that ratio; therefore also the pyramid of which the square
ABCD is the base is half of the pyramid set up from the square circumscribed about the circle.
And the pyramid set up from the square about the circle is greater than the cone, for it encloses it.
Therefore the pyramid of which the square
ABCD is the base and the vertex is the same with that of the cone is greater than the half of the cone.
Let the circumferences
AB,
BC,
CD,
DA be bisected at the points
E,
F,
G,
H, and let
AE,
EB,
BF,
FC,
CG,
GD,
DH,
HA be joined; therefore also each of the triangles
AEB,
BFC,
CGD,
DHA is greater than the half part of that segment of the circle
ABCD which is about it.
Now, on each of the triangles
AEB,
BFC,
CGD,
DHA let pyramids be set up which have the same vertex as the cone; therefore also each of the pyramids so set up is, in the same manner, greater than the half part of that segment of the cone which is about it.
Thus, by bisecting the circumferences that are left, joining straight lines, setting up on each of the triangles a pyramid which has the same vertex as the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone exceeds the third part of the cylinder. [
X. 1]
Let such be left, and let them be the segments on
AE,
EB,
BF,
FC,
CG,
GD,
DH,
HA; therefore the remainder, the pyramid of which the polygon
AEBFCGDH is the base and the vertex the same with that of the cone, is greater than a third part of the cylinder.
But the pyramid of which the polygon
AEBFCGDH is the base and the vertex the same with that of the cone is a third part of the prism of which the polygon
AEBFCGDH is the base and the height is the same with that of the cylinder; therefore the prism of which the polygon
AEBFCGDH is the base and the height is the same with that of the cylinder is greater than the cylinder of which the circle
ABCD is the base.
But it is also less, for it is enclosed by it: which is impossible.
Therefore the cylinder is not less than triple of the cone.
But it was proved that neither is it greater than triple; therefore the cylinder is triple of the cone; hence the cone is a third part of the cylinder.
Therefore etc. Q. E. D.
PROPOSITION 11.
Cones and cylinders which are of the same height are to one another as their bases.
Let there be cones and cylinders of the same height, let the circles
ABCD,
EFGH be their bases,
KL,
MN their axes and
AC,
EG the diameters of their bases; I say that, as the circle
ABCD is to the circle
EFGH, so is the cone
AL to the cone
EN.
For, if not, then, as the circle
ABCD is to the circle
EFGH, so will the cone
AL be either to some solid less than the cone
EN or to a greater.
First, let it be in that ratio to a less solid
O, and let the solid
X be equal to that by which the solid
O is less than the cone
EN; therefore the cone
EN is equal to the solids
O,
X.
Let the square
EFGH be inscribed in the circle
EFGH; therefore the square is greater than the half of the circle.
Let there be set up from the square
EFGH a pyramid of equal height with the cone; therefore the pyramid so set up is greater than the half of the cone, inasmuch as, if we circumscribe a square about the circle, and set up from it a pyramid of equal height with the cone, the inscribed pyramid is half of the circumscribed pyramid, for they are to one another as their bases, [
XII. 6] while the cone is less than the circumscribed pyramid.
Let the circumferences
EF,
FG,
GH,
HE be bisected at the points
P,
Q,
R,
S, and let
HP,
PE,
EQ,
QF,
FR,
RG,
GS,
SH be joined.
Therefore each of the triangles
HPE,
EQF,
FRG,
GSH is greater than the half of that segment of the circle which is about it.
On each of the triangles
HPE,
EQF,
FRG,
GSH let there be set up a pyramid of equal height with the cone; therefore, also, each of the pyramids so set up is greater than the half of that segment of the cone which is about it.
Thus, bisecting the circumferences which are left, joining straight lines, setting up on each of the triangles pyramids of equal height with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the solid
X. [
X. 1]
Let such be left, and let them be the segments on
HP,
PE,
EQ,
QF,
FR,
RG,
GS,
SH; therefore the remainder, the pyramid of which the polygon
HPEQFRGS is the base and the height the same with that of the cone, is greater than the solid
O.
Let there also be inscribed in the circle
ABCD the polygon
DTAUBVCW similar and similarly situated to the polygon
HPEQFRGS, and on it let a pyramid be set up of equal height with the cone
AL.
