Elements, Book 13

By Euclid

Edition: 0.0.0-dev | March 03, 2014

Authority: SCTA

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BOOK XIII.

BOOK XIII. PROPOSITIONS.

PROPOSITION 1.

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.

2 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 .

3 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 .

4 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 ]

5 And CE is the rectangle AB , BC , and FH the square on AC ; therefore CE is equal to FH .

6 And, since BA is double of AD , while BA is equal to KA , and AD to AH , therefore KA is also double of AH .

7 But, as KA is to AH , so is CK to CH ; [ VI. 1 ] therefore CK is double of CH .

10 But CE was also proved equal to HF ; therefore the whole square AE is equal to the gnomon MNO .

11 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 .

12 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 .

13 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 .

PROPOSITION 2.

15 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.

16 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 .

17 Let the squares AF , CG be described on AB , CD respectively, let the figure in AF be drawn, and let BE be drawn through.

18 Now, since the square on BA is five times the square on AC , AF is five times AH .

20 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 .

21 But the gnomon MNO was also proved quadruple of AH ; therefore the gnomon MNO is equal to CG .

22 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 ]

23 But LH , HB are also double of HB ; therefore KB is equal to LH , HB .

24 But the whole gnomon MNO was also proved equal to the whole CG ; therefore the remainder HF is equal to BG .

25 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 .

27 But DC is greater than CB ; therefore CB is also greater than BD .

28 Therefore, when the straight line CD is cut in extreme and mean ratio, CB is the greater segment.

LEMMA.

30 That the double of AC is greater than BC is to be proved thus.

32 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 .

33 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 ]

35 Similarly we can prove that neither is a straight line less than CB double of CA ; for the absurdity is much greater.

36 Therefore the double of AC is greater than CB . Q. E. D.

PROPOSITION 3.

37 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.

38 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 .

39 For let the square AE be described on AB , and let the figure be drawn double.

40 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 .

41 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 .

42 But RS is quadruple of FG ; therefore CE is also quadruple of FG .

43 Again, since AD is equal to DC , HK is also equal to KF .

44 Hence the square GF is also equal to the square HL .

45 Therefore GK is equal to KL , that is, MN to NE ; hence MF is also equal to FE .

46 But MF is equal to CG ; therefore CG is also equal to FE .

47 Let CN be added to each; therefore the gnomon OPQ is equal to CE .

48 But CE was proved quadruple of GF ; therefore the gnomon OPQ is also quadruple of the square FG .

49 Therefore the gnomon OPQ and the square FG are five times FG .

50 But the gnomon OPQ and the square FG are the square DN .

51 And DN is the square on DB , and GF the square on DC .

52 Therefore the square on DB is five times the square on DC . Q. E. D.

PROPOSITION 4.

53 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.

54 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 .

55 For let the square ADEB be described on AB , and let the figure be drawn.

56 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 ]

57 And AK is the rectangle AB , BC , and HG the square on AC ; therefore AK is equal to HG .

58 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 .

59 But AK , CE are the gnomon LMN and the square CK ; therefore the gnomon LMN and the square CK are double of AK .

60 But, further, AK was also proved equal to HG ; therefore the gnomon LMN and the squares CK , HG are triple of the square HG .

61 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 .

62 Therefore the squares on AB , BC are triple of the square on AC . Q. E. D.

PROPOSITION 5.

63 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.

64 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 .

65 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.

66 For let the square AE be described on AB , and let the figure be drawn.

67 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]

68 And CE is the rectangle AB , BC , and CH the square on AC ; therefore CE is equal to HC .

69 But HE is equal to CE , and DH is equal to HC ; therefore DH is also equal to HE .

71 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 .

72 Therefore, as DB is to BA , so is BA to AD . [ VI. 17 ]

73 And DB is greater than BA ; therefore BA is also greater than AD . [ V. 14 ]

74 Therefore DB has been cut in extreme and mean ratio at A , and AB is the greater segment. Q. E. D.

PROPOSITION 6.

75 If a rational straight line be cut in extreme and mean ratio , each of the segments is the irrational straight line called apotome.

76 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.

77 For let BA be produced, and let AD be made half of BA .

78 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 ]

79 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 ]

80 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.

