BOOK XIII.
HISTORICAL NOTE.
I have already given, in the note to
IV. 10, the evidence upon which the construction of the five regular solids is attributed to the Pythagoreans. Some of them, the cube, the tetrahedron (which is nothing but a pyramid), and the octahedron (which is only a double pyramid with a square base), cannot but have been known to the Egyptians. And it appears that dodecahedra have been found, of bronze or other material, which may belong to periods earlier than Pythagoras' time by some centuries (for references see Cantor's
Geschichte der Mathematik I_{3}, pp. 175-6).
It is true that the author of the scholium No. I to Eucl. XIII. says that the Book is about “the five so-called Platonic figures, which however do not belong to Plato, three of the aforesaid five figures being due to the Pythagoreans, namely the cube, the pyramid and the dodecahedron, while the octahedron and the icosahedron are due to Theaetetus.”
This statement (taken probably from Geminus) may perhaps rest on the fact that Theaetetus was the first to write at any length about the two last-mentioned solids. We are told indeed by Suidas (S. V.
Θεαίτητος) that Theaetetus “first wrote on the ’five solids’ as they are called.”
This no doubt means that Theaetetus was the first to write a complete and systematic treatise on all the regular solids; it does not exclude the possibility that Hippasus or others had already written on the dodecahedron. The fact that Theaetetus wrote upon the regular solids agrees very well with the evidence which we possess of his contributions to the theory of irrationals, the connexion between which and the investigation of the regular solids is seen in Euclid's Book XIII.
Theaetetus flourished about 380 B.C., and his work on the regular solids was soon followed by another, that of Aristaeus, an elder contemporary of Euclid, who also wrote an important book on
Solid Loci, i.e. on conics treated as loci. This Aristaeus (known as “the elder”
) wrote in the period about 320 B.C. We hear of his
Comparison of the five regular solids from Hypsicles (2nd cent. B.C.), the writer of the short book commonly included in the editions of the
Elements as Book XIV. Hypsicles gives in this Book some six propositions supplementing Eucl. XIII.; and he introduces the second of the propositions (Heiberg's Euclid, Vol. v. p. 6) as follows:
“
The same circle circumscribes both the pentagon of the dodecahedron and the triangle of the icosahedron when both are inscribed in the same sphere. This is proved by Aristaeus in the book entitled
Comparison of the five figures.”
Hypsicles proceeds (pp. 7 sqq.) to give a proof of this theorem. Allman pointed out (
Greek Geometry from Thales to Euclid, 1889, pp. 201-2) that this proof depends on eight theorems, six of which appear in Euclid's Book XIII. (in Propositions
8,
10,
12,
15, 16 with Por., 17); two other propositions not mentioned by Allman are also used, namely
XIII. 4 and 9. This seems, as Allman says, to confirm the inference of Bretschneider (p. 171) that, as Aristaeus' work was the newest and latest in which, before Euclid's time, this subject was treated, we have in Eucl. XIII. at least a partial recapitulation of the contents of the treatise of Aristaeus.
After Euclid, Apollonius wrote on the comparison of the dodecahedron and the icosahedron inscribed in one and the same sphere. This we also learn from Hypsicles, who says in the next words following those about Aristaeus above quoted: “But it is proved by Apollonius in the second edition of his
Comparison of the dodecahedron with the icosahedron that, as the surface of the dodecahedron is to the surface of the icosahedron [inscribed in the same sphere], so is the dodecahedron itself [i.e. its volume] to the icosahedron, because the perpendicular is the same from the centre of the sphere to the pentagon of the dodecahedron and to the triangle of the icosahedron.”
BOOK XIII. PROPOSITIONS.
PROPOSITION 1.
If a straight line be cut in extreme and mean ratio,
the square on the greater segment added to the half of the whole is five times the square on the half.
For let the straight line
AB be cut in extreme and mean ratio at the point
C, and let
AC be the greater segment; let the straight line
AD be produced in a straight line with
CA, and let
AD be made half of
AB; I say that the square on
CD is five times the square on
AD.
For let the squares
AE,
DF be described on
AB,
DC, and let the figure in
DF be drawn; let
FC be carried through to
G.
Now, since
AB has been cut in extreme and mean ratio at
C, therefore the rectangle
AB,
BC is equal to the square on
AC. [
VI. Def. 3,
VI. 17]
And
CE is the rectangle
AB,
BC, and
FH the square on
AC; therefore
CE is equal to
FH.
And, since
BA is double of
AD, while
BA is equal to
KA, and
AD to
AH, therefore
KA is also double of
AH.
But, as
KA is to
AH, so is
CK to
CH; [
VI. 1] therefore
CK is double of
CH.
But
LH,
HC are also double of
CH.
Therefore
KC is equal to
LH,
HC.
But
CE was also proved equal to
HF; therefore the whole square
AE is equal to the gnomon
MNO.
And, since
BA is double of
AD, the square on
BA is quadruple of the square on
AD, that is,
AE is quadruple of
DH.
But
AE is equal to the gnomon
MNO; therefore the gnomon
MNO is also quadruple of
AP; therefore the whole
DF is five times
AP.
And
DF is the square on
DC, and
AP the square on
DA; therefore the square on
CD is five times the square on
DA.
Therefore etc. Q. E. D.
PROPOSITION 2.
If the square on a straight line be five times the square on a segment of it,
then,
when the double of the said segment is cut in extreme and mean ratio,
the greater segment is the remaining part of the original straight line.
For let the square on the straight line
AB be five times the square on the segment
AC of it, and let
CD be double of
AC; I say that, when
CD is cut in extreme and mean ratio, the greater segment is
CB.
Let the squares
AF,
CG be described on
AB,
CD respectively, let the figure in
AF be drawn, and let
BE be drawn through.
Now, since the square on
BA is five times the square on
AC,
AF is five times
AH.
Therefore the gnomon
MNO is quadruple of
AH.
And, since
DC is double of
CA, therefore the square on
DC is quadruple of the square on
CA, that is,
CG is quadruple of
AH.
But the gnomon
MNO was also proved quadruple of
AH; therefore the gnomon
MNO is equal to
CG.
And, since
DC is double of
CA, while
DC is equal to
CK, and
AC to
CH, therefore
KB is also double of
BH. [
VI. 1]
But
LH,
HB are also double of
HB; therefore
KB is equal to
LH,
HB.
But the whole gnomon
MNO was also proved equal to the whole
CG; therefore the remainder
HF is equal to
BG.
And
BG is the rectangle
CD,
DB, for
CD is equal to
DG; and
HF is the square on
CB; therefore the rectangle
CD,
DB is equal to the square on
CB.
Therefore, as
DC is to
CB, so is
CB to
BD.
But
DC is greater than
CB; therefore
CB is also greater than
BD.
Therefore, when the straight line
CD is cut in extreme and mean ratio,
CB is the greater segment.
Therefore etc. Q. E. D.
LEMMA.
That the double of
AC is greater than
BC is to be proved thus.
If not, let
BC be, if possible, double of
CA.
Therefore the square on
BC is quadruple of the square on
CA; therefore the squares on
BC,
CA are five times the square on
CA.
But, by hypothesis, the square on
BA is also five times the square on
CA; therefore the square on
BA is equal to the squares on
BC,
CA: which is impossible. [
II. 4]
Therefore
CB is not double of
AC.
Similarly we can prove that neither is a straight line less than
CB double of
CA; for the absurdity is much greater.
Therefore the double of
AC is greater than
CB. Q. E. D.
PROPOSITION 3.
If a straight line be cut in extreme and mean ratio,
the square on the lesser segment added to the half of the greater segment is five times the square on the half of the greater segment.