Since then, as the square on
AC is to the square on
EG, so is the polygon
DTAUBVCW to the polygon
HPEQFRGS, [
XII. 1] while, as the square on
AC is to the square on
EG, so is the circle
ABCD to the circle
EFGH, [
XII. 2] therefore also, as the circle
ABCD is to the circle
EFGH, so is the polygon
DTAUBVCW to the polygon
HPEQFRGS.
But, as the circle
ABCD is to the circle
EFGH, so is the cone
AL to the solid
O, and, as the polygon
DTAUBVCW is to the polygon
HPEQFRGS, so is the pyramid of which the polygon
DTAUBVCW is the base and the point
L the vertex to the pyramid of which the polygon
HPEQFRGS is the base and the point
N the vertex. [
XII. 6]
Therefore also, as the cone
AL is to the solid
O, so is the pyramid of which the polygon
DTAUBVCW is the base and the point
L the vertex to the pyramid of which the polygon
HPEQFRGS is the base and the point
N the vertex; [
V. 11] therefore, alternately, as the cone
AL is to the pyramid in it, so is the solid
O to the pyramid in the cone
EN. [
V. 16]
But the cone
AL is greater than the pyramid in it; therefore the solid
O is also greater than the pyramid in the cone
EN.
But it is also less: which is absurd.
Therefore the cone
AL is not to any solid less than the cone
EN as the circle
ABCD is to the circle
EFGH.
Similarly we can prove that neither is the cone
EN to any solid less than the cone
AL as the circle
EFGH is to the circle
ABCD.
I say next that neither is the cone
AL to any solid greater than the cone
EN as the circle
ABCD is to the circle
EFGH.
For, if possible, let it be in that ratio to a greater solid
O; therefore, inversely, as the circle
EFGH is to the circle
ABCD, so is the solid
O to the cone
AL.
But, as the solid
O is to the cone
AL, so is the cone
EN to some solid less than the cone
AL; therefore also, as the circle
EFGH is to the circle
ABCD, so is the cone
EN to some solid less than the cone
AL: which was proved impossible.
Therefore the cone
AL is not to any solid greater than the cone
EN as the circle
ABCD is to the circle
EFGH.
But it was proved that neither is it in this ratio to a less solid; therefore, as the circle
ABCD is to the circle
EFGH, so is the cone
AL to the cone
EN.
But, as the cone is to the cone, so is the cylinder to the cylinder, for each is triple of each; [
XII. 10]
Therefore also, as the circle
ABCD is to the circle
EFGH, so are the cylinders on them which are of equal height.
Therefore etc. Q. E. D.
PROPOSITION 12.
Similar cones and cylinders are to one another in the triplicate ratio of the diameters in their bases.
Let there be similar cones and cylinders, let the circles
ABCD,
EFGH be their bases,
BD,
FH the diameters of the bases, and
KL,
MN the axes of the cones and cylinders; I say that the cone of which the circle
ABCD is the base and the point
L the vertex has to the cone of which the circle
EFGH is the base and the point
N the vertex the ratio triplicate of that which
BD has to
FH.
For, if the cone
ABCDL has not to the cone
EFGHN the ratio triplicate of that which
BD has to
FH, the cone
ABCDL will have that triplicate ratio either to some solid less than the cone
EFGHN or to a greater.
First, let it have that triplicate ratio to a less solid
O.
Let the square
EFGH be inscribed in the circle
EFGH; [
IV. 6] therefore the square
EFGH is greater than the half of the circle
EFGH.
Now let there be set up on the square
EFGH a pyramid having the same vertex with the cone; therefore the pyramid so set up is greater than the half part of the cone.
Let the circumferences
EF,
FG,
GH,
HE be bisected at the points
P,
Q,
R,
S, and let
EP,
PF,
FQ,
QG,
GR,
RH,
HS,
SE be joined.
Therefore each of the triangles
EPF,
FQG,
GRH,
HSE is also greater than the half part of that segment of the circle
EFGH which is about it.
Now on each of the triangles
EPF,
FQG,
GRH,
HSE let a pyramid be set up having the same vertex with the cone; therefore each of the pyramids so set up is also greater than the half part of that segment of the cone which is about it.
Thus, bisecting the circumferences so left, joining straight lines, setting up on each of the triangles pyramids having the same vertex with the cone, and doing this continually, we shall leave some segments of the cone which will be less than the excess by which the cone
EFGHN exceeds the solid
O. [
X. 1]
Let such be left, and let them be the segments on
EP,
PF,
FQ,
QG,
GR,
RH,
HS,
SE; therefore the remainder, the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex, is greater than the solid
O.
