If a straight-line is cut in extreme and mean ratio then the square on the greater piece, added to half of the whole, is five times the square on the half.
For let the straight-line AB have been cut in extreme and mean ratio at point C, and let AC be the greater piece.
And let the straight-line AD have been produced in a straight-line with CA.
And let AD be made (equal to) half of AB.
I say that the (square) on CD is five times the (square) on DA.
For let the squares AE and DF have been described on AB and DC (respectively).
And let the figure in DF have been drawn.
And let FC have been drawn across to G.
And since AB has been cut in extreme and mean ratio at C, the (rectangle contained) by AB and BC is thus equal to the (square) on AC [Def. 6.3] [Prop. 6.17].
And CE is the (rectangle contained) by AB and BC, and FH the (square) on AC.
Thus, CE (is) equal to FH.
And since BA is double AD, and BA (is) equal to KA, and AD to AH, KA (is) thus also double AH.
And as KA (is) to AH, so CK (is) to CH [Prop. 6.1].
Thus, CK (is) double CH.
And LH plus HC is also double CH [Prop. 1.43].
Thus, KC (is) equal to LH plus HC.
And CE was also shown (to be) equal to HF.
Thus, the whole square AE is equal to the gnomon MNO.
And since BA is double AD, the (square) on BA is four times the (square) on AD —that is to say, AE (is four times) DH.
And AE (is) equal to gnomon MNO.
And, thus, gnomon MNO is also four times AP.
Thus, the whole of DF is five times AP.
And DF is the (square) on DC, and AP the (square) on DA.
Thus, the (square) on CD is five times the (square) on DA.
Thus, if a straight-line is cut in extreme and mean ratio then the square on the greater piece, added to half of the whole, is five times the square on the half.
(Which is) the very thing it was required to show.
Proposition 2
If the square on a straight-line is five times the (square) on a piece of it, and double the aforementioned piece is cut in extreme and mean ratio, then the greater piece 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 piece of it, AC.
And let CD be double AC.
I say that if CD is cut in extreme and mean ratio then the greater piece is CB.
For let the squares AF and CG have been described on each of AB and CD (respectively).
And let the figure in AF have been drawn.
And let BE have been drawn across.
And since the (square) on BA is five times the (square) on AC, AF is five times AH.
Thus, gnomon MNO (is) four times AH.
And since DC is double CA, the (square) on DC is thus four times the (square) on CA —that is to say, CG (is four times) AH.
And the gnomon MNO was also shown (to be) four times AH.
Thus, gnomon MNO (is) equal to CG.
And since DC is double CA, and DC (is) equal to CK, and AC to CH, [ KC (is) thus also double CH ], (and) KB (is) also double BH [Prop. 6.1].
And LH plus HB is also double HB [Prop. 1.43].
Thus, KB (is) equal to LH plus HB.
And the whole gnomon MNO was also shown (to be) equal to the whole of CG.
Thus, the remainder HF is also equal to (the remainder) BG.
And BG is the (rectangle contained) by CD and DB.
For CD (is) equal to DG.
And HF (is) the square on CB.
Thus, the (rectangle contained) by CD and DB is equal to the (square) on CB.
Thus, as DC is to CB, so CB (is) to BD [Prop. 6.17].
And DC (is) greater than CB (see lemma).
Thus, CB (is) also greater than BD [Prop. 5.14].
Thus, if the straight-line CD is cut in extreme and mean ratio then the greater piece is CB.
Thus, if the square on a straight-line is five times the (square) on a piece of itself, and double the aforementioned piece is cut in extreme and mean ratio, then the greater piece is the remaining part of the original straight-line. (Which is) the very thing it was required to show.
Lemma
And it can be shown that double AC (i.e., DC) is greater than BC, as follows.
For if (double AC is) not (greater than BC), if possible, let BC be double CA.
Thus, the (square) on BC (is) four times the (square) on CA.
Thus, the (sum of) the (squares) on BC and CA (is) five times the (square) on CA.
And the (square) on BA was assumed (to be) five times the (square) on CA.
Thus, the (square) on BA is equal to the (sum of) the (squares) on BC and CA.
The very thing (is) impossible [Prop. 2.4].
Thus, CB is not double AC.
So, similarly, we can show that a (straight-line) less than CB is not double AC either.
For (in this case) the absurdity is much [greater].
Thus, double AC is greater than CB.
(Which is) the very thing it was required to show.
Proposition 3
If a straight-line is cut in extreme and mean ratio then the square on the lesser piece added to half of the greater piece is five times the square on half of the greater piece.
For let some straight-line AB have been cut in extreme and mean ratio at point C.
And let AC be the greater piece.
And let AC have been cut in half at D.
I say that the (square) on BD is five times the (square) on DC.
For let the square AE have been described on AB.
And let the figure have been drawn double.
Since AC is double DC, the (square) on AC (is) thus four times the (square) on DC —that is to say, RS (is four times) FG.
And since the (rectangle contained) by AB and BC is equal to the (square) on AC [Def. 6.3] [Prop. 6.17], and CE is the (rectangle contained) by AB and BC, CE is thus equal to RS.
And RS (is) four times FG.
Thus, CE (is) also four times FG.
Again, since AD is equal to DC, HK is also equal to KF.
Hence, square GF is also equal to square HL.
Thus, GK (is) equal to KL ---that is to say, MN to NE.
Hence, MF is also equal to FE.
But, MF is equal to CG.
Thus, CG is also equal to FE.
Let CN have been added to both.
Thus, gnomon OPQ is equal to CE.
But, CE was shown (to be) equal to four times GF.
Thus, gnomon OPQ is also four times square FG.
Thus, gnomon OPQ plus square FG is five times FG.
But, gnomon OPQ plus square FG is (square) DN.
And DN is the (square) on DB, and GF the (square) on DC.
Thus, the (square) on DB is five times the (square) on DC.
(Which is) the very thing it was required to show.
Proposition 4
If a straight-line is cut in extreme and mean ratio then the sum of the squares on the whole and the lesser piece is three times the square on the greater piece.
Let AB be a straight-line, and let it have been cut in extreme and mean ratio at C, and let AC be the greater piece.
I say that the (sum of the squares) on AB and BC is three times the (square) on CA.
For let the square ADEB have been described on AB, and let the (remainder of the) figure have been drawn.