For let any straight line
AB be cut in extreme and mean ratio at the point
C, let
AC be the greater segment, and let
AC be bisected at
D; I say that the square on
BD is five times the square on
DC.
For let the square
AE be described on
AB, and let the figure be drawn double.
Since
AC is double of
DC, therefore the square on
AC is quadruple of the square on
DC, that is,
RS is quadruple of
FG.
And, since the rectangle
AB,
BC is equal to the square on
AC, and
CE is the rectangle
AB,
BC, therefore
CE is equal to
RS.
But
RS is quadruple of
FG; therefore
CE is also quadruple of
FG.
Again, since
AD is equal to
DC,
HK is also equal to
KF.
Hence the square
GF is also equal to the square
HL.
Therefore
GK is equal to
KL, that is,
MN to
NE; hence
MF is also equal to
FE.
But
MF is equal to
CG; therefore
CG is also equal to
FE.
Let
CN be added to each; therefore the gnomon
OPQ is equal to
CE.
But
CE was proved quadruple of
GF; therefore the gnomon
OPQ is also quadruple of the square
FG.
Therefore the gnomon
OPQ and the square
FG are five times
FG.
But the gnomon
OPQ and the square
FG are the square
DN.
And
DN is the square on
DB, and
GF the square on
DC.
Therefore the square on
DB is five times the square on
DC. Q. E. D.
PROPOSITION 4.
If a straight line be cut in extreme and mean ratio,
the square on the whole and the square on the lesser segment together are triple of the square on the greater segment.
Let
AB be a straight line, let it be cut in extreme and mean ratio at
C, and let
AC be the greater segment; I say that the squares on
AB,
BC are triple of the square on
CA.
For let the square
ADEB be described on
AB, and let the figure be drawn.
Since then
AB has been cut in extreme and mean ratio at
C, and
AC is the greater segment, therefore the rectangle
AB,
BC is equal to the square on
AC. [
VI. Def. 3,
VI. 17]
And
AK is the rectangle
AB,
BC, and
HG the square on
AC; therefore
AK is equal to
HG.
And, since
AF is equal to
FE, let
CK be added to each; therefore the whole
AK is equal to the whole
CE; therefore
AK,
CE are double of
AK.
But
AK,
CE are the gnomon
LMN and the square
CK; therefore the gnomon
LMN and the square
CK are double of
AK.
But, further,
AK was also proved equal to
HG; therefore the gnomon
LMN and the squares
CK,
HG are triple of the square
HG.
And the gnomon
LMN and the squares
CK,
HG are the whole square
AE and
CK, which are the squares on
AB,
BC, while
HG is the square on
AC.
Therefore the squares on
AB,
BC are triple of the square on
AC. Q. E. D.
PROPOSITION 5.
If a straight line be cut in extreme and mean ratio,
and there be added to it a straight line equal to the greater segment,
the whole straight line has been cut in extreme and mean ratio,
and the original straight line is the greater segment.
For let the straight line
AB be cut in extreme and mean ratio at the point
C, let
AC be the greater segment, and let
AD be equal to
AC.
I say that the straight line
DB has been cut in extreme and mean ratio at
A, and the original straight line
AB is the greater segment.
For let the square
AE be described on
AB, and let the figure be drawn.
Since
AB has been cut in extreme and mean ratio at
C, therefore the rectangle
AB,
BC is equal to the square on
AC. [
VI. Def. 3, VI. 17]
And
CE is the rectangle
AB,
BC, and
CH the square on
AC; therefore
CE is equal to
HC.
But
HE is equal to
CE, and
DH is equal to
HC; therefore
DH is also equal to
HE.
Therefore the whole
DK is equal to the whole
AE.
And
DK is the rectangle
BD,
DA, for
AD is equal to
DL; and
AE is the square on
AB; therefore the rectangle
BD,
DA is equal to the square on
AB.
Therefore, as
DB is to
BA, so is
BA to
AD. [
VI. 17]
And
DB is greater than
BA; therefore
BA is also greater than
AD. [
V. 14]
Therefore
DB has been cut in extreme and mean ratio at
A, and
AB is the greater segment. Q. E. D.
PROPOSITION 6.
If a rational straight line be cut in extreme and mean ratio,
each of the segments is the irrational straight line called apotome.
Let
AB be a rational straight line, let it be cut in extreme and mean ratio at
C, and let
AC be the greater segment; I say that each of the straight lines
AC,
CB is the irrational straight line called apotome.
For let
BA be produced, and let
AD be made half of
BA.
Since then the straight line
AB has been cut in extreme and mean ratio, and to the greater segment
AC is added
AD which is half of
AB, therefore the square on
CD is five times the square on
DA. [
XIII. 1]
Therefore the square on
CD has to the square on
DA the ratio which a number has to a number; therefore the square on
CD is commensurable with the square on
DA. [
X. 6]
But the square on
DA is rational, for
DA is rational, being half of
AB which is rational; therefore the square on
CD is also rational; [
X. Def. 4] therefore
CD is also rational.
And, since the square on
CD has not to the square on
DA the ratio which a square number has to a square number, therefore
CD is incommensurable in length with
DA; [
X. 9] therefore
CD,
DA are rational straight lines commensurable in square only; therefore
AC is an apotome. [
X. 73]
Again, since
AB has been cut in extreme and mean ratio, and
AC is the greater segment, therefore the rectangle
AB,
BC is equal to the square on
AC. [
VI. Def. 3,
VI. 17]
Therefore the square on the apotome
AC, if applied to the rational straight line
AB, produces
BC as breadth.
But the square on an apotome, if applied to a rational straight line, produces as breadth a first apotome; [
X. 97] therefore
CB is a first apotome.
And
CA was also proved to be an apotome.
Therefore etc. Q. E. D.
PROPOSITION 7.
If three angles of an equilateral pentagon, taken either in order or not in order,
be equal,
the pentagon will be equiangular.
For in the equilateral pentagon
ABCDE let, first, three angles taken in order, those at
A,
B,
C, be equal to one another; I say that the pentagon
ABCDE is equiangular.
For let
AC,
BE,
FD be joined.
Now, since the two sides
CB,
BA are equal to the two sides
BA,
AE respectively, and the angle
CBA is equal to the angle
BAE, therefore the base
AC is equal to the base
BE, the triangle
ABC is equal to the triangle
ABE, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend, [
I. 4] that is, the angle
BCA to the angle
BEA, and the angle
ABE to the angle
CAB; hence the side
AF is also equal to the side
BF. [
I. 6]
But the whole
AC was also proved equal to the whole
BE; therefore the remainder
FC is also equal to the remainder
FE.
But
CD is also equal to
DE.
Therefore the two sides
FC,
CD are equal to the two sides
FE,
ED; and the base
FD is common to them; therefore the angle
FCD is equal to the angle
FED. [
I. 8]
But the angle
BCA was also proved equal to the angle
AEB; therefore the whole angle
BCD is also equal to the whole angle
AED.
But, by hypothesis, the angle
BCD is equal to the angles at
A,
B; therefore the angle
AED is also equal to the angles at
A,
B.
Similarly we can prove that the angle
CDE is also equal to the angles at
A,
B,
C; therefore the pentagon
ABCDE is equiangular.
Next, let the given equal angles not be angles taken in order, but let the angles at the points
A,
C,
D be equal; I say that in this case too the pentagon
ABCDE is equiangular.
For let
BD be joined.
Then, since the two sides
BA,
AE are equal to the two sides
BC,
CD, and they contain equal angles, therefore the base
BE is equal to the base
BD, the triangle
ABE is equal to the triangle
BCD, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [
I. 4] therefore the angle
AEB is equal to the angle
CDB.