Let there be also inscribed in the circle
ABCD the polygon
ATBUCVDW similar and similarly situated to the polygon
EPFQGRHS, and let there be set up on the polygon
ATBUCVDW a pyramid having the same vertex with the cone; of the triangles containing the pyramid of which the polygon
ATBUCVDW is the base and the point
L the vertex let
LBT be one, and of the triangles containing the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex let
NFP be one; and let
KT,
MP be joined.
Now, since the cone
ABCDL is similar to the cone
EFGHN, therefore, as
BD is to
FH, so is the axis
KL to the axis
MN. [
XI. Def. 24]
But, as
BD is to
FH, so is
BK to
FM; therefore also, as
BK is to
FM, so is
KL to
MN.
And, alternately, as
BK is to
KL, so is
FM to
MN. [
V. 16]
And the sides are proportional about equal angles, namely the angles
BKL,
FMN; therefore the triangle
BKL is similar to the triangle
FMN. [
VI. 6]
Again, since, as
BK is to
KT, so is
FM to
MP, and they are about equal angles, namely the angles
BKT,
FMP, inasmuch as, whatever part the angle
BKT is of the four right angles at the centre
K, the same part also is the angle
FMP of the four right angles at the centre
M; since then the sides are proportional about equal angles, therefore the triangle
BKT is similar to the triangle
FMP. [
VI. 6]
Again, since it was proved that, as
BK is to
KL, so is
FM to
MN, while
BK is equal to
KT, and
FM to
PM, therefore, as
TK is to
KL, so is
PM to
MN; and the sides are proportional about equal angles, namely the angles
TKL,
PMN, for they are right; therefore the triangle
LKT is similar to the triangle
NMP. [
VI. 6]
And since, owing to the similarity of the triangles
LKB,
NMF, as
LB is to
BK, so is
NF to
FM, and, owing to the similarity of the triangles
BKT,
FMP, as
KB is to
BT, so is
MF to
FP, therefore,
ex aequali, as
LB is to
BT, so is
NF to
FP. [
V. 22]
Again since, owing to the similarity of the triangles
LTK,
NPM, as
LT is to
TK, so is
NP to
PM, and, owing to the similarity of the triangles
TKB,
PMF, as
KT is to
TB, so is
MP to
PF; therefore,
ex aequali, as
LT is to
TB, so is
NP to
PF. [
V. 22]
But it was also proved that, as
TB is to
BL, so is
PF to
FN.
Therefore,
ex aequali, as
TL is to
LB, so is
PN to
NF. [
V. 22]
Therefore in the triangles
LTB,
NPF the sides are proportional; therefore the triangles
LTB,
NPF are equiangular; [
VI. 5] hence they are also similar. [
VI. Def. I]
Therefore the pyramid of which the triangle
BKT is the base and the point
L the vertex is also similar to the pyramid of which the triangle
FMP is the base and the point
N the vertex, for they are contained by similar planes equal in multitude. [
XI. Def. 9]
But similar pyramids which have triangular bases are to one another in the triplicate ratio of their corresponding sides. [
XII. 8]
Therefore the pyramid
BKTL has to the pyramid
FMPN the ratio triplicate of that which
BK has to
FM.
Similarly, by joining straight lines from
A,
W,
D,
V,
C,
U to
K, and from
E,
S,
H,
R,
G,
Q to
M, and setting up on each of the triangles pyramids which have the same vertex with the cones, we can prove that each of the similarly arranged pyramids will also have to each similarly arranged pyramid the ratio triplicate of that which the corresponding side
BK has to the corresponding side
FM, that is, which
BD has to
FH.
And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [
V. 12] therefore also, as the pyramid
BKTL is to the pyramid
FMPN, so is the whole pyramid of which the polygon
ATBUCVDW is the base and the point
L the vertex to the whole pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex; hence also the pyramid of which
ATBUCVDW is the base and the point
L the vertex has to the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex the ratio triplicate of that which
BD has to
FH.