Therefore, since AB has been cut in extreme and mean ratio at C, and AC is the greater piece, the (rectangle contained) by AB and BC is thus equal to the (square) on AC [Def. 6.3] [Prop. 6.17].
And AK is the (rectangle contained) by AB and BC, and HG the (square) on AC.
Thus, AK is equal to HG.
And since AF is equal to FE [Prop. 1.43], let CK have been added to both.
Thus, the whole of AK is equal to the whole of CE.
Thus, AK plus CE is double AK.
But, AK plus CE is the gnomon LMN plus the square CK.
Thus, gnomon LMN plus square CK is double AK.
But, indeed, AK was also shown (to be) equal to HG.
Thus, gnomon LMN plus [square CK is double HG.
Hence, gnomon LMN plus] the squares CK and HG is three times the square HG.
And gnomon LMN plus the squares CK and HG is the whole of AE plus CK —which are the squares on AB and BC (respectively)—and GH (is) the square on AC.
Thus, the (sum of the) squares on AB and BC is three times the square on AC.
(Which is) the very thing it was required to show.
Proposition 5
If a straight-line is cut in extreme and mean ratio, and a (straight-line) equal to the greater piece is added to it, then the whole straight-line has been cut in extreme and mean ratio, and the original straight-line is the greater piece.
For let the straight-line AB have been cut in extreme and mean ratio at point C.
And let AC be the greater piece.
And let AD be [made] equal to AC.
I say that the straight-line DB has been cut in extreme and mean ratio at A, and that the original straight-line AB is the greater piece.
For let the square AE have been described on AB, and let the (remainder of the) figure have been drawn.
And since AB has been cut in extreme and mean ratio at C, the (rectangle contained) by AB and BC is thus equal to the (square) on AC [Def. 6.3] [Prop. 6.17].
And CE is the (rectangle contained) by AB and BC, and CH the (square) on AC.
But, HE is equal to CE [Prop. 1.43], and DH equal to HC.
Thus, DH is also equal to HE.
[Let HB have been added to both.]
Thus, the whole of DK is equal to the whole of AE.
And DK is the (rectangle contained) by BD and DA.
For AD (is) equal to DL.
And AE (is) the (square) on AB.
Thus, the (rectangle contained) by BD and DA is equal to the (square) on AB.
Thus, as DB (is) to BA, so BA (is) to AD [Prop. 6.17].
And DB (is) greater than BA.
Thus, BA (is) also greater than AD [Prop. 5.14].
Thus, DB has been cut in extreme and mean ratio at A, and the greater piece is AB.
(Which is) the very thing it was required to show.
Proposition 6
If a rational straight-line is cut in extreme and mean ratio then each of the pieces is that irrational (straight-line) called an apotome.
Let AB be a rational straight-line cut in extreme and mean ratio at C, and let AC be the greater piece.
I say that AC and CB is each that irrational (straight-line) called an apotome.
For let BA have been produced, and let AD be made (equal) to half of BA.
Therefore, since the straight-line AB has been cut in extreme and mean ratio at C, and AD, which is half of AB, has been added to the greater piece AC, the (square) on CD is thus five times the (square) on DA [Prop. 13.1].
Thus, the (square) on CD has to the (square) on DA the ratio which a number (has) to a number.
The (square) on CD (is) thus commensurable with the (square) on DA [Prop. 10.6].
And the (square) on DA (is) rational.
For DA [is] rational, being half of AB, which is rational.
Thus, the (square) on CD (is) also rational [Def. 10.4].
Thus, CD is also rational.
And since the (square) on CD does not have to the (square) on DA the ratio which a square number (has) to a square number, CD (is) thus incommensurable in length with DA [Prop. 10.9].
Thus, CD and DA are rational (straight-lines which are) commensurable in square only.
Thus, AC is an apotome [Prop. 10.73].
Again, since AB has been cut in extreme and mean ratio, and AC is the greater piece, the (rectangle contained) by AB and BC is thus equal to the (square) on AC [Def. 6.3] [Prop. 6.17].
Thus, the (square) on the apotome AC, applied to the rational (straight-line) AB, makes BC as width.
And the (square) on an apotome, applied to a rational (straight-line), makes a first apotome as width [Prop. 10.97].
Thus, CB is a first apotome.
And CA was also shown (to be) an apotome.
Thus, if a rational straight-line is cut in extreme and mean ratio then each of the pieces is that irrational (straight-line) called an apotome.
Proposition 7
If three angles, either consecutive or not consecutive, of an equilateral pentagon are equal then the pentagon will be equiangular.
For let three angles of the equilateral pentagon ABCDE —first of all, the consecutive (angles) at A, B, and C —-be equal to one another.
I say that pentagon ABCDE is equiangular.
For let AC, BE, and FD have been joined.
And since the two (straight-lines) CB and BA are equal to the two (straight-lines) BA and AE, respectively, and angle CBA is equal to angle BAE, base AC is thus equal to base BE, and triangle ABC equal to triangle ABE, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4], (that is), BCA (equal) to BEA, and ABE to CAB.
And hence side AF is also equal to side BF [Prop. 1.6].
And the whole of AC was also shown (to be) equal to the whole of BE.
Thus, the remainder FC is also equal to the remainder FE.
And CD is also equal to DE.
So, the two (straight-lines) FC and CD are equal to the two (straight-lines) FE and ED (respectively).
And FD is their common base.
Thus, angle FCD is equal to angle FED [Prop. 1.8].
And BCA was also shown (to be) equal to AEB.
And thus the whole of BCD (is) equal to the whole of AED.
But, (angle) BCD was assumed (to be) equal to the angles at A and B.
Thus, (angle) AED is also equal to the angles at A and B.
So, similarly, we can show that angle CDE is also equal to the angles at A, B, C.
Thus, pentagon ABCDE is equiangular.
And so let consecutive angles not be equal, but let the (angles) at points A, C, and D be equal.
I say that pentagon ABCDE is also equiangular in this case.
For let BD have been joined.
And since the two (straight-lines) BA and AE are equal to the (straight-lines) BC and CD, and they contain equal angles, base BE is thus equal to base BD, and triangle ABE is equal to triangle BCD, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4].
Thus, angle AEB is equal to (angle) CDB.
And angle BED is also equal to (angle) BDE, since side BE is also equal to side BD [Prop. 1.5].