But the angle
BED is also equal to the angle
BDE, since the side
BE is also equal to the side
BD. [
I. 5]
Therefore the whole angle
AED is equal to the whole angle
CDE.
But the angle
CDE is, by hypothesis, equal to the angles at
A,
C; therefore the angle
AED is also equal to the angles at
A,
C.
For the same reason the angle
ABC is also equal to the angles at
A,
C,
D.
Therefore the pentagon
ABCDE is equiangular. Q. E. D.
PROPOSITION 8.
If in an equilateral and equiangular pentagon straight lines subtend two angles taken in order,
they cut one another in extreme and mean ratio,
and their greater segments are equal to the side of the pentagon.
For in the equilateral and equiangular pentagon
ABCDE let the straight lines
AC,
BE, cutting one another at the point
H, subtend two angles taken in order, the angles at
A,
B; I say that each of them has been cut in extreme and mean ratio at the point
H, and their greater segments are equal to the side of the pentagon.
For let the circle
ABCDE be circumscribed about the pentagon
ABCDE. [
IV. 14]
Then, since the two straight lines
EA,
AB are equal to the two
AB,
BC, and they contain equal angles, therefore the base
BE is equal to the base
AC, the triangle
ABE is equal to the triangle
ABC, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend. [
I. 4]
Therefore the angle
BAC is equal to the angle
ABE; therefore the angle
AHE is double of the angle
BAH. [
I. 32]
But the angle
EAC is also double of the angle
BAC, inasmuch as the circumference
EDC is also double of the circumference
CB; [
III. 28,
VI. 33] therefore the angle
HAE is equal to the angle
AHE; hence the straight line
HE is also equal to
EA, that is, to
AB. [
I. 6]
And, since the straight line
BA is equal to
AE, the angle
ABE is also equal to the angle
AEB. [
I. 5]
But the angle
ABE was proved equal to the angle
BAH; therefore the angle
BEA is also equal to the angle
BAH.
And the angle
ABE is common to the two triangles
ABE and
ABH; therefore the remaining angle
BAE is equal to the remaining angle
AHB; [
I. 32] therefore the triangle
ABE is equiangular with the triangle
ABH; therefore, proportionally, as
EB is to
BA, so is
AB to
BH. [
VI. 4]
But
BA is equal to
EH; therefore, as
BE is to
EH, so is
EH to
HB.
And
BE is greater than
EH; therefore
EH is also greater than
HB. [
V. 14]
Therefore
BE has been cut in extreme and mean ratio at
H, and the greater segment
HE is equal to the side of the pentagon.
Similarly we can prove that
AC has also been cut in extreme and mean ratio at
H, and its greater segment
CH is equal to the side of the pentagon. Q. E. D.
PROPOSITION 9.
If the side of the hexagon and that of the decagon inscribed in the same circle be added together,
the whole straight line has been cut in extreme and mean ratio,
and its greater segment is the side of the hexagon.
Let
ABC be a circle; of the figures inscribed in the circle
ABC let
BC be the side of a decagon,
CD that of a hexagon, and let them be in a straight line; I say that the whole straight line
BD has been cut in extreme and mean ratio, and
CD is its greater segment.
For let the centre of the circle, the point
E, be taken, let
EB,
EC,
ED be joined, and let
BE be carried through to
A.
Since
BC is the side of an equilateral decagon, therefore the circumference
ACB is five times the circumference
BC; therefore the circumference
AC is quadruple of
CB.
But, as the circumference
AC is to
CB, so is the angle
AEC to the angle
CEB; [
VI. 33] therefore the angle
AEC is quadruple of the angle
CEB.
And, since the angle
EBC is equal to the angle
ECB, [
I. 5] therefore the angle
AEC is double of the angle
ECB. [
I. 32]
And, since the straight line
EC is equal to
CD, for each of them is equal to the side of the hexagon inscribed in the circle
ABC, [
IV. 15, Por.] the angle
CED is also equal to the angle
CDE; [
I. 5] therefore the angle
ECB is double of the angle
EDC. [
I. 32]
But the angle
AEC was proved double of the angle
ECB; therefore the angle
AEC is quadruple of the angle
EDC.
But the angle
AEC was also proved quadruple of the angle
BEC; therefore the angle
EDC is equal to the angle
BEC.
But the angle
EBD is common to the two triangles
BEC and
BED; therefore the remaining angle
BED is also equal to the remaining angle
ECB; [
I. 32] therefore the triangle
EBD is equiangular with the triangle
EBC.
Therefore, proportionally, as
DB is to
BE, so is
EB to
BC. [
VI. 4]
But
EB is equal to
CD.
Therefore, as
BD is to
DC, so is
DC to
CB.
And
BD is greater than
DC; therefore
DC is also greater than
CB.
Therefore the straight line
BD has been cut in extreme and mean ratio, and
DC is its greater segment. Q. E. D.
PROPOSITION 10.
If an equilateral pentagon be inscribed in a circle,
the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.
Let
ABCDE be a circle, and let the equilateral pentagon
ABCDE be inscribed in the circle
ABCDE.
I say that the square on the side of the pentagon
ABCDE is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the circle
ABCDE.
For let the centre of the circle, the point
F, be taken, let
AF be joined and carried through to the point
G, let
FB be joined, let
FH be drawn from
F perpendicular to
AB and be carried through to
K, let
AK,
KB be joined, let
FL be again drawn from
F perpendicular to
AK, and be carried through to
M, and let
KN be joined.
Since the circumference
ABCG is equal to the circumference
AEDG, and in them
ABC is equal to
AED, therefore the remainder, the circumference
CG, is equal to the remainder
GD.
But
CD belongs to a pentagon; therefore
CG belongs to a decagon.
And, since
FA is equal to
FB, and
FH is perpendicular, therefore the angle
AFK is also equal to the angle
KFB. [
I. 5,
I. 26]
Hence the circumference
AK is also equal to
KB; [
III. 26] therefore the circumference
AB is double of the circumference
BK; therefore the straight line
AK is a side of a decagon.
For the same reason
AK is also double of
KM.
Now, since the circumference
AB is double of the circumference
BK, while the circumference
CD is equal to the circumference
AB, therefore the circumference
CD is also double of the circumference
BK.
But the circumference
CD is also double of
CG; therefore the circumference
CG is equal to the circumference
BK.
But
BK is double of
KM, since
KA is so also; therefore
CG is also double of
KM.
But, further, the circumference
CB is also double of the circumference
BK, for the circumference
CB is equal to
BA.
Therefore the whole circumference
GB is also double of
BM; hence the angle
GFB is also double of the angle
BFM. [
VI. 33]
But the angle
GFB is also double of the angle
FAB, for the angle
FAB is equal to the angle
ABF.
Therefore the angle
BFN is also equal to the angle
FAB.
But the angle
ABF is common to the two triangles
ABF and
BFN; therefore the remaining angle
AFB is equal to the remaining angle
BNF; [
I. 32] therefore the triangle
ABF is equiangular with the triangle
BFN.
Therefore, proportionally, as the straight line
AB is to
BF, so is
FB to
BN; [
VI. 4] therefore the rectangle
AB,
BN is equal to the square on
BF. [
VI. 17]
Again, since
AL is equal to
LK, while
LN is common and at right angles, therefore the base
KN is equal to the base
AN; [
I. 4] therefore the angle
LKN is also equal to the angle
LAN.
But the angle
LAN is equal to the angle
KBN; therefore the angle
LKN is also equal to the angle
KBN.
And the angle at
A is common to the two triangles
AKB and
AKN.
Therefore the remaining angle
AKB is equal to the remaining angle
KNA; [
I. 32] therefore the triangle
KBA is equiangular with the triangle
KNA.