But, by hypothesis, the cone of which the circle
ABCD
is the base and the point
L the vertex has also to the solid
O the ratio triplicate of that which
BD has to
FH; therefore, as the cone of which the circle
ABCD is the base and the point
L the vertex is to the solid
O, so is the pyramid of which the polygon
ATBUCVDW is the base and
L the vertex to the pyramid of which the polygon
EPFQGRHS is the base and the point
N the vertex; therefore, alternately, as the cone of which the circle
ABCD is the base and
L the vertex is to the pyramid contained in it of which the polygon
ATBUCVDW is the base and
L the vertex, so is the solid
O to the pyramid of which the polygon
EPFQGRHS is the base and
N the vertex. [
V. 16]
But the said cone is greater than the pyramid in it; for it encloses it.
Therefore the solid
O is also greater than the pyramid of which the polygon
EPFQGRHS is the base and
N the vertex.
But it is also less: which is impossible.
Therefore the cone of which the circle
ABCD is the base and
L the vertex has not to any solid less than the cone of which the circle
EFGH is the base and the point
N the vertex the ratio triplicate of that which
BD has to
FH:
Similarly we can prove that neither has the cone
EFGHN to any solid less than the cone
ABCDL the ratio triplicate of that which
FH has to
BD.
I say next that neither has the cone
ABCDL to any solid greater than the cone
EFGHN the ratio triplicate of that which
BD has to
FH.
For, if possible, let it have that ratio to a greater solid
O.
Therefore, inversely, the solid
O has to the cone
ABCDL the ratio triplicate of that which
FH has to
BD.
But, as the solid
O is to the cone
ABCDL, so is the cone
EFGHN to some solid less than the cone
ABCDL.
Therefore the cone
EFGHN also has to some solid less than the cone
ABCDL the ratio triplicate of that which
FH has to
BD: which was proved impossible.
Therefore the cone
ABCDL has not to any solid greater than the cone
EFGHN the ratio triplicate of that which
BD has to
FH.
But it was proved that neither has it this ratio to a less solid than the cone
EFGHN.
Therefore the cone
ABCDL has to the cone
EFGHN the ratio triplicate of that which
BD has to
FH.
But, as the cone is to the cone, so is the cylinder to the cylinder, for the cylinder which is on the same base as the cone and of equal height with it is triple of the cone; [
XII. 10] therefore the cylinder also has to the cylinder the ratio triplicate of that which
BD has to
FH.
Therefore etc. Q. E. D.
PROPOSITION 13.
If a cylinder be cut by a plane which is parallel to its opposite planes, then, as the cylinder is to the cylinder, so will the axis be to the axis.
For let the cylinder
AD be cut by the plane
GH which is parallel to the opposite planes
AB,
CD, and let the plane
GH meet the axis at the point
K; I say that, as the cylinder
BG is to the cylinder
GD, so is the axis
EK to the axis
KF.
For let the axis
EF be produced in both directions to the points
L,
M, and let there be set out any number whatever of axes
EN,
NL equal to the axis
EK, and any number whatever
FO,
OM equal to
FK; and let the cylinder
PW on the axis
LM be conceived of which the circles
PQ,
VW are the bases.
Let planes be carried through the points
N,
O parallel to
AB,
CD and to the bases of the cylinder
PW, and let them produce the circles
RS,
TU about the centres
N,
O.
Then, since the axes
LN,
NE,
EK are equal to one another, therefore the cylinders
QR,
RB,
BG are to one another as their bases. [
XII. 11]
But the bases are equal; therefore the cylinders
QR,
RB,
BG are also equal to one another.
Since then the axes
LN,
NE,
EK are equal to one another, and the cylinders
QR,
RB,
BG are also equal to one another, and the multitude of the former is equal to the multitude of the latter, therefore, whatever multiple the axis
KL is of the axis
EK, the same multiple also will the cylinder
QG be of the cylinder
GB.
For the same reason, whatever multiple the axis
MK is of the axis
KF, the same multiple also is the cylinder
WG of the cylinder
GD.
And, if the axis
KL is equal to the axis
KM, the cylinder
QG will also be equal to the cylinder
GW, if the axis is greater than the axis, the cylinder will also be greater than the cylinder, and if less, less.
Thus, there being four magnitudes, the axes
EK,
KF and the cylinders
BG,
GD, there have been taken equimultiples of the axis
EK and of the cylinder
BG, namely the axis
LK and the cylinder
QG, and equimultiples of the axis
KF and of the cylinder
GD, namely the axis
KM and the cylinder
GW; and it has been proved that, if the axis
KL is in excess of the axis
KM, the cylinder
QG is also in excess of the cylinder
GW, if equal, equal, and if less, less.