Thus, the whole angle AED is also equal to the whole (angle) CDE.
But, (angle) CDE was assumed (to be) equal to the angles at A and C.
Thus, angle AED is also equal to the (angles) at A and C.
So, for the same (reasons), (angle) ABC is also equal to the angles at A, C, and D.
Thus, pentagon ABCDE is equiangular.
(Which is) the very thing it was required to show.
Proposition 8
If straight-lines subtend two consecutive angles of an equilateral and equiangular pentagon then they cut one another in extreme and mean ratio, and their greater pieces are equal to the sides of the pentagon.
For let the two straight-lines, AC and BE, cutting one another at point H, have subtended two consecutive angles, at A and B (respectively), of the equilateral and equiangular pentagon ABCDE.
I say that each of them has been cut in extreme and mean ratio at point H, and that their greater pieces are equal to the sides of the pentagon.
For let the circle ABCDE have been circumscribed about pentagon ABCDE [Prop. 4.14].
And since the two straight-lines EA and AB are equal to the two (straight-lines) AB and BC (respectively), and they contain equal angles, the base BE is thus equal to the base AC, and triangle ABE is equal to triangle ABC, and the remaining angles will be equal to the remaining angles, respectively, which the equal sides subtend [Prop. 1.4].
And EAC is also double BAC, inasmuch as circumference EDC is also double circumference CB [Prop. 3.28] [Prop. 6.33].
Thus, angle HAE (is) equal to (angle) AHE.
Hence, straight-line HE is also equal to (straight-line) EA —that is to say, to (straight-line) AB [Prop. 1.6].
And since straight-line BA is equal to AE, angle ABE is also equal to AEB [Prop. 1.5].
But, ABE was shown (to be) equal to BAH.
Thus, BEA is also equal to BAH.
And (angle) ABE is common to the two triangles ABE and ABH.
Thus, the remaining angle BAE is equal to the remaining (angle) AHB [Prop. 1.32].
Thus, triangle ABE is equiangular to triangle ABH.
Thus, proportionally, as EB is to BA, so AB (is) to BH [Prop. 6.4].
And BA (is) equal to EH.
Thus, as BE (is) to EH, so EH (is) to HB.
And BE (is) greater than EH.
EH (is) thus also greater than HB [Prop. 5.14].
Thus, BE has been cut in extreme and mean ratio at H, and the greater piece HE is equal to the side of the pentagon.
So, similarly, we can show that AC has also been cut in extreme and mean ratio at H, and that its greater piece CH is equal to the side of the pentagon.
(Which is) the very thing it was required to show.
Proposition 9
If the side of a hexagon and of a decagon inscribed in the same circle are added together then the whole straight-line has been cut in extreme and mean ratio (at the junction point), and its greater piece is the side of the hexagon.
Let ABC be a circle.
And of the figures inscribed in circle ABC, let BC be the side of a decagon, and CD (the side) of a hexagon.
And let them be (laid down) straight-on (to one another).
I say that the whole straight-line BD has been cut in extreme and mean ratio (at C), and that CD is its greater piece.
For let the center of the circle, point E, have been found [Prop. 3.1], and let EB, EC, and ED have been joined, and let BE have been drawn across to A.
Since BC is a side on an equilateral decagon, circumference ACB (is) thus five times circumference BC.
Thus, circumference AC (is) four times CB.
And as circumference AC (is) to CB, so angle AEC (is) to CEB [Prop. 6.33].
Thus, (angle) AEC (is) four times CEB.
And since angle EBC (is) equal to ECB [Prop. 1.5], angle AEC is thus double ECB [Prop. 1.32].
And since straight-line EC is equal to CD —for each of them is equal to the side of the hexagon [inscribed] in circle ABC [Prop. 4.15 corr.] — angle CED is also equal to angle CDE [Prop. 1.5].
Thus, angle ECB (is) double EDC [Prop. 1.32].
But, AEC was shown (to be) double ECB.
Thus, AEC (is) four times EDC.
And AEC was also shown (to be) four times BEC.
Thus, EDC (is) equal to BEC.
And angle EBD (is) common to the two triangles BEC and BED.
Thus, the remaining (angle) BED is equal to the (remaining angle) ECB [Prop. 1.32].
Thus, triangle EBD is equiangular to triangle EBC.
Thus, proportionally, as DB is to BE, so EB (is) to BC [Prop. 6.4].
And EB (is) equal to CD.
Thus, as BD is to DC, so DC (is) to CB.
And BD (is) greater than DC.
Thus, DC (is) also greater than CB [Prop. 5.14].
Thus, the straight-line BD has been cut in extreme and mean ratio [at C ], and DC is its greater piece.
(Which is), the very thing it was required to show.
Proposition 10
If an equilateral pentagon is inscribed in a circle then the square on the side of the pentagon is (equal to) the (sum of the squares) on the (sides) of the hexagon and of the decagon inscribed in the same circle.
Let ABCDE be a circle.
And let the equilateral pentagon ABCDE have been inscribed in circle ABCDE.
I say that the square on the side of pentagon ABCDE is the (sum of the squares) on the sides of the hexagon and of the decagon inscribed in circle ABCDE.
For let the center of the circle, point F, have been found [Prop. 3.1].
And, AF being joined, let it have been drawn across to point G.
And let FB have been joined.
And let FH have been drawn from F perpendicular to AB.
And let it have been drawn across to K.
And let AK and KB have been joined.
And, again, let FL have been drawn from F perpendicular to AK.
And let it have been drawn across to M.
And let KN have been joined.
Since circumference ABCG is equal to circumference AEDG, of which ABC is equal to AED, the remaining circumference CG is thus equal to the remaining (circumference) GD.
And CD (is the side) of the pentagon.
CG (is) thus (the side) of the decagon.
And since FA is equal to FB, and FH is perpendicular (to AB), angle AFK (is) thus also equal to KFB [Prop. 1.5] [Prop. 1.26].
Hence, circumference AK is also equal to KB [Prop. 3.26].
Thus, circumference AB (is) double circumference BK.
Thus, straight-line AK is the side of the decagon.
So, for the same (reasons, circumference) AK is also double KM.
And since circumference AB is double circumference BK, and circumference CD (is) equal to circumference AB, circumference CD (is) thus also double circumference BK.
And circumference CD is also double CG.