81 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 ]

82 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 ]

83 Therefore the square on the apotome AC , if applied to the rational straight line AB , produces BC as breadth.

84 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.

PROPOSITION 7.

87 If three angles of an equilateral pentagon, taken either in order or not in order , be equal , the pentagon will be equiangular.

88 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.

90 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 ]

91 But the whole AC was also proved equal to the whole BE ; therefore the remainder FC is also equal to the remainder FE .

93 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 ]

94 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 .

95 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 .

96 Similarly we can prove that the angle CDE is also equal to the angles at A , B , C ; therefore the pentagon ABCDE is equiangular.

97 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.

99 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 .

100 But the angle BED is also equal to the angle BDE , since the side BE is also equal to the side BD . [ I. 5 ]

101 Therefore the whole angle AED is equal to the whole angle CDE .

102 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 .

103 For the same reason the angle ABC is also equal to the angles at A , C , D .

104 Therefore the pentagon ABCDE is equiangular. Q. E. D.

PROPOSITION 8.

105 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.

106 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.

107 For let the circle ABCDE be circumscribed about the pentagon ABCDE . [ IV. 14 ]

108 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 ]

109 Therefore the angle BAC is equal to the angle ABE ; therefore the angle AHE is double of the angle BAH . [ I. 32 ]

110 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 ]

111 And, since the straight line BA is equal to AE , the angle ABE is also equal to the angle AEB . [ I. 5 ]

112 But the angle ABE was proved equal to the angle BAH ; therefore the angle BEA is also equal to the angle BAH .

113 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 ]

114 But BA is equal to EH ; therefore, as BE is to EH , so is EH to HB .

115 And BE is greater than EH ; therefore EH is also greater than HB . [ V. 14 ]

116 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.

117 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.

118 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.

119 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.

120 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 .

121 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 .

122 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 .

123 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 ]

124 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 ]

125 But the angle AEC was proved double of the angle ECB ; therefore the angle AEC is quadruple of the angle EDC .

126 But the angle AEC was also proved quadruple of the angle BEC ; therefore the angle EDC is equal to the angle BEC .

127 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 .

128 Therefore, proportionally, as DB is to BE , so is EB to BC . [ VI. 4 ]

131 And BD is greater than DC ; therefore DC is also greater than CB .

132 Therefore the straight line BD has been cut in extreme and mean ratio, and DC is its greater segment. Q. E. D.

PROPOSITION 10.

133 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.

134 Let ABCDE be a circle, and let the equilateral pentagon ABCDE be inscribed in the circle ABCDE .

135 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 .

136 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.

137 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 .

138 But CD belongs to a pentagon; therefore CG belongs to a decagon.

139 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 ]

140 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.

142 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 .

143 But the circumference CD is also double of CG ; therefore the circumference CG is equal to the circumference BK .

144 But BK is double of KM , since KA is so also; therefore CG is also double of KM .

145 But, further, the circumference CB is also double of the circumference BK , for the circumference CB is equal to BA .

146 Therefore the whole circumference GB is also double of BM ; hence the angle GFB is also double of the angle BFM . [ VI. 33 ]

147 But the angle GFB is also double of the angle FAB , for the angle FAB is equal to the angle ABF .

148 Therefore the angle BFN is also equal to the angle FAB .

149 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 .

150 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 ]

151 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 .

152 But the angle LAN is equal to the angle KBN ; therefore the angle LKN is also equal to the angle KBN .

153 And the angle at A is common to the two triangles AKB and AKN .

154 Therefore the remaining angle AKB is equal to the remaining angle KNA ; [ I. 32 ] therefore the triangle KBA is equiangular with the triangle KNA .

155 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 ]

156 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 .

157 And BA is a side of the pentagon, BF of the hexagon [ IV. 15, Por. ], and AK of the decagon.

PROPOSITION 11.

159 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.

160 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.

161 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 .

163 But BF is also rational; therefore the whole BK is rational.

164 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 .

165 And, if we join AD , we conclude that the angles at L are right, and CD is double of CL .

166 For the same reason the angles at M are also right, and AC is double of CM .

167 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 .

168 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 .

169 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 .

170 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 .

171 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 .

172 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 .

173 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 .

174 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 .