Therefore, as the axis
EK is to the axis
KF, so is the cylinder
BG to the cylinder
GD. [
V. Def. 5] Q. E. D.
PROPOSITION 14.
Cones and cylinders which are on equal bases are to one another as their heights.
For let
EB,
FD be cylinders on equal bases, the circles
AB,
CD; I say that, as the cylinder
EB is to the cylinder
FD, so is the axis
GH to the axis
KL.
For let the axis
KL be produced to the point
N, let
LN be made equal to the axis
GH, and let the cylinder
CM be conceived about
LN as axis.
Since then the cylinders
EB,
CM are of the same height, they are to one another as their bases. [
XII. 11]
But the bases are equal to one another; therefore the cylinders
EB,
CM are also equal.
And, since the cylinder
FM has been cut by the plane
CD which is parallel to its opposite planes, therefore, as the cylinder
CM is to the cylinder
FD, so is the axis
LN to the axis
KL. [
XII. 13]
But the cylinder
CM is equal to the cylinder
EB, and the axis
LN to the axis
GH; therefore, as the cylinder
EB is to the cylinder
FD, so is the axis
GH to the axis
KL.
But, as the cylinder
EB is to the cylinder
FD, so is the cone
ABG to the cone
CDK. [
XII. 10]
Therefore also, as the axis
GH is to the axis
KL, so is the cone
ABG to the cone
CDK and the cylinder
EB to the cylinder
FD. Q. E. D.
PROPOSITION 15.
In equal cones and cylinders the bases are reciprocally proportional to the heights; and those cones and cylinders in which the bases are reciprocally proportional to the heights are equal.
Let there be equal cones and cylinders of which the circles
ABCD,
EFGH are the bases; let
AC,
EG be the diameters of the bases, and
KL,
MN the axes, which are also the heights of the cones or cylinders; let the cylinders
AO,
EP be completed.
I say that in the cylinders
AO,
EP the bases are reciprocally proportional to the heights, that is, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL.
For the height
LK is either equal to the height
MN or not equal.
First, let it be equal.
Now the cylinder
AO is also equal to the cylinder
EP.
But cones and cylinders which are of the same height are to one another as their bases; [
XII. 11] therefore the base
ABCD is also equal to the base
EFGH.
Hence also, reciprocally, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL.
Next, let the height
LK not be equal to
MN, but let
MN be greater; from the height
MN let
QN be cut off equal to
KL, through the point
Q let the cylinder
EP be cut by the plane
TUS parallel to the planes of the circles
EFGH,
RP, and let the cylinder
ES be conceived erected from the circle
EFGH as base and with height
NQ.
Now, since the cylinder
AO is equal to the cylinder
EP, therefore, as the cylinder
AO is to the cylinder
ES, so is the cylinder
EP to the cylinder
ES. [
V. 7]
But, as the cylinder
AO is to the cylinder
ES, so is the base
ABCD to the base
EFGH, for the cylinders
AO,
ES are of the same height; [
XII. 11] and, as the cylinder
EP is to the cylinder
ES, so is the height
MN to the height
QN, for the cylinder
EP has been cut by a plane which is parallel to its opposite planes. [
XII. 13]
Therefore also, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
QN. [
V. 11]
But the height
QN is equal to the height
KL; therefore, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL.
Therefore in the cylinders
AO,
EP the bases are reciprocally proportional to the heights.
Next, in the cylinders
AO,
EP let the bases be reciprocally proportional to the heights, that is, as the base
ABCD is to the base
EFGH, so let the height
MN be to the height
KL; I say that the cylinder
AO is equal to the cylinder
EP.
For, with the same construction, since, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
KL, while the height
KL is equal to the height
QN, therefore, as the base
ABCD is to the base
EFGH, so is the height
MN to the height
QN
But, as the base
ABCD is to the base
EFGH, so is the cylinder
AO to the cylinder
ES, for they are of the same height; [
XII. 11] and, as the height
MN is to
QN, so is the cylinder
EP to the cylinder
ES; [
XII. 13] therefore, as the cylinder
AO is to the cylinder
ES, so is the cylinder
EP to the cylinder
ES. [
V. 11]
Therefore the cylinder
AO is equal to the cylinder
EP. [
V. 9]
And the same is true for the cones also. Q. E. D.
PROPOSITION 16.
Given two circles about the same centre,
to inscribe in the greater circle an equilateral polygon with an even number of sides which does not touch the lesser circle.