Thus, circumference CG (is) equal to circumference BK.
But, BK is double KM, since KA (is) also (double KM).
Thus, (circumference) CG is also double KM.
But, indeed, circumference CB is also double circumference BK.
For circumference CB (is) equal to BA.
Thus, the whole circumference GB is also double BM.
Hence, angle GFB [is] also double angle BFM [Prop. 6.33].
And GFB (is) also double FAB.
For FAB (is) equal to ABF.
Thus, BFN is also equal to FAB.
And angle ABF (is) common to the two triangles ABF and BFN.
Thus, the remaining (angle) AFB is equal to the remaining (angle) BNF [Prop. 1.32].
Thus, triangle ABF is equiangular to triangle BFN.
Thus, proportionally, as straight-line AB (is) to BF, so FB (is) to BN [Prop. 6.4].
Thus, the (rectangle contained) by AB and BN is equal to the (square) on BF [Prop. 6.17].
Again, since AL is equal to LK, and LN is common and at right-angles (to KA), base KN is thus equal to base AN [Prop. 1.4].
And, thus, angle LKN is equal to angle LAN.
But, LAN is equal to KBN [Prop. 3.29] [Prop. 1.5].
Thus, LKN is also equal to KBN.
And the (angle) at A (is) common to the two triangles AKB and AKN.
Thus, the remaining (angle) AKB is equal to the remaining (angle) KNA [Prop. 1.32].
Thus, triangle KBA is equiangular to triangle KNA.
Thus, proportionally, as straight-line BA is to AK, so KA (is) to AN [Prop. 6.4].
Thus, the (rectangle contained) by BA and AN is equal to the (square) on AK [Prop. 6.17].
And the (rectangle contained) by AB and BN was also shown (to be) equal to the (square) on BF.
Thus, the (rectangle contained) by AB and BN plus the (rectangle contained) by BA and AN, which is the (square) on BA [Prop. 2.2], is equal to the (square) on BF plus the (square) on AK.
And BA is the side of the pentagon, and BF (the side) of the hexagon [Prop. 4.15 corr.], and AK (the side) of the decagon.
Thus, the square on the side of the pentagon (inscribed in a circle) is (equal to) the (sum of the squares) on the (sides) of the hexagon and of the decagon inscribed in the same circle.
Proposition 11
If an equilateral pentagon is inscribed in a circle which has a rational diameter then the side of the pentagon is that irrational (straight-line) called minor.
For let the equilateral pentagon ABCDE have been inscribed in the circle ABCDE which has a rational diameter.
I say that the side of pentagon [ ABCDE ] is that irrational (straight-line) called minor.
For let the center of the circle, point F, have been found [Prop. 3.1].
And let AF and FB have been joined.
And let them have been drawn across to points G and H (respectively).
And let AC have been joined.
And let FK made (equal) to the fourth part of AF.
And AF (is) rational.
FK (is) thus also rational.
And BF is also rational.
Thus, the whole of BK is rational.
And since circumference ACG is equal to circumference ADG, of which ABC is equal to AED, the remainder CG is thus equal to the remainder GD.
And if we join AD then the angles at L are inferred (to be) right-angles, and CD (is inferred to be) double CL [Prop. 1.4].
So, for the same (reasons), the (angles) at M are also right-angles, and AC (is) double CM.
Therefore, since angle ALC (is) equal to AMF, and (angle) LAC (is) common to the two triangles ACL and AMF, the remaining (angle) ACL is thus equal to the remaining (angle) MFA [Prop. 1.32].
Thus, triangle ACL is equiangular to triangle AMF.
Thus, proportionally, as LC (is) to CA, so MF (is) to FA [Prop. 6.4].
And (we can take) the doubles of the leading (magnitudes).
Thus, as double LC (is) to CA, so double MF (is) to FA.
And as double MF (is) to FA, so MF (is) to half of FA.
And, thus, as double LC (is) to CA, so MF (is) to half of FA.
And (we can take) the halves of the following (magnitudes).
Thus, as double LC (is) to half of CA, so MF (is) to the fourth of FA.
And DC is double LC, and CM half of CA, and FK the fourth part of FA.
Thus, as DC is to CM, so MF (is) to FK.
Via composition, as the sum of DCM (i.e., DC and CM) (is) to CM, so MK (is) to KF [Prop. 5.18].
And, thus, as the (square) on the sum of DC and CM (is) to the (square) on CM, so the (square) on MK (is) to the (square) on KF.
And since the greater piece of a (straight-line) subtending two sides of a pentagon, such as AC, (which is) cut in extreme and mean ratio is equal to the side of the pentagon [Prop. 13.8] --- that is to say, to DC --- and the square on the greater piece added to half of the whole is five times the (square) on half of the whole [Prop. 13.1], and CM (is) half of the whole, AC, thus the (square) on DC and CM, (taken) as one, is five times the (square) on CM.
And the (square) on DC and CM, (taken) as one, (is) to the (square) on CM, so the (square) on MK was shown (to be) to the (square) on KF.
Thus, the (square) on MK (is) five times the (square) on KF.
And the square on KF (is) rational.
For the diameter (is) rational.
Thus, the (square) on MK (is) also rational.
Thus, MK is rational [in square only].
And since BF is four times FK, BK is thus five times KF.
Thus, the (square) on BK (is) twenty-five times the (square) on KF.
And the (square) on MK (is) five times the square on KF.
Thus, the (square) on BK (is) five times the (square) on KM.
Thus, the (square) on BK does not have to the (square) on KM the ratio which a square number (has) to a square number.
Thus, BK is incommensurable in length with KM [Prop. 10.9].
And each of them is a rational (straight-line).
Thus, BK and KM are rational (straight-lines which are) commensurable in square only.
And if from a rational (straight-line) a rational (straight-line) is subtracted, which is commensurable in square only with the whole, then the remainder is that irrational (straight-line called) an apotome [Prop. 10.73].
Thus, MB is an apotome, and MK its attachment.
So, I say that (it is) also a fourth (apotome).
So, let the (square) on N be (made) equal to that (magnitude) by which the (square) on BK is greater than the (square) on KM.
Thus, the square on BK is greater than the (square) on KM by the (square) on N.
And since KF is commensurable (in length) with FB then, via composition, KB is also commensurable (in length) with FB [Prop. 10.15].