Therefore, proportionally, as the straight line
BA is to
AK, so is
KA to
AN; [
VI. 4] therefore the rectangle
BA,
AN is equal to the square on
AK. [
VI. 17]
But the rectangle
AB,
BN was also proved equal to the square on
BF; therefore the rectangle
AB,
BN together with the rectangle
BA,
AN, that is, the square on
BA [
II. 2], is equal to the square on
BF together with the square on
AK.
And
BA is a side of the pentagon,
BF of the hexagon [
IV. 15, Por.], and
AK of the decagon.
Therefore etc. Q. E. D.
PROPOSITION 11.
If in a circle which has its diameter rational an equilateral pentagon be inscribed,
the side of the pentagon is the irrational straight line called minor.
For in the circle
ABCDE which has its diameter rational let the equilateral pentagon
ABCDE be inscribed; I say that the side of the pentagon is the irrational straight line called minor.
For let the centre of the circle, the point
F, be taken, let
AF,
FB be joined and carried through to the points,
G,
H, let
AC be joined, and let
FK be made a fourth part of
AF.
Now
AF is rational; therefore
FK is also rational.
But
BF is also rational; therefore the whole
BK is rational.
And, since the circumference
ACG is equal to the circumference
ADG, and in them
ABC is equal to
AED, therefore the remainder
CG is equal to the remainder
GD.
And, if we join
AD, we conclude that the angles at
L are right, and
CD is double of
CL.
For the same reason the angles at
M are also right, and
AC is double of
CM.
Since then the angle
ALC is equal to the angle
AMF, and the angle
LAC is common to the two triangles
ACL and
AMF, therefore the remaining angle
ACL is equal to the remaining angle
MFA; [
I. 32] therefore the triangle
ACL is equiangular with the triangle
AMF; therefore, proportionally, as
LC is to
CA, so is
MF to
FA.
And the doubles of the antecedents may be taken; therefore, as the double of
LC is to
CA, so is the double of
MF to
FA.
But, as the double of
MF is to
FA, so is
MF to the half of
FA; therefore also, as the double of
LC is to
CA, so is
MF to the half of
FA.
And the halves of the consequents may be taken; therefore, as the double of
LC is to the half of
CA, so is
MF to the fourth of
FA.
And
DC is double of
LC,
CM is half of
CA, and
FK a fourth part of
FA; therefore, as
DC is to
CM, so is
MF to
FK.
Componendo also, as the sum of
DC,
CM is to
CM, so is
MK to
KF; [
V. 18] therefore also, as the square on the sum of
DC,
CM is to the square on
CM, so is the square on
MK to the square on
KF.
And since, when the straight line subtending two sides of the pentagon, as
AC, is cut in extreme and mean ratio, the greater segment is equal to the side of the pentagon, that is, to
DC, [
XIII. 8] while the square on the greater segment added to the half of the whole is five times the square on the half of the whole, [
XIII. 1] and
CM is half of the whole
AC, therefore the square on
DC,
CM taken as one straight line is five times the square on
CM.
But it was proved that, as the square on
DC,
CM taken as one straight line is to the square on
CM, so is the square on
MK to the square on
KF; therefore the square on
MK is five times the square on
KF.
But the square on
KF is rational, for the diameter is rational; therefore the square on
MK is also rational; therefore
MK is rational
And, since
BF is quadruple of
FK, therefore
BK is five times
KF; therefore the square on
BK is twenty-five times the square on
KF.
But the square on
MK is five times the square on
KF; therefore the square on
BK is five times the square on
KM; therefore the square on
BK has not to the square on
KM the ratio which a square number has to a square number; therefore
BK is incommensurable in length with
KM. [
X. 9]
And each of them is rational.
Therefore
BK,
KM are rational straight lines commensurable in square only.
But, if from a rational straight line there be subtracted a rational straight line which is commensurable with the whole in square only, the remainder is irrational, namely an apotome; therefore
MB is an apotome and
MK the annex to it. [
X. 73]
I say next that
MB is also a fourth apotome.
Let the square on
N be equal to that by which the square on
BK is greater than the square on
KM; therefore the square on
BK is greater than the square on
KM by the square on
N.
And, since
KF is commensurable with
FB,
componendo also,
KB is commensurable with
FB. [
X. 15]
But
BF is commensurable with
BH; therefore
BK is also commensurable with
BH. [
X. 12]
And, since the square on
BK is five times the square on
KM, therefore the square on
BK has to the square on
KM the ratio which 5 has to 1.
Therefore,
convertendo, the square on
BK has to the square on
N the ratio which 5 has to 4 [
V. 19, Por.], and this is not the ratio which a square number has to a square number; therefore
BK is incommensurable with
N; [
X. 9] therefore the square on
BK is greater than the square on
KM by the square on a straight line incommensurable with
BK.
Since then the square on the whole
BK is greater than the square on the annex
KM by the square on a straight line incommensurable with
BK, and the whole
BK is commensurable with the rational straight line,
BH, set out, therefore
MB is a fourth apotome. [
X. Deff. III. 4]
But the rectangle contained by a rational straight line and a fourth apotome is irrational, and its square root is irrational, and is called minor. [
X. 94]
But the square on
AB is equal to the rectangle
HB,
BM, because, when
AH is joined, the triangle
ABH is equiangular with the triangle
ABM, and, as
HB is to
BA, so is
AB to
BM.
Therefore the side
AB of the pentagon is the irrational straight line called minor. Q. E. D.
PROPOSITION 12.
If an equilateral triangle be inscribed in a circle,
the square on the side of the triangle is triple of the square on the radius of the circle.
Let
ABC be a circle, and let the equilateral triangle
ABC be inscribed in it; I say that the square on one side of the triangle
ABC is triple of the square on the radius of the circle.
For let the centre
D of the circle
ABC be taken, let
AD be joined and carried through to
E, and let
BE be joined.
Then, since the triangle
ABC is equilateral, therefore the circumference
BEC is a third part of the circumference of the circle
ABC.
Therefore the circumference
BE is a sixth part of the circumference of the circle; therefore the straight line
BE belongs to a hexagon; therefore it is equal to the radius
DE. [
IV. 15, Por.]
And, since
AE is double of
DE, the square on
AE is quadruple of the square on
ED, that is, of the square on
BE.
But the square on
AE is equal to the squares on
AB,
BE; [
III. 31,
I. 47] therefore the squares on
AB,
BE are quadruple of the square on
BE.
Therefore,
separando, the square on
AB is triple of the square on
BE.
But
BE is equal to
DE; therefore the square on
AB is triple of the square on
DE.
Therefore the square on the side of the triangle is triple of the square on the radius. Q. E. D.
PROPOSITION 13.
To construct a pyramid,
to comprehend it in a given sphere,
and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
Let the diameter
AB of the given sphere be set out, and let it be cut at the point
C so that
AC is double of
CB; let the semicircle
ADB be described on
AB, let
CD be drawn from the point
C at right angles to
AB, and let
DA be joined; let the circle
EFG which has its radius equal to
DC be set out, let the equilateral triangle
EFG be inscribed in the circle
EFG, [
IV. 2] let the centre of the circle, the point
H, be taken, [
III. 1] let
EH,
HF,
HG be joined; from the point
H let
HK be set up at right angles to the plane of the circle
EFG, [
XI. 12] let
HK equal to the straight line
AC be cut off from
HK, and let
KE,
KF,
KG be joined.
Now, since
KH is at right angles to the plane of the circle
EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle
EFG. [
XI. Def. 3]
But each of the straight lines
HE,
HF,
HG meets it: therefore
HK is at right angles to each of the straight lines
HE,
HF,
HG.