Let
ABCD,
EFGH be the two given circles about the same centre
K; thus it is required to inscribe in the greater circle
ABCD an equilateral polygon with an even number of sides which does not touch the circle
EFGH.
For let the straight line
BKD be drawn through the centre
K, and from the point
G let
GA be drawn at right angles to the straight line
BD and carried through to
C; therefore
AC touches the circle
EFGH. [
III. 16, Por.]
Then, bisecting the circumference
BAD, bisecting the half of it, and doing this continually, we shall leave a circumference less than
AD. [
X. 1]
Let such be left, and let it be
LD; from
L let
LM be drawn perpendicular to
BD and carried through to
N, and let
LD,
DN be joined; therefore
LD is equal to
DN. [
III. 3,
I. 4]
Now, since
LN is parallel to
AC, and
AC touches the circle
EFGH, therefore
LN does not touch the circle
EFGH; therefore
LD,
DN are far from touching the circle
EFGH.
If then we fit into the circle
ABCD straight lines equal to the straight line
LD and placed continuously, there will be inscribed in the circle
ABCD an equilateral polygon with an even number of sides which does not touch the lesser circle
EFGH. Q. E. F.
PROPOSITION 17.
Given two spheres about the same centre,
to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.
Let two spheres be conceived about the same centre
A; thus it is required to inscribe in the greater sphere a polyhedral solid which does not touch the lesser sphere at its surface.
Let the spheres be cut by any plane through the centre; then the sections will be circles, inasmuch as the sphere was produced by the diameter remaining fixed and the semicircle being carried round it; [
XI. Def. 14] hence, in whatever position we conceive the semicircle to be, the plane carried through it will produce a circle on the circumference of the sphere.
And it is manifest that this circle is the greatest possible, inasmuch as the diameter of the sphere, which is of course the diameter both of the semicircle and of the circle, is greater than all the straight lines drawn across in the circle or the sphere.
Let then
BCDE be the circle in the greater sphere, and
FGH the circle in the lesser sphere; let two diameters in them,
BD,
CE, be drawn at right angles to one another; then, given the two circles
BCDE,
FGH about the same centre, let there be inscribed in the greater circle
BCDE an equilateral polygon with an even number of sides which does not touch the lesser circle
FGH, let
BK,
KL,
LM
ME be its sides in the quadrant
BE. let
KA be joined and carried through to
N, let
AO be set up from the point
A at right angles to the plane of the circle
BCDE, and let it meet the surface of the sphere at
O, and through
AO and each of the straight lines
BD,
KN let planes be carried; they will then make greatest circles on the surface of the sphere, for the reason stated.
Let them make such, and in them let
BOD,
KON be the semicircles on
BD,
KN.
Now, since
OA is at right angles to the plane of the circle
BCDE, therefore all the planes through
OA are also at right angles to the plane of the circle
BCDE; [
XI. 18] hence the semicircles
BOD,
KON are also at right angles to the plane of the circle
BCDE.
And, since the semicircles
BED,
BOD,
KON are equal, for they are on the equal diameters
BD,
KN, therefore the quadrants
BE,
BO,
KO are also equal to one another.
Therefore there are as many straight lines in the quadrants
BO,
KO equal to the straight lines
BK,
KL,
LM,
ME as there are sides of the polygon in the quadrant
BE.
Let them be inscribed, and let them be
BP,
PQ,
QR,
RO and
KS,
ST,
TU,
UO, let
SP,
TQ,
UR be joined, and from
P,
S let perpendiculars be drawn to the plane of the circle
BCDE; [
XI. 11] these will fall on
BD,
KN, the common sections of the planes, inasmuch as the planes of
BOD,
KON are also at right angles to the plane of the circle
BCDE. [cf.
XI. Def. 4]
Let them so fall, and let them be
PV,
SW, and let
WV be joined.