But, BF is commensurable (in length) with BH.
Thus, BK is also commensurable (in length) with BH [Prop. 10.12].
And since the (square) on BK is five times the (square) on KM, the (square) on BK thus has to the (square) on KM the ratio which 5 (has) to one.
Thus, via conversion, the (square) on BK has to the (square) on N the ratio which 5 (has) to 4 [Prop. 5.19 corr.], which is not (that) of a square (number) to a square (number).
BK is thus incommensurable (in length) with N [Prop. 10.9].
Thus, the square on BK is greater than the (square) on KM by the (square) on (some straight-line which is) incommensurable (in length) with ( BK).
Therefore, since the square on the whole, BK, is greater than the (square) on the attachment, KM, by the (square) on (some straight-line which is) incommensurable (in length) with ( BK), and the whole, BK, is commensurable (in length) with the (previously) laid down rational (straight-line) BH, MB is thus a fourth apotome [Def. 10.14].
And the rectangle contained by a rational (straight-line) and a fourth apotome is irrational, and its square-root is that irrational (straight-line) called minor [Prop. 10.94].
And the square on AB is the rectangle contained by HB and BM, on account of joining AH, (so that) triangle ABH becomes equiangular with triangle ABM [Prop. 6.8], and (proportionally) as HB is to BA, so AB (is) to BM.
Thus, the side AB of the pentagon is that irrational (straight-line) called minor.
(Which is) the very thing it was required to show.
Proposition 12
If an equilateral triangle is inscribed in a circle then the square on the side of the triangle is three times the (square) on the radius of the circle.
Let there be a circle ABC, and let the equilateral triangle ABC have been inscribed in it [Prop. 4.2].
I say that the square on one side of triangle ABC is three times the (square) on the radius of circle ABC.
For let the center, D, of circle ABC have been found [Prop. 3.1].
And AD (being) joined, let it have been drawn across to E.
And let BE have been joined.
And since triangle ABC is equilateral, circumference BEC is thus the third part of the circumference of circle ABC.
Thus, circumference BE is the sixth part of the circumference of the circle.
Thus, straight-line BE is (the side) of a hexagon.
Thus, it is equal to the radius DE [Prop. 4.15 corr.].
And since AE is double DE, the (square) on AE is four times the (square) on ED ---that is to say, of the (square) on BE.
And the (square) on AE (is) equal to the (sum of the squares) on AB and BE [Prop. 3.31] [Prop. 1.47].
Thus, the (sum of the squares) on AB and BE is four times the (square) on BE.
Thus, via separation, the (square) on AB is three times the (square) on BE.
And BE (is) equal to DE.
Thus, the (square) on AB is three times the (square) on DE.
Thus, the square on the side of the triangle is three times the (square) on the radius [of the circle].
(Which is) the very thing it was required to show.
Proposition 13
To construct a (regular) pyramid (i.e., a tetrahedron), and to enclose (it) in a given sphere, and to show 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 laid out, and let it have been cut at point C such that AC is double CB [Prop. 6.10].
And let the semi-circle ADB have been drawn on AB.
And let CD have been drawn from point C at right-angles to AB.
And let DA have been joined.
And let the circle EFG be laid down having a radius equal to DC, and let the equilateral triangle EFG have been inscribed in circle EFG [Prop. 4.2].
And let the center of the circle, point H, have been found [Prop. 3.1].
And let EH, HF, and HG have been joined.
And let HK have been set up, at point H, at right-angles to the plane of circle EFG [Prop. 11.12].
And let HK, equal to the straight-line AC, have been cut off from HK.
And let KE, KF, and KG have been joined.
And since KH is at right-angles to the plane of circle EFG, it will thus also make right-angles with all of the straight-lines joining it (which are) also in the plane of circle EFG [Def. 11.3].
And HE, HF, and HG each join it.
Thus, HK is at right-angles to each of HE, HF, and HG.
And since AC is equal to HK, and CD to HE, and they contain right-angles, the base DA is thus equal to the base KE [Prop. 1.4].
So, for the same (reasons), KF and KG is each equal to DA.
Thus, the three (straight-lines) KE, KF, and KG are equal to one another.
And since AC is double CB, AB (is) thus triple BC.
And as AB (is) to BC, so the (square) on AD (is) to the (square) on DC, as will be shown later [see lemma].
Thus, the (square) on AD (is) three times the (square) on DC.
And the (square) on FE is also three times the (square) on EH [Prop. 13.12], and DC is equal to EH.
Thus, DA (is) also equal to EF.
But, DA was shown (to be) equal to each of KE, KF, and KG.
Thus, EF, FG, and GE are equal to KE, KF, and KG, respectively.
Thus, the four triangles EFG, KEF, KFG, and KEG are equilateral.
Thus, a pyramid, whose base is triangle EFG, and apex the point K, has been constructed from four equilateral triangles.
So, it is also necessary to enclose it in the given sphere, and to show 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 have been produced in a straight-line with KH, and let HL be made equal to CB.
And since as AC (is) to CD, so CD (is) to CB [Prop. 6.8 corr.], and AC (is) equal to KH, and CD to HE, and CB to HL, thus as KH is to HE, so EH (is) to HL.
Thus, the (rectangle contained) by KH and HL is equal to the (square) on EH [Prop. 6.17].
And each of the angles KHE and EHL is a right-angle.
Thus, the semi-circle drawn on KL will also pass through E [inasmuch as if we join EL then the angle LEK becomes a right-angle, on account of triangle ELK becoming equiangular to each of the triangles ELH and EHK [Prop. 6.8] [Prop. 3.31] ].
So, if KL remains (fixed), and the semi-circle is carried around, and again established at the same (position) from which it began to be moved, it will also pass through points F and G, (because) if FL and LG are joined, the angles at F and G will similarly become right-angles.
And the pyramid will have been enclosed by the given sphere.
For the diameter, KL, 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.
So, I say 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 CB, AB is thus triple BC.
Thus, via conversion, BA is one and a half times AC.
And as BA (is) to AC, so the (square) on BA (is) to the (square) on AD [inasmuch as if DB is joined then as BA is to AD, so DA (is) to AC, on account of the similarity of triangles DAB and DAC.
And as the first is to the third (of four proportional magnitudes), so the (square) on the first (is) to the (square) on the second.] Thus, 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.