And, since
AC is equal to
HK, and
CD to
HE, and they contain right angles, therefore the base
DA is equal to the base
KE. [
I. 4]
For the same reason each of the straight lines
KF,
KG is also equal to
DA; therefore the three straight lines
KE,
KF,
KG are equal to one another.
And, since
AC is double of
CB, therefore
AB is triple of
BC.
But, as
AB is to
BC, so is the square on
AD to the square on
DC, as will be proved afterwards.
Therefore the square on
AD is triple of the square on
DC.
But the square on
FE is also triple of the square on
EH, [
XIII. 12] and
DC is equal to
EH; therefore
DA is also equal to
EF.
But
DA was proved equal to each of the straight lines
KE,
KF,
KG; therefore each of the straight lines
EF,
FG,
GE is also equal to each of the straight lines
KE,
KF,
KG; therefore the four triangles
EFG,
KEF,
KFG,
KEG are equilateral.
Therefore a pyramid has been constructed out of four equilateral triangles, the triangle
EFG being its base and the point
K its vertex.
It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
For let the straight line
HL be produced in a straight line with
KH, and let
HL be made equal to
CB.
Now, since, as
AC is to
CD, so is
CD to
CB, [
VI. 8, Por.] while
AC is equal to
KH,
CD to
HE, and
CB to
HL, therefore, as
KH is to
HE, so is
EH to
HL; therefore the rectangle
KH,
HL is equal to the square on
EH. [
VI. 17]
And each of the angles
KHE.
EHL is right; therefore the semicircle described on
KL will pass through
E also. [cf.
VI. 8,
III. 31.]
If then,
KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points
F,
G, since, if
FL,
LG be joined, the angles at
F,
G similarly become right angles; and the pyramid will be comprehended in the given sphere.
For
KL, the diameter of the sphere, is equal to the diameter
AB of the given sphere, inasmuch as
KH was made equal to
AC, and
HL to
CB.
I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid
For, since
AC is double of
CB, therefore
AB is triple of
BC; and,
convertendo,
BA is one and a half times
AC.
But, as
BA is to
AC, so is the square on
BA to the square on
AD.
Therefore the square on
BA is also one and a half times the square on
AD.
And
BA is the diameter of the given sphere, and
AD is equal to the side of the pyramid.
Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q. E. D.
LEMMA.
It is to be proved that, as
AB is to
BC, so is the square on
AD to the square on
DC.
For let the figure of the semicircle be set out, let
DB be joined, let the square
EC be described on
AC, and let the parallelogram
FB be completed.
Since then, because the triangle
DAB is equiangular with the triangle
DAC, as
BA is to
AD, so is
DA to
AC, [
VI. 8,
VI. 4] therefore the rectangle
BA,
AC is equal to the square on
AD. [
VI. 17]
And since, as
AB is to
BC, so is
EB to
BF, [
VI. 1] and
EB is the rectangle
BA,
AC, for
EA is equal to
AC, and
BF is the rectangle
AC,
CB, therefore, as
AB is to
BC, so is the rectangle
BA,
AC to the rectangle
AC,
CB.
And the rectangle
BA,
AC is equal to the square on
AD, and the rectangle
AC,
CB to the square on
DC, for the perpendicular
DC is a mean proportional between the segments
AC,
CB of the base, because the angle
ADB is right. [
VI. 8, Por.]
Therefore, as
AB is to
BC, so is the square on
AD to the square on
DC. Q. E. D.
PROPOSITION 14.
To construct an octahedron and comprehend it in a sphere,
as in the preceding case;
and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.
Let the diameter
AB of the given sphere be set out, and let it be bisected at
C; let the semicircle
ADB be described on
AB, let
CD be drawn from
C at right angles to
AB, let
DB be joined; let the square
EFGH, having each of its sides equal to
DB, be set out, let
HF,
EG be joined, from the point
K let the straight line
KL be set up at right angles to the plane of the square
EFGH [
XI. 12], and let it be carried through to the other side of the plane, as
KM; from the straight lines
KL,
KM let
KL,
KM be respectively cut off equal to one of the straight lines
EK,
FK,
GK,
HK, and let
LE,
LF,
LG,
LH,
ME,
MF,
MG,
MH be joined.
Then, since
KE is equal to
KH, and the angle
EKH is right, therefore the square on
HE is double of the square on
EK. [
I. 47]
Again, since
LK is equal to
KE, and the angle
LKE is right, therefore the square on
EL is double of the square on
EK. [
id.]
But the square on
HE was also proved double of the square on
EK; therefore the square on
LE is equal to the square on
EH; therefore
LE is equal to
EH.
For the same reason
LH is also equal to
HE; therefore the triangle
LEH is equilateral.
Similarly we can prove that each of the remaining triangles of which the sides of the square
EFGH are the bases, and the points
L,
M the vertices, is equilateral; therefore an octahedron has been constructed which is contained by eight equilateral triangles.
It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.
For, since the three straight lines
LK,
KM,
KE are equal to one another, therefore the semicircle described on
LM will also pass through
E.
And for the same reason, if,
LM remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points
F,
G,
H, and the octahedron will have been comprehended in a sphere.
I say next that it is also comprehended in the given sphere.
For, since
LK is equal to
KM, while
KE is common, and they contain right angles, therefore the base
LE is equal to the base
EM. [
I. 4]
And, since the angle
LEM is right, for it is in a semicircle, [
III. 31] therefore the square on
LM is double of the square on
LE. [
I. 47]
Again, since
AC is equal to
CB,
AB is double of
BC.
But, as
AB is to
BC, so is the square on
AB to the square on
BD; therefore the square on
AB is double of the square on
BD.
But the square on
LM was also proved double of the square on
LE.
And the square on
DB is equal to the square on
LE, for
EH was made equal to
DB.
Therefore the square on
AB is also equal to the square on
LM; therefore
AB is equal to
LM.
And
AB is the diameter of the given sphere; therefore
LM is equal to the diameter of the given sphere.
Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double of the square on the side of the octahedron. Q. E. D.
PROPOSITION 15.
To construct a cube and comprehend it in a sphere,
like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.
Let the diameter
AB of the given sphere be set out, and let it be cut at
C so that
AC is double of
CB; let the semicircle
ADB be described on
AB, let
CD be drawn from
C at right angles to
AB, and let
DB be joined; let the square
EFGH having its side equal to
DB be set out, from
E,
F,
G,
H let
EK,
FL,
GM,
HN be drawn at right angles to the plane of the square
EFGH, from
EK,
FL,
GM,
HN let
EK,
FL,
GM,
HN respectively be cut off equal to one of the straight lines
EF,
FG,
GH,
HE, and let
KL,
LM,
MN,
NK be joined; therefore the cube
FN has been constructed which is contained by six equal squares.
It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.
For let
KG,
EG be joined.
Then, since the angle
KEG is right, because
KE is also at right angles to the plane
EG and of course to the straight line
EG also, [
XI. Def. 3] therefore the semicircle described on
KG will also pass through the point
E.
Again, since
GF is at right angles to each of the straight lines
FL,
FE,
GF is also at right angles to the plane
FK; hence also, if we join
FK,
GF will be at right angles to
FK; and for this reason again the semicircle described on
GK will also pass through
F.
Similarly it will also pass through the remaining angular points of the cube.
If then,
KG remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, the cube will be comprehended in a sphere.
I say next that it is also comprehended in the given sphere.
For, since
GF is equal to
FE, and the angle at
F is right, therefore the square on
EG is double of the square on
EF.
But
EF is equal to
EK; therefore the square on
EG is double of the square on
EK; hence the squares on
GE,
EK, that is the square on
GK [
I. 47], is triple of the square on
EK.
And, since
AB is triple of
BC, while, as
AB is to
BC, so is the square on
AB to the square on
BD, therefore the square on
AB is triple of the square on
BD.