Now since, in the equal semicircles
BOD,
KON, equal straight lines
BP,
KS have been cut off, and the perpendiculars
PV,
SW have been drawn, therefore
PV is equal to
SW, and
BV to
KW. [
III. 27,
I. 26]
But the whole
BA is also equal to the whole
KA; therefore the remainder
VA is also equal to the remainder
WA; therefore, as
BV is to
VA, so is
KW to
WA; therefore
WV is parallel to
KB. [
VI. 2]
And, since each of the straight lines
PV,
SW is at right angles to the plane of the circle
BCDE, therefore
PV is parallel to
SW. [
XI. 6]
But it was also proved equal to it; therefore
WV,
SP are also equal and parallel. [
I. 33]
And, since
WV is parallel to
SP, while
WV is parallel to
KB, therefore
SP is also parallel to
KB. [
XI. 9]
And
BP,
KS join their extremities; therefore the quadrilateral
KBPS is in one plane, inasmuch as, if two straight lines be parallel, and points be taken at random on each of them, the straight line joining the points is in the same plane with the parallels. [
XI. 7]
For the same reason each of the quadrilaterals
SPQT,
TQRU is also in one plane.
But the triangle
URO is also in one plane. [
XI. 2]
If then we conceive straight lines joined from the points
P,
S,
Q,
T,
R,
U to
A, there will be constructed a certain polyhedral solid figure between the circumferences
BO,
KO, consisting of pyramids of which the quadrilaterals
KBPS,
SPQT,
TQRU and the triangle
URO are the bases and the point
A the vertex.
And, if we make the same construction in the case of each of the sides
KL,
LM,
ME as in the case of
BK, and further in the case of the remaining three quadrants, there will be constructed a certain polyhedral figure inscribed in the sphere and contained by pyramids, of which the said quadrilaterals and the triangle
URO, and the others corresponding to them, are the bases and the point
A the vertex.
I say that the said polyhedron will not touch the lesser sphere at the surface on which the circle
FGH is.
Let
AX be drawn from the point
A perpendicular to the plane of the quadrilateral
KBPS, and let it meet the plane at the point
X; [
XI. 11] let
XB,
XK be joined.
Then, since
AX is at right angles to the plane of the quadrilateral
KBPS, therefore it is also at right angles to all the straight lines which meet it and are in the plane of the quadrilateral. [
XI. Def. 3]
Therefore
AX is at right angles to each of the straight lines
BX,
XK.
And, since
AB is equal to
AK, the square on
AB is also equal to the square on
AK.
And the squares on
AX,
XB are equal to the square on
AB, for the angle at
X is right; [
I. 47] and the squares on
AX,
XK are equal to the square on
AK. [
id.]
Therefore the squares on
AX,
XB are equal to the squares on
AX,
XK.
Let the square on
AX be subtracted from each; therefore the remainder, the square on
BX, is equal to the remainder, the square on
XK; therefore
BX is equal to
XK.
Similarly we can prove that the straight lines joined from
X to
P,
S are equal to each of the straight lines
BX,
XK.
Therefore the circle described with centre
X and distance one of the straight lines
XB,
XK will pass through
P,
S also, and
KBPS will be a quadrilateral in a circle.
Now, since
KB is greater than
WV, while
WV is equal to
SP, therefore
KB is greater than
SP.
But
KB is equal to each of the straight lines
KS,
BP; therefore each of the straight lines
KS,
BP is greater than
SP.
And, since
KBPS is a quadrilateral in a circle, and
KB,
BP,
KS are equal, and
PS less, and
BX is the radius of the circle, therefore the square on
KB is greater than double of the square on
BX.
Let
KZ be drawn from
K perpendicular to
BV.
Then, since
BD is less than double of
DZ, and, as
BD is to
DZ, so is the rectangle
DB,
BZ to the rectangle
DZ,
ZB, if a square be described upon
BZ and the parallelogram on
ZD be completed, then the rectangle
DB,
BZ is also less than double of the rectangle
DZ,
ZB.
And, if
KD be joined, the rectangle
DB,
BZ is equal to the square on
BK, and the rectangle
DZ,
ZB equal to the square on
KZ; [
III. 31,
VI. 8 and Por.] therefore the square on
KB is less than double of the square on
KZ.
But the square on
KB is greater than double of the square on
BX; therefore the square on
KZ is greater than the square on
BX.
And, since
BA is equal to
KA, the square on
BA is equal to the square on
AK.
And the squares on
BX,
XA are equal to the square on
BA, and the squares on
KZ,
ZA equal to the square on
KA; [
I. 47] therefore the squares on
BX,
XA are equal to the squares on
KZ,
ZA, and of these the square on
KZ is greater than the square on
BX; therefore the remainder, the square on
ZA, is less than the square on
XA.
Therefore
AX is greater than
AZ; therefore
AX is much greater than
AG.
And
AX is the perpendicular on one base of the polyhedron, and
AG on the surface of the lesser sphere; hence the polyhedron will not touch the lesser sphere on its surface.