Thus, the square on the diameter of the sphere is one and a half times the (square) on the side of the pyramid.
(Which is) the very thing it was required to show.
Lemma
It must be shown that as AB is to BC, so the (square) on AD (is) to the (square) on DC.
For, let the figure of the semi-circle have been set out, and let DB have been joined.
And let the square EC have been described on AC.
And let the parallelogram FB have been completed.
Therefore, since, on account of triangle DAB being equiangular to triangle DAC [Prop. 6.8] [Prop. 6.4], (proportionally) as BA is to AD, so DA (is) to AC, the (rectangle contained) by BA and AC is thus equal to the (square) on AD [Prop. 6.17].
And since as AB is to BC, so EB (is) to BF [Prop. 6.1].
And EB is the (rectangle contained) by BA and AC —for EA (is) equal to AC.
And BF the (rectangle contained) by AC and CB.
Thus, as AB (is) to BC, so the (rectangle contained) by BA and AC (is) to the (rectangle contained) by AC and CB.
And the (rectangle contained) by BA and AC is equal to the (square) on AD, and the (rectangle contained) by AC and CB (is) equal to the (square) on DC.
For the perpendicular DC is the mean proportional to the pieces of the base, AC and CB, on account of ADB being a right-angle [Prop. 6.8 corr.].
Thus, as AB (is) to BC, so the (square) on AD (is) to the (square) on DC.
(Which is) the very thing it was required to show.
Proposition 14
To construct an octahedron, and to enclose (it) in a (given) sphere, like in the preceding (proposition), and to show that the square on the diameter of the sphere is double the (square) on the side of the octahedron.
Let the diameter AB of the given sphere be laid out, and let it have been cut in half at C.
And let the semi-circle ADB have been drawn on AB.
And let CD be drawn from C at right-angles to AB.
And let DB have been joined.
And let the square EFGH, having each of its sides equal to DB, be laid out.
And let HF and EG have been joined.
And let the straight-line KL have been set up, at point K, at right-angles to the plane of square EFGH [Prop. 11.12].
And let it have been drawn across on the other side of the plane, like KM.
And let KL and KM, equal to one of EK, FK, GK, and HK, have been cut off from KL and KM, respectively.
And let LE, LF, LG, LH, ME, MF, MG, and MH have been joined.
And since KE is equal to KH, and angle EKH is a right-angle, the (square) on the HE is thus double the (square) on EK [Prop. 1.47].
Again, since LK is equal to KE, and angle LKE is a right-angle, the (square) on EL is thus double the (square) on EK [Prop. 1.47].
And the (square) on HE was also shown (to be) double the (square) on EK.
Thus, the (square) on LE is equal to the (square) on EH.
Thus, LE is equal to EH.
So, for the same (reasons), LH is also equal to HE.
Triangle LEH is thus equilateral.
So, similarly, we can show that each of the remaining triangles, whose bases are the sides of the square EFGH, and apexes the points L and M, are equilateral.
Thus, an octahedron contained by eight equilateral triangles has been constructed.
So, it is also necessary to enclose it by the given sphere, and to show that the square on the diameter of the sphere is double the (square) on the side of the octahedron.
For since the three (straight-lines) LK, KM, and KE are equal to one another, the semi-circle drawn on LM will thus also pass through E.
And, for the same (reasons), if LM remains (fixed), and the semi-circle is carried around, and again established at the same (position) from which it began to be moved, then it will also pass through points F, G, and H, and the octahedron will have been enclosed by a sphere.
So, I say that (it is) also (enclosed) by the given (sphere).
For since LK is equal to KM, and KE (is) common, and they contain right-angles, the base LE is thus equal to the base EM [Prop. 1.4].
And since angle LEM is a right-angle—for (it is) in a semi-circle [Prop. 3.31] —the (square) on LM is thus double the (square) on LE [Prop. 1.47].
Again, since AC is equal to CB, AB is double BC.
And as AB (is) to BC, so the (square) on AB (is) to the (square) on BD [Prop. 6.8] [Def. 5.9].
Thus, the (square) on AB is double the (square) on BD.
And the (square) on LM was also shown (to be) double the (square) on LE.
And the (square) on DB is equal to the (square) on LE.
For EH was made equal to DB.
Thus, the (square) on AB (is) also equal to the (square) on LM.
Thus, AB (is) equal to LM.
And AB is the diameter of the given sphere.
Thus, LM is equal to the diameter of the given sphere.
Thus, the octahedron has been enclosed by the given sphere, and it has been simultaneously proved that the square on the diameter of the sphere is double the (square) on the side of the octahedron.
(Which is) the very thing it was required to show.
Proposition 15
To construct a cube, and to enclose (it) in a sphere, like in the (case of the) pyramid, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube.
Let the diameter AB of the given sphere be laid out, and let it have been cut at C such that AC is double CB.
And let the semi-circle ADB have been drawn on AB.
And let CD have been drawn from C at right-angles to AB.
And let DB have been joined.
And let the square EFGH, having (its) side equal to DB, be laid out.
And let EK, FL, GM, and HN have been drawn from (points) E, F, G, and H, (respectively), at right-angles to the plane of square EFGH.
And let EK, FL, GM, and HN, equal to one of EF, FG, GH, and HE, have been cut off from EK, FL, GM, and HN, respectively.
And let KL, LM, MN, and NK have been joined.
Thus, a cube contained by six equal squares has been constructed.
So, it is also necessary to enclose it by the given sphere, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube.
For let KG and EG have been joined.
And since angle KEG is a right-angle—on account of KE also being at right-angles to the plane EG, and manifestly also to the straight-line EG [Def. 11.3] —the semi-circle drawn on KG will thus also pass through point E.
Again, since GF is at right-angles to each of FL and FE, GF is thus also at right-angles to the plane FK [Prop. 11.4].
Hence, if we also join FK then GF will also be at right-angles to FK.
And, again, on account of this, the semi-circle drawn on GK will also pass through point F.
Similarly, it will also pass through the remaining (angular) points of the cube.
So, if KG remains (fixed), and the semi-circle is carried around, and again established at the same (position) from which it began to be moved, then the cube will have been enclosed by a sphere.
So, I say that (it is) also (enclosed) by the given (sphere).
For since GF is equal to FE, and the angle at F is a right-angle, the (square) on EG is thus double the (square) on EF [Prop. 1.47].