But the square on
GK was also proved triple of the square on
KE.
And
KE was made equal to
DB; therefore
KG is also equal to
AB.
And
AB is the diameter of the given sphere; therefore
KG is also equal to the diameter of the given sphere.
Therefore the cube has been comprehended in the given sphere; and it has been demonstrated at the same time that the square on the diameter of the sphere is triple of the square on the side of the cube. Q. E. D.
PROPOSITION 16.
To construct an icosahedron and comprehend it in a sphere,
like the aforesaid figures; and to prove that the side of the icosahedron is the irrational straight line called minor.
Let the diameter
AB of the given sphere be set out, and let it be cut at
C so that
AC is quadruple of
CB, let the semicircle
ADB be described on
AB, let the straight line
CD be drawn from
C at right angles to
AB, and let
DB be joined; let the circle
EFGHK be set out and let its radius be equal to
DB, let the equilateral and equiangular pentagon
EFGHK be inscribed in the circle
EFGHK, let the circumferences
EF,
FG,
GH,
HK,
KE be bisected at the points
L,
M,
N,
O,
P, and let
LM,
MN,
NO,
OP,
PL,
EP be joined.
Therefore the pentagon
LMNOP is also equilateral, and the straight line
EP belongs to a decagon.
Now from the points
E,
F,
G,
H,
K let the straight lines
EQ,
FR,
GS,
HT,
KU be set up at right angles to the plane of the circle, and let them be equal to the radius of the circle
EFGHK, let
QR,
RS,
ST,
TU,
UQ,
QL,
LR,
RM,
MS,
SN,
NT,
TO,
OU,
UP,
PQ be joined.
Now, since each of the straight lines
EQ,
KU is at right angles to the same plane, therefore
EQ is parallel to
KU. [
XI. 6]
But it is also equal to it; and the straight lines joining those extremities of equal and parallel straight lines which are in the same direction are equal and parallel. [
I. 33]
Therefore
QU is equal and parallel to
EK.
But
EK belongs to an equilateral pentagon; therefore
QU also belongs to the equilateral pentagon inscribed in the circle
EFGHK.
For the same reason each of the straight lines
QR,
RS,
ST,
TU also belongs to the equilateral pentagon inscribed in the circle
EFGHK; therefore the pentagon
QRSTU is equilateral.
And, since
QE belongs to a hexagon, and
EP to a decagon, and the angle
QEP is right, therefore
QP belongs to a pentagon; for the square on the side of the pentagon is equal to the square on the side of the hexagon and the square on the side of the decagon inscribed in the same circle. [
XIII. 10]
For the same reason
PU is also a side of a pentagon.
But
QU also belongs to a pentagon; therefore the triangle
QPU is equilateral.
For the same reason each of the triangles
QLR,
RMS,
SNT,
TOU is also equilateral.
And, since each of the straight lines
QL,
QP was proved to belong to a pentagon, and
LP also belongs to a pentagon, therefore the triangle
QLP is equilateral.
For the same reason each of the triangles
LRM,
MSN,
NTO,
OUP is also equilateral.
Let the centre of the circle
EFGHK the point
V, be taken; from
V let
VZ be set up at right angles to the plane of the circle, let it be produced in the other direction, as
VX, let there be cut off
VW, the side of a hexagon, and each of the straight lines
VX,
WZ, being sides of a decagon, and let
QZ,
QW,
UZ,
EV,
LV,
LX,
XM be joined.
Now, since each of the straight lines
VW,
QE is at right angles to the plane of the circle, therefore
VW is parallel to
QE. [
XI. 6]
But they are also equal; therefore
EV,
QW are also equal and parallel. [
I. 33]
But
EV belongs to a hexagon; therefore
QW also belongs to a hexagon.
And, since
QW belongs to a hexagon, and
WZ to a decagon, and the angle
QWZ is right, therefore
QZ belongs to a pentagon. [
XIII. 10]
For the same reason
UZ also belongs to a pentagon, inasmuch as, if we join
VK,
WU, they will be equal and opposite, and
VK, being a radius, belongs to a hexagon; [
IV. 15, Por.] therefore
WU also belongs to a hexagon.
But
WZ belongs to a decagon, and the angle
UWZ is right; therefore
UZ belongs to a pentagon. [
XIII. 10]
But
QU also belongs to a pentagon; therefore the triangle
QUZ is equilateral.
For the same reason each of the remaining triangles of which the straight lines
QR,
RS,
ST,
TU are the bases, and the point
Z the vertex, is also equilateral.
Again, since
VL belongs to a hexagon, and
VX to a decagon, and the angle
LVX is right, therefore
LX belongs to a pentagon. [
XIII. 10]
For the same reason, if we join
MV, which belongs to a hexagon,
MX is also inferred to belong to a pentagon.
But
LM also belongs to a pentagon; therefore the triangle
LMX is equilateral.
Similarly it can be proved that each of the remaining triangles of which
MN,
NO,
OP,
PL are the bases, and the point
X the vertex, is also equilateral.
Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles.
It is next required to comprehend it in the given sphere, and to prove that the side of the icosahedron is the irrational straight line called minor.
For, since
VW belongs to a hexagon, and
WZ to a decagon, therefore
VZ has been cut in extreme and mean ratio at
W, and
VW is its greater segment; [
XIII. 9] therefore, as
ZV is to
VW, so is
VW to
WZ.
But
VW is equal to
VE, and
WZ to
VX; therefore, as
ZV is to
VE, so is
EV to
VX.
And the angles
ZVE,
EVX are right; therefore, if we join the straight line
EZ, the angle
XEZ will be right because of the similarity of the triangles
XEZ,
VEZ.
For the same reason, since, as
ZV is to
VW, so is
VW to
WZ, and
ZV is equal to
XW, and
VW to
WQ, therefore, as
XW is to
WQ, so is
QW to
WZ.
And for this reason again, if we join
QX, the angle at
Q will be right; [
VI. 8] therefore the semicircle described on
XZ will also pass through
Q. [
III. 31]
And if,
XZ remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through
Q and the remaining angular points of the icosahedron, and the icosahedron will have been comprehended in a sphere.
I say next that it is also comprehended in the given sphere.
For let
VW be bisected at
A'.
Then, since the straight line
VZ has been cut in extreme and mean ratio at
W, and
ZW is its lesser segment, therefore the square on
ZW added to the half of the greater segment, that is
WA', is five times the square on the half of the greater segment; [
XIII. 3] therefore the square on
ZA' is five times the square on .
And
ZX is double of
ZA', and
VW double of ; therefore the square on
ZX is five times the square on
WV.
And, since
AC is quadruple of
CB, therefore
AB is five times
BC.
But, as
AB is to
BC, so is the square on
AB to the square on
BD; [
VI. 8,
V. Def. 9] therefore the square on
AB is five times the square on
BD.
But the square on
ZX was also proved to be five times the square on
VW.
And
DB is equal to
VW, for each of them is equal to the radius of the circle
EFGHK; therefore
AB is also equal to
XZ.
And
AB is the diameter of the given sphere; therefore
XZ is also equal to the diameter of the given sphere.
Therefore the icosahedron has been comprehended in the given sphere
I say next that the side of the icosahedron is the irrational straight line called minor.
For, since the diameter of the sphere is rational, and the square on it is five times the square on the radius of the circle
EFGHK, therefore the radius of the circle
EFGHK is also rational; hence its diameter is also rational.
But, if an equilateral pentagon be inscribed in a circle which has its diameter rational, the side of the pentagon is the irrational straight line called minor. [
XIII. 11]
And the side of the pentagon
EFGHK is the side of the icosahedron.