175 But the square on KF is rational, for the diameter is rational; therefore the square on MK is also rational; therefore MK is rational

176 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 .

177 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 ]

179 Therefore BK , KM are rational straight lines commensurable in square only.

180 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 ]

182 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 .

183 And, since KF is commensurable with FB , componendo also, KB is commensurable with FB . [ X. 15 ]

184 But BF is commensurable with BH ; therefore BK is also commensurable with BH . [ X. 12 ]

185 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.

186 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 .

187 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 ]

188 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 ]

189 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 .

190 Therefore the side AB of the pentagon is the irrational straight line called minor. Q. E. D.

PROPOSITION 12.

191 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.

192 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.

193 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.

194 Then, since the triangle ABC is equilateral, therefore the circumference BEC is a third part of the circumference of the circle ABC .

195 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. ]

196 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 .

197 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 .

198 Therefore, separando , the square on AB is triple of the square on BE .

199 But BE is equal to DE ; therefore the square on AB is triple of the square on DE .

200 Therefore the square on the side of the triangle is triple of the square on the radius. Q. E. D.

PROPOSITION 13.

201 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.

202 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.

203 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 ]

204 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 .

205 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 ]

206 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.

207 And, since AC is double of CB , therefore AB is triple of BC .

208 But, as AB is to BC , so is the square on AD to the square on DC , as will be proved afterwards.

209 Therefore the square on AD is triple of the square on DC .

210 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 .

211 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.

212 Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex.

213 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.

214 For let the straight line HL be produced in a straight line with KH , and let HL be made equal to CB .

215 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 ]

216 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 .]

217 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.

218 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 .

219 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

220 For, since AC is double of CB , therefore AB is triple of BC ; and, convertendo , BA is one and a half times AC .

221 But, as BA is to AC , so is the square on BA to the square on AD .

222 Therefore the square on BA is also one and a half times the square on AD .

223 And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid.

224 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.

225 It is to be proved that, as AB is to BC , so is the square on AD to the square on DC .

226 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.

227 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 ]

228 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 .

229 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. ]

230 Therefore, as AB is to BC , so is the square on AD to the square on DC . Q. E. D.

PROPOSITION 14.

231 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.

232 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.

233 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 ]

234 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. ]

235 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 .

236 For the same reason LH is also equal to HE ; therefore the triangle LEH is equilateral.

237 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.

238 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.

239 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 .

240 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.

241 I say next that it is also comprehended in the given sphere.

242 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 ]

243 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 ]

244 Again, since AC is equal to CB , AB is double of BC .

245 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 .

246 But the square on LM was also proved double of the square on LE .

247 And the square on DB is equal to the square on LE , for EH was made equal to DB .

248 Therefore the square on AB is also equal to the square on LM ; therefore AB is equal to LM .

249 And AB is the diameter of the given sphere; therefore LM is equal to the diameter of the given sphere.

250 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.

251 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.

252 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.

253 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.

255 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 .

256 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 .

257 Similarly it will also pass through the remaining angular points of the cube.

258 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.

259 I say next that it is also comprehended in the given sphere.

260 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 .

261 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 .

262 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 .

263 But the square on GK was also proved triple of the square on KE .

264 And KE was made equal to DB ; therefore KG is also equal to AB .

265 And AB is the diameter of the given sphere; therefore KG is also equal to the diameter of the given sphere.

266 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.

267 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.

268 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.

269 Therefore the pentagon LMNOP is also equilateral, and the straight line EP belongs to a decagon.

270 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.

271 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 ]

272 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 ]

274 But EK belongs to an equilateral pentagon; therefore QU also belongs to the equilateral pentagon inscribed in the circle EFGHK .

275 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.

276 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 ]

277 For the same reason PU is also a side of a pentagon.

278 But QU also belongs to a pentagon; therefore the triangle QPU is equilateral.

279 For the same reason each of the triangles QLR , RMS , SNT , TOU is also equilateral.

280 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.

281 For the same reason each of the triangles LRM , MSN , NTO , OUP is also equilateral.

282 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.

283 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 ]

284 But they are also equal; therefore EV , QW are also equal and parallel. [ I. 33 ]

285 But EV belongs to a hexagon; therefore QW also belongs to a hexagon.

286 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 ]

287 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.