And EF (is) equal to EK.
Thus, the (square) on EG is double the (square) on EK.
Hence, the (sum of the squares) on GE and EK —that is to say, the (square) on GK [Prop. 1.47] —is three times the (square) on EK.
And since AB is three times BC, and as AB (is) to BC, so the (square) on AB (is) to the (square) on BD [Prop. 6.8] [Def. 5.9], the (square) on AB (is) thus three times the (square) on BD.
And the (square) on GK was also shown (to be) three times the (square) on KE.
And KE was made equal to DB.
Thus, KG (is) also equal to AB.
And AB is the radius of the given sphere.
Thus, KG is also equal to the diameter of the given sphere.
Thus, the cube has been enclosed by the given sphere.
And it has simultaneously been shown that the square on the diameter of the sphere is three times the (square) on the side of the cube.
(Which is) the very thing it was required to show.
Proposition 16
To construct an icosahedron, and to enclose (it) in a sphere, like the aforementioned figures, and to show that the side of the icosahedron is that irrational (straight-line) called minor.
Let the diameter AB of the given sphere be laid out, and let it have been cut at C such that AC is four times CB [Prop. 6.10].
And let the semi-circle ADB have been drawn on AB.
And let the straight-line CD have been drawn from C at right-angles to AB.
And let DB have been joined.
And let the circle EFGHK be set down, and let its radius be equal to DB.
And let the equilateral and equiangular pentagon EFGHK have been inscribed in circle EFGHK [Prop. 4.11].
And let the circumferences EF, FG, GH, HK, and KE have been cut in half at points L, M, N, O, and P (respectively).
And let LM, MN, NO, OP, PL, and EP have been joined.
Thus, pentagon LMNOP is also equilateral, and EP (is) the side of the decagon (inscribed in the circle).
And let the straight-lines EQ, FR, GS, HT, and KU, which are equal to the radius of circle EFGHK, have been set up at right-angles to the plane of the circle, at points E, F, G, H, and K (respectively).
And let QR, RS, ST, TU, UQ, QL, LR, RM, MS, SN, NT, TO, OU, UP, and PQ have been joined.
And since EQ and KU are each at right-angles to the same plane, EQ is thus parallel to KU [Prop. 11.6].
And it is also equal to it.
And straight-lines joining equal and parallel (straight-lines) on the same side are (themselves) equal and parallel [Prop. 1.33].
Thus, QU is equal and parallel to EK.
And EK (is the side) of an equilateral pentagon (inscribed in circle EFGHK).
Thus, QU (is) also the side of an equilateral pentagon inscribed in circle EFGHK.
So, for the same (reasons), QR, RS, ST, and TU are also the sides of an equilateral pentagon inscribed in circle EFGHK.
Pentagon QRSTU (is) thus equilateral.
And side QE is (the side) of a hexagon (inscribed in circle EFGHK), and EP (the side) of a decagon, and (angle) QEP is a right-angle, thus QP is (the side) of a pentagon (inscribed in the same circle).
For the square on the side of a pentagon is (equal to the sum of) the (squares) on (the sides of) a hexagon and a decagon inscribed in the same circle [Prop. 13.10].
So, for the same (reasons), PU is also the side of a pentagon.
And QU is also (the side) of a pentagon.
Thus, triangle QPU is equilateral.
So, for the same (reasons), (triangles) QLR, RMS, SNT, and TOU are each also equilateral.
And since QL and QP were each shown (to be the sides) of a pentagon, and LP is also (the side) of a pentagon, triangle QLP is thus equilateral.
So, for the same (reasons), triangles LRM, MSN, NTO, and OUP are each also equilateral.
Let the center, point V, of circle EFGHK have been found [Prop. 3.1].
And let VZ have been set up, at (point) V, at right-angles to the plane of the circle.
And let it have been produced on the other side (of the circle), like VX.
And let VW have been cut off (from XZ so as to be equal to the side) of a hexagon, and each of VX and WZ (so as to be equal to the side) of a decagon.
And let QZ, QW, UZ, EV, LV, LX, and XM havebeen joined.
And since VW and QE are each at right-angles to the plane of the circle, VW is thus parallel to QE [Prop. 11.6].
And they are also equal.
EV and QW are thus equal and parallel (to one another) [Prop. 1.33].
And EV (is the side) of a hexagon.
Thus, QW (is) also (the side) of a hexagon.
And since QW is (the side) of a hexagon, and WZ (the side) of a decagon, and angle QWZ is a right-angle [Def. 11.3] [Prop. 1.29], QZ is thus (the side) of a pentagon [Prop. 13.10].
So, for the same (reasons), UZ is also (the side) of a pentagon—inasmuch as, if we join VK and WU then they will be equal and opposite.
And VK, being (equal) to the radius (of the circle), is (the side) of a hexagon [Prop. 4.15 corr.].
Thus, WU (is) also the side of a hexagon.
And WZ (is the side) of a decagon, and (angle) UWZ (is) a right-angle.
Thus, UZ (is the side) of a pentagon [Prop. 13.10].
And QU is also (the side) of a pentagon.
Triangle QUZ is thus equilateral.
So, for the same (reasons), each of the remaining triangles, whose bases are the straight-lines QR, RS, ST, and TU, and apexes the point Z, are also equilateral.
Again, since VL (is the side) of a hexagon, and VX (the side) of a decagon, and angle LVX is a right-angle, LX is thus (the side) of a pentagon [Prop. 13.10].
So, for the same (reasons), if we join MV, which is (the side) of a hexagon, MX is also inferred (to be the side) of a pentagon.
And LM is also (the side) of a pentagon.
Thus, triangle LMX is equilateral.
So, similarly, it can be shown that each of the remaining triangles, whose bases are the (straight-lines) MN, NO, OP, and PL, and apexes the point X, are also equilateral.
Thus, an icosahedron contained by twenty equilateral triangles has been constructed.
So, it is also necessary to enclose it in the given sphere, and to show that the side of the icosahedron is that irrational (straight-line) called minor.
For, since VW is (the side) of a hexagon, and WZ (the side) of a decagon, VZ has thus been cut in extreme and mean ratio at W, and VW is its greater piece [Prop. 13.9].
Thus, as ZV is to VW, so VW (is) to WZ.