Therefore, given two spheres about the same centre, a polyhedral solid has been inscribed in the greater sphere which does not touch the lesser sphere at its surface. Q. E. F.
PORISM.
But if in another sphere also a polyhedral solid be inscribed similar to the solid in the sphere
BCDE, the polyhedral solid in the sphere
BCDE has to the polyhedral solid in the other sphere the ratio triplicate of that which the diameter of the sphere
BCDE has to the diameter of the other sphere.
For, the solids being divided into their pyramids similar in multitude and arrangement, the pyramids will be similar.
But similar pyramids are to one another in the triplicate ratio of their corresponding sides; [
XII. 8, Por.] therefore the pyramid of which the quadrilateral
KBPS is the base, and the point
A the vertex, has to the similarly arranged pyramid in the other sphere the ratio triplicate of that which the corresponding side has to the corresponding side, that is, of that which the radius
AB of the sphere about
A as centre has to the radius of the other sphere.
Similarly also each pyramid of those in the sphere about
A as centre has to each similarly arranged pyramid of those in the other sphere the ratio triplicate of that which
AB has to the radius of the other sphere.
And, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [
V. 12] hence the whole polyhedral solid in the sphere about
A as centre has to the whole polyhedral solid in the other sphere the ratio triplicate of that which
AB has to the radius of the other sphere, that is, of that which the diameter
BD has to the diameter of the other sphere. Q. E. D.
PROPOSITION 18.
Spheres are to one another in the triplicate ratio of their respective diameters.
Let the spheres
ABC,
DEF be conceived, and let
BC,
EF be their diameters; I say that the sphere
ABC has to the sphere
DEF the ratio triplicate of that which
BC has to
EF.
For, if the sphere
ABC has not to the sphere
DEF the ratio triplicate of that which
BC has to
EF, then the sphere
ABC will have either to some less sphere than the sphere
DEF, or to a greater, the ratio triplicate of that which
BC has to
EF.
First, let it have that ratio to a less sphere
GHK, let
DEF be conceived about the same centre with
GHK, let there be inscribed in the greater sphere
DEF a polyhedral solid which does not touch the lesser sphere
GHK at its surface, [
XII. 17] and let there also be inscribed in the sphere
ABC a polyhedral solid similar to the polyhedral solid in the sphere
DEF; therefore the polyhedral solid in
ABC has to the polyhedral solid in
DEF the ratio triplicate of that which
BC has to
EF. [
XII. 17, Por.]
But the sphere
ABC also has to the sphere
GHK the ratio triplicate of that which
BC has to
EF; therefore, as the sphere
ABC is to the sphere
GHK, so is the polyhedral solid in the sphere
ABC to the polyhedral solid in the sphere
DEF; and, alternately, as the sphere
ABC is to the polyhedron in it, so is the sphere
GHK to the polyhedral solid in the sphere
DEF. [
V. 16]
But the sphere
ABC is greater than the polyhedron in it; therefore the sphere
GHK is also greater than the polyhedron in the sphere
DEF.
But it is also less, for it is enclosed by it.
Therefore the sphere
ABC has not to a less sphere than the sphere
DEF the ratio triplicate of that which the diameter
BC has to
EF.
Similarly we can prove that neither has the sphere
DEF to a less sphere than the sphere
ABC the ratio triplicate of that which
EF has to
BC.
I say next that neither has the sphere
ABC to any greater sphere than the sphere
DEF the ratio triplicate of that which
BC has to
EF.
For, if possible, let it have that ratio to a greater,
LMN; therefore, inversely, the sphere
LMN has to the sphere
ABC the ratio triplicate of that which the diameter
EF has to the diameter
BC.
But, inasmuch as
LMN is greater than
DEF, therefore, as the sphere
LMN is to the sphere
ABC, so is the sphere
DEF to some less sphere than the sphere
ABC, as was before proved. [
XII. 2, Lemma]
Therefore the sphere
DEF also has to some less sphere than the sphere
ABC the ratio triplicate of that which
EF has to
BC: which was proved impossible.
Therefore the sphere
ABC has not to any sphere greater than the sphere
DEF the ratio triplicate of that which
BC has to
EF.
But it was proved that neither has it that ratio to a less sphere.
Therefore the sphere
ABC has to the sphere
DEF the ratio triplicate of that which
BC has to
EF. Q. E. D.