288 But WZ belongs to a decagon, and the angle UWZ is right; therefore UZ belongs to a pentagon. [ XIII. 10 ]

289 But QU also belongs to a pentagon; therefore the triangle QUZ is equilateral.

290 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.

291 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 ]

292 For the same reason, if we join MV , which belongs to a hexagon, MX is also inferred to belong to a pentagon.

293 But LM also belongs to a pentagon; therefore the triangle LMX is equilateral.

294 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.

295 Therefore an icosahedron has been constructed which is contained by twenty equilateral triangles.

296 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.

297 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 .

298 But VW is equal to VE , and WZ to VX ; therefore, as ZV is to VE , so is EV to VX .

299 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 .

300 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 .

301 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 ]

302 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.

303 I say next that it is also comprehended in the given sphere.

305 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 .

306 And ZX is double of ZA' , and VW double of ; therefore the square on ZX is five times the square on WV .

307 And, since AC is quadruple of CB , therefore AB is five times BC .

308 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 .

309 But the square on ZX was also proved to be five times the square on VW .

310 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 .

311 And AB is the diameter of the given sphere; therefore XZ is also equal to the diameter of the given sphere.

312 Therefore the icosahedron has been comprehended in the given sphere

313 I say next that the side of the icosahedron is the irrational straight line called minor.

314 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.

315 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 ]

316 And the side of the pentagon EFGHK is the side of the icosahedron.

317 Therefore the side of the icosahedron is the irrational straight line called minor.

PORISM.

318 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.

319 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.

320 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.

321 I say that the pentagon UBWCV is equilateral, and in one plane, and is further equiangular.

323 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 ]

324 But PN is equal to NB , and PR to RU ; therefore the squares on BN , NR are triple of the square on RU .

325 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 .

326 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 .

327 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 .

328 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 .

331 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.

332 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 .

333 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 .

334 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 .]

335 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 .

336 But every straight line is in one plane; [ XI. 1 ] therefore the pentagon UBWCV is in one plane.

338 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 ]

339 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 .

340 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 .

341 But BC is also double of BN ; therefore BV is equal to BC .

342 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 ]

343 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.

344 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.

345 And it was also proved equilateral; therefore the pentagon BUVCW is equilateral and equiangular, and it is on one side BC of the cube.

346 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.

347 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.

348 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 ]

349 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.

351 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 ]

352 But NS is equal to XZ , inasmuch as NP is also equal to PZ , and XP to PS .

353 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 .

354 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 .

355 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 ]

356 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.

357 And Z is the centre of the sphere which comprehends the cube; therefore the point U is on the surface of the sphere.

358 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.

359 I say next that the side of the dodecahedron is the irrational straight line called apotome.

360 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.

361 [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 .

362 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.]

363 But RS is equal to UV ; therefore, when NO is cut in extreme and mean ratio, UV is the greater segment.

364 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.

365 [But if a rational line be cut in extreme and mean ratio, each of the segments is an irrational apotome.]

366 Therefore UV , being a side of the dodecahedron, is an irrational apotome. [ XIII. 6 ]

PORISM.

367 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.

368 To set out the sides of the five figures and to compare them with one another.

369 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.

370 Then, since AD is double of DB , therefore AB is triple of BD .

371 Convertendo , therefore, BA is one and a half times AD .

372 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 .

373 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 ]

374 And AB is the diameter of the sphere; therefore AF is equal to the side of the pyramid.

375 Again, since AD is double of DB , therefore AB is triple of BD .

376 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 .

377 But the square on the diameter of the sphere is also triple of the square on the side of the cube. [ XIII. 15 ]

378 And AB is the diameter of the sphere; therefore BF is the side of the cube.

379 And, since AC is equal to CB , therefore AB is double of BC .

380 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 .

381 But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [ XIII. 14 ]

382 And AB is the diameter of the given sphere; therefore BE is the side of the octahedron.

383 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 .

384 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 .

385 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 .

386 But HC is equal to CB ; therefore the square on BC is five times the square on CK .

387 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 .

388 Therefore BC is triple of CD ; therefore the square on BC is nine times the square on CD .

389 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 .

390 Let CL be made equal to CK , from L let LM be drawn at right angles to AB , and let MB be joined.

391 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 .

392 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. ]