Therefore the side of the icosahedron is the irrational straight line called minor.
PORISM.
From this it is manifest that the square on the diameter of the sphere is five times the square on the radius of the circle from which the icosahedron has been described, and that the diameter of the sphere is composed of the side of the hexagon and two of the sides of the decagon inscribed in the same circle. Q. E. D.
PROPOSITION 17.
To construct a dodecahedron and comprehend it in a sphere,
like the aforesaid figures, and to prove that the side of the dodecahedron is the irrational straight line called apotome.
Let
ABCD,
CBEF, two planes of the aforesaid cube at right angles to one another, be set out, let the sides
AB,
BC,
CD,
DA,
EF,
EB,
FC be bisected at
G,
H,
K,
L,
M,
N,
O respectively, let
GK,
HL,
MH,
NO be joined, let the straight lines
NP,
PO,
HQ be cut in extreme and mean ratio at the points
R,
S,
T respectively, and let
RP,
PS,
TQ be their greater segments; from the points
R,
S,
T let
RU,
SV,
TW be set up at right angles to the planes of the cube towards the outside of the cube, let them be made equal to
RP,
PS,
TQ, and let
UB,
BW,
WC,
CV,
VU be joined.
I say that the pentagon
UBWCV is equilateral, and in one plane, and is further equiangular.
For let
RB,
SB,
VB be joined.
Then, since the straight line
NP has been cut in extreme and mean ratio at
R, and
RP is the greater segment, therefore the squares on
PN,
NR are triple of the square on
RP. [
XIII. 4]
But
PN is equal to
NB, and
PR to
RU; therefore the squares on
BN,
NR are triple of the square on
RU.
But the square on
BR is equal to the squares on
BN,
NR; [
I. 47] therefore the square on
BR is triple of the square on
RU; hence the squares on
BR,
RU are quadruple of the square on
RU.
But the square on
BU is equal to the squares on
BR,
RU; therefore the square on
BU is quadruple of the square on
RU; therefore
BU is double of
RU.
But
VU is also double of
UR, inasmuch as
SR is also double of
PR, that is, of
RU; therefore
BU is equal to
UV.
Similarly it can be proved that each of the straight lines
BW,
WC,
CV is also equal to each of the straight lines
BU,
UV.
Therefore the pentagon
BUVCW is equilateral.
I say next that it is also in one plane.
For let
PX be drawn from
P parallel to each of the straight lines
RU,
SV and towards the outside of the cube, and let
XH,
HW be joined; I say that
XHW is a straight line.
For, since
HQ has been cut in extreme and mean ratio at
T, and
QT is its greater segment, therefore, as
HQ is to
QT, so is
QT to
TH.
But
HQ is equal to
HP, and
QT to each of the straight lines
TW,
PX; therefore, as
HP is to
PX, so is
WT to
TH.
And
HP is parallel to
TW, for each of them is at right angles to the plane
BD; [
XI. 6] and
TH is parallel to
PX, for each of them is at right angles to the plane
BF. [
id.]
But if two triangles, as
XPH,
HTW, which have two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining straight lines will be in a straight line; [
VI. 32] therefore
XH is in a straight line with
HW.
But every straight line is in one plane; [
XI. 1] therefore the pentagon
UBWCV is in one plane.
I say next that it is also equiangular.
For, since the straight line
NP has been cut in extreme and mean ratio at
R, and
PR is the greater segment, while
PR is equal to
PS, therefore
NS has also been cut in extreme and mean ratio at
P, and
NP is the greater segment; [
XIII. 5] therefore the squares on
NS,
SP are triple of the square on
NP. [
XIII. 4]
But
NP is equal to
NB, and
PS to
SV; therefore the squares on
NS,
SV are triple of the square on
NB; hence the squares on
VS,
SN,
NB are quadruple of the square on
NB.
But the square on
SB is equal to the squares on
SN,
NB; therefore the squares on
BS,
SV, that is, the square on
BV —for the angle
VSB is right—is quadruple of the square on
NB; therefore
VB is double of
BN.
But
BC is also double of
BN; therefore
BV is equal to
BC.
And, since the two sides
BU,
UV are equal to the two sides
BW,
WC, and the base
BV is equal to the base
BC, therefore the angle
BUV is equal to the angle
BWC. [
I. 8]
Similarly we can prove that the angle
UVC is also equal to the angle
BWC; therefore the three angles
BWC,
BUV,
UVC are equal to one another.
But if in an equilateral pentagon three angles are equal to one another, the pentagon will be equiangular, [
XIII. 7] therefore the pentagon
BUVCW is equiangular.
And it was also proved equilateral; therefore the pentagon
BUVCW is equilateral and equiangular, and it is on one side
BC of the cube.
Therefore, if we make the same construction in the case of each of the twelve sides of the cube, a solid figure will have been constructed which is contained by twelve equilateral and equiangular pentagons, and which is called a dodecahedron.
It is then required to comprehend it in the given sphere, and to prove that the side of the dodecahedron is the irrational straight line called apotome.
For let
XP be produced, and let the produced straight line be
XZ; therefore
PZ meets the diameter of the cube, and they bisect one another, for this has been proved in the last theorem but one of the eleventh book. [
XI. 38]
Let them cut at
Z; therefore
Z is the centre of the sphere which comprehends the cube, and
ZP is half of the side of the cube.
Let
UZ be joined.
Now, since the straight line
NS has been cut in extreme and mean ratio at
P, and
NP is its greater segment, therefore the squares on
NS,
SP are triple of the square on
NP. [
XIII. 4]
But
NS is equal to
XZ, inasmuch as
NP is also equal to
PZ, and
XP to
PS.
But further
PS is also equal to
XU, since it is also equal to
RP; therefore the squares on
ZX,
XU are triple of the square on
NP.
But the square on
UZ is equal to the squares on
ZX,
XU; therefore the square on
UZ is triple of the square on
NP.
But the square on the radius of the sphere which comprehends the cube is also triple of the square on the half of the side of the cube, for it has previously been shown how to construct a cube and comprehend it in a sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube. [
XIII. 15]
But, if whole is so related to whole, so is half to half also; and
NP is half of the side of the cube; therefore
UZ is equal to the radius of the sphere which comprehends the cube.
And
Z is the centre of the sphere which comprehends the cube; therefore the point
U is on the surface of the sphere.
Similarly we can prove that each of the remaining angles of the dodecahedron is also on the surface of the sphere; therefore the dodecahedron has been comprehended in the given sphere.
I say next that the side of the dodecahedron is the irrational straight line called apotome.
For since, when
NP has been cut in extreme and mean ratio,
RP is the greater segment, and, when
PO has been cut in extreme and mean ratio,
PS is the greater segment, therefore, when the whole
NO is cut in extreme and mean ratio,
RS is the greater segment.
[Thus, since, as
NP is to
PR, so is
PR to
RN, the same is true of the doubles also, for parts have the same ratio as their equimultiples; [
V. 15] therefore as
NO is to
RS, so is
RS to the sum of
NR,
SO.
But
NO is greater than
RS; therefore
RS is also greater than the sum of
NR,
SO; therefore
NO has been cut in extreme and mean ratio, and
RS is its greater segment.]
But
RS is equal to
UV; therefore, when
NO is cut in extreme and mean ratio,
UV is the greater segment.
And, since the diameter of the sphere is rational, and the square on it is triple of the square on the side of the cube, therefore
NO, being a side of the cube, is rational.
[But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.]
Therefore
UV, being a side of the dodecahedron, is an irrational apotome. [
XIII. 6]
PORISM.
From this it is manifest that, when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. Q. E. D.
PROPOSITION 18.
To set out the sides of the five figures and to compare them with one another.
Let
AB, the diameter of the given sphere, be set out, and let it be cut at
C so that
AC is equal to
CB, and at
D so that
AD is double of
DB; let the semicircle
AEB be described on
AB, from
C,
D let
CE,
DF be drawn at right angles to
AB, and let
AF,
FB,
EB be joined.
Then, since
AD is double of
DB, therefore
AB is triple of
BD.
Convertendo, therefore,
BA is one and a half times
AD.
But, as
BA is to
AD, so is the square on
BA to the square on
AF, [
V. Def. 9,
VI. 8] for the triangle
AFB is equiangular with the triangle
AFD; therefore the square on
BA is one and a half times the square on
AF.
But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [
XIII. 13]
And
AB is the diameter of the sphere; therefore
AF is equal to the side of the pyramid.
Again, since
AD is double of
DB, therefore
AB is triple of
BD.
But, as
AB is to
BD, so is the square on
AB to the square on
BF; [
VI. 8,
V. Def. 9] therefore the square on
AB is triple of the square on
BF.
But the square on the diameter of the sphere is also triple of the square on the side of the cube. [
XIII. 15]
And
AB is the diameter of the sphere; therefore
BF is the side of the cube.
And, since
AC is equal to
CB, therefore
AB is double of
BC.
But, as
AB is to
BC, so is the square on
AB to the square on
BE; therefore the square on
AB is double of the square on
BE.
But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [
XIII. 14]
And
AB is the diameter of the given sphere; therefore
BE is the side of the octahedron.
Next, let
AG be drawn from the point
A at right angles to the straight line
AB, let
AG be made equal to
AB, let
GC be joined, and from
H let
HK be drawn perpendicular to
AB.
Then, since
GA is double of
AC, for
GA is equal to
AB, and, as
GA is to
AC, so is
HK to
KC, therefore
HK is also double of
KC.
Therefore the square on
HK is quadruple of the square on
KC; therefore the squares on
HK,
KC, that is, the square on
HC, is five times the square on
KC.
But
HC is equal to
CB; therefore the square on
BC is five times the square on
CK.
And, since
AB is double of
CB, and, in them,
AD is double of
DB, therefore the remainder
BD is double of the remainder
DC.
Therefore
BC is triple of
CD; therefore the square on
BC is nine times the square on
CD.
But the square on
BC is five times the square on
CK; therefore the square on
CK is greater than the square on
CD; therefore
CK is greater than
CD.
Let
CL be made equal to
CK, from
L let
LM be drawn at right angles to
AB, and let
MB be joined.
Now, since the square on
BC is five times the square on
CK, and
AB is double of
BC, and
KL double of
CK, therefore the square on
AB is five times the square on
KL.
But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [
XIII. 16, Por.]
And
AB is the diameter of the sphere; therefore
KL is the radius of the circle from which the icosahedron has been described; therefore
KL is a side of the hexagon in the said circle. [
IV. 15, Por.]
And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [
XIII. 16, Por.] and
AB is the diameter of the sphere, while
KL is a side of the hexagon, and
AK is equal to
LB, therefore each of the straight lines
AK,
LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.
And, since
LB belongs to a decagon, and
ML to a hexagon, for
ML is equal to
KL, since it is also equal to
HK, being the same distance from the centre, and each of the straight lines
HK,
KL is double of
KC, therefore
MB belongs to a pentagon. [
XIII. 10]
But the side of the pentagon is the side of the icosahedron; [
XIII. 16] therefore
MB belongs to the icosahedron.
Now, since
FB is a side of the cube, let it be cut in extreme and mean ratio at
N, and let
NB be the greater segment; therefore
NB is a side of the dodecahedron. [
XIII. 17, Por.]
And, since the square on the diameter of the sphere was proved to be one and a half times the square on the side
AF of the pyramid, double of the square on the side
BE of the octahedron and triple of the side
FB of the cube, therefore, of parts of which the square on the diameter of the sphere contains six, the square on the side of the pyramid contains four, the square on the side of the octahedron three, and the square on the side of the cube two.
Therefore the square on the side of the pyramid is fourthirds of the square on the side of the octahedron, and double of the square on the side of the cube; and the square on the side of the octahedron is one and a half times the square on the side of the cube.
The said sides, therefore, of the three figures, I mean the pyramid, the octahedron and the cube, are to one another in rational ratios.
But the remaining two, I mean the side of the icosahedron and the side of the dodecahedron, are not in rational ratios either to one another or to the aforesaid sides; for they are irrational, the one being minor [
XIII. 16] and the other an apotome [
XIII. 17].
That the side
MB of the icosahedron is greater than the side
NB of the dodecahedron we can prove thus.
For, since the triangle
FDB is equiangular with the triangle
FAB, [
VI. 8] proportionally, as
DB is to
BF, so is
BF to
BA. [
VI. 4]
And, since the three straight lines are proportional, as the first is to the third, so is the square on the first to the square on the second; [
V. Def. 9,
VI. 20, Por.] therefore, as
DB is to
BA, so is the square on
DB to the square on
BF; therefore, inversely, as
AB is to
BD, so is the square on
FB to the square on
BD.
But
AB is triple of
BD; therefore the square on
FB is triple of the square on
BD.
But the square on
AD is also quadruple of the square on
DB, for
AD is double of
DB; therefore the square on
AD is greater than the square on
FB; therefore
AD is greater than
FB; therefore
AL is by far greater than
FB.
And, when
AL is cut in extreme and mean ratio,
KL is the greater segment, inasmuch as
LK belongs to a hexagon, and
KA to a decagon; [
XIII. 9] and, when
FB is cut in extreme and mean ratio,
NB is the greater segment; therefore
KL is greater than
NB.
But
KL is equal to
LM; therefore
LM is greater than
NB.
Therefore
MB, which is a side of the icosahedron, is by far greater than
NB which is a side of the dodecahedron. Q. E. D.
I say next that
no other figure,
besides the said five figures,
can be constructed which is contained by equilateral and equiangular figures equal to one another.
For a solid angle cannot be constructed with two triangles, or indeed planes.
With three triangles the angle of the pyramid is constructed, with four the angle of the octahedron, and with five the angle of the icosahedron; but a solid angle cannot be formed by six equilateral and equiangular triangles placed together at one point, for, the angle of the equilateral triangle being two-thirds of a right angle, the six will be equal to four right angles: which is impossible, for any solid angle is contained by angles less than four right angles. [
XI. 21]
For the same reason, neither can a solid angle be constructed by more than six plane angles.
By three squares the angle of the cube is contained, but by four it is impossible for a solid angle to be contained, for they will again be four right angles.
By three equilateral and equiangular pentagons the angle of the dodecahedron is contained; but by four such it is impossible for any solid angle to be contained, for, the angle of the equilateral pentagon being a right angle and a fifth, the four angles will be greater than four right angles: which is impossible.
Neither again will a solid angle be contained by other polygonal figures by reason of the same absurdity.
Therefore etc. Q. E. D.
LEMMA.
But that
the angle of the equilateral and equiangular pentagon is a right angle and a fifth we must prove thus.
Let
ABCDE be an equilateral and equiangular pentagon, let the circle
ABCDE be circumscribed about it, let its centre
F be taken, and let
FA,
FB,
FC,
FD,
FE be joined.
Therefore they bisect the angles of the pentagon at
A,
B,
C,
D,
E.
And, since the angles at
F are equal to four right angles and are equal, therefore one of them, as the angle
AFB, is one right angle less a fifth; therefore the remaining angles
FAB,
ABF consist of one right angle and a fifth.
But the angle
FAB is equal to the angle
FBC; therefore the whole angle
ABC of the pentagon consists of one right angle and a fifth. Q. E. D.