And VW (is) equal to VE, and WZ to VX.
Thus, as ZV is to VE, so EV (is) to VX.
And angles ZVE and EVX are right-angles.
Thus, if we join straight-line EZ then angle XEZ will be a right-angle, on account of the similarity of triangles XEZ and VEZ [Prop. 6.8].
So, for the same (reasons), since as ZV is to VW, so VW (is) to WZ, and ZV (is) equal to XW, and VW to WQ, thus as XW is to WQ, so QW (is) to WZ.
And, again, on account of this, if we join QX then the angle at Q will be a right-angle [Prop. 6.8].
Thus, the semi-circle drawn on XZ will also pass through Q [Prop. 3.31].
And if XZ remains fixed, and the semi-circle is carried around, and again established at the same (position) from which it began to be moved, then it will also pass through (point) Q, and (through) the remaining (angular) points of the icosahedron.
And the icosahedron will have been enclosed by a sphere.
So, I say that (it is) also (enclosed) by the given (sphere).
For let VW have been cut in half at a.
And since the straight-line VZ has been cut in extreme and mean ratio at W, and ZW is its lesser piece, then the square on ZW added to half of the greater piece, Wa, is five times the (square) on half of the greater piece [Prop. 13.3].
Thus, the (square) on Za is five times the (square) on aW.
And ZX is double Za, and VW double aW.
Thus, the (square) on ZX is five times the (square) on WV.
And since AC is four times CB, AB is thus five times BC.
And as AB (is) to BC, so the (square) on AB (is) to the (square) on BD [Prop. 6.8] [Def. 5.9].
Thus, the (square) on AB is five times the (square) on BD.
And the (square) on ZX was also shown (to be) five times the (square) on VW.
And DB is equal to VW.
For each of them is equal to the radius of circle EFGHK.
Thus, AB (is) also equal to XZ.
And AB is the diameter of the given sphere.
Thus, XZ is equal to the diameter of the given sphere.
Thus, the icosahedron has been enclosed by the given sphere.
So, I say that the side of the icosahedron is that 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 circle EFGHK, the radius of circle EFGHK is thus also rational.
Hence, its diameter is also rational.
And if an equilateral pentagon is inscribed in a circle having a rational diameter then the side of the pentagon is that irrational (straight-line) called minor [Prop. 13.11].
And the side of pentagon EFGHK is (the side) of the icosahedron.
Thus, the side of the icosahedron is that irrational (straight-line) called minor.
Corollary
So, (it is) clear, from this, 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 the diameter of the sphere is the sum of (the side) of the hexagon, and two of (the sides) of the decagon, inscribed in the same circle.
Proposition 17
To construct a dodecahedron, and to enclose (it) in a sphere, like the aforementioned figures, and to show that the side of the dodecahedron is that irrational (straight-line) called an apotome.
Let two planes of the aforementioned cube [Prop. 13.15], ABCD and CBEF, (which are) at right-angles to one another, be laid out.
And let the sides AB, BC, CD, DA, EF, EB, and FC have each been cut in half at points G, H, K, L, M, N, and O (respectively).
And let GK, HL, MH, and NO have been joined.
And let NP, PO, and HQ have each been cut in extreme and mean ratio at points R, S, and T (respectively).
And let their greater pieces be RP, PS, and TQ (respectively).
And let RU, SV, and TW have been set up on the exterior side of the cube, at points R, S, and T (respectively), at right-angles to the planes of the cube.
And let them be made equal to RP, PS, and TQ.
And let UB, BW, WC, CV, and VU have been joined.
I say that the pentagon UBWCV is equilateral, and in one plane, and, further, equiangular.
For let RB, SB, and VB have been joined.
And since the straight-line NP has been cut in extreme and mean ratio at R, and RP is the greater piece, the (sum of the squares) on PN and NR is thus three times the (square) on RP [Prop. 13.4].
And PN (is) equal to NB, and PR to RU.
Thus, the (sum of the squares) on BN and NR is three times the (square) on RU.
And the (square) on BR is equal to the (sum of the squares) on BN and NR [Prop. 1.47].
Thus, the (square) on BR is three times the (square) on RU.
Hence, the (sum of the squares) on BR and RU is four times the (square) on RU.
And the (square) on BU is equal to the (sum of the squares) on BR and RU [Prop. 1.47].
Thus, the (square) on BU is four times the (square) on UR.
Thus, BU is double RU.
And VU is also double UR, inasmuch as SR is also double PR —that is to say, RU.
Thus, BU (is) equal to UV.
So, similarly, it can be shown that each of BW, WC, CV is equal to each of BU and UV.
Thus, pentagon BUVCW is equilateral.
So, I say that it is also in one plane.
For let PX have been drawn from P, parallel to each of RU and SV, on the exterior side of the cube.
And let XH and HW have been 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 piece, thus as HQ is to QT, so QT (is) to TH.
And HQ (is) equal to HP, and QT to each of TW and PX.
Thus, as HP is to PX, so WT (is) to TH.
And HP is parallel to TW.
For of each of them is at right-angles to the plane BD [Prop. 11.6].
And TH (is parallel) to PX.
For each of them is at right-angles to the plane BF [Prop. 11.6].
And if two triangles, like XPH and HTW, having two sides proportional to two sides, are placed together at a single angle such that their corresponding sides are also parallel then the remaining sides will be straight-on (to one another) [Prop. 6.32].
Thus, XH is straight-on to HW.
And every straight-line is in one plane [Prop. 11.1].
Thus, pentagon UBWCV is in one plane.
So, I say 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 piece [thus as the sum of NP and PR is to PN, so NP (is) to PR ], and PR (is) equal to PS [thus as SN is to NP, so NP (is) to PS ], NS has thus also been cut in extreme and mean ratio at P, and NP is the greater piece [Prop. 13.5].
Thus, the (sum of the squares) on NS and SP is three times the (square) on NP [Prop. 13.4].
And NP (is) equal to NB, and PS to SV.
Thus, the (sum of the) squares on NS and SV is three times the (square) on NB.
Hence, the (sum of the squares) on VS, SN, and NB is four times the (square) on NB.
And the (square) on SB is equal to the (sum of the squares) on SN and NB [Prop. 1.47].
Thus, the (sum of the squares) on BS and SV —that is to say, the (square) on BV [for angle VSB (is) a right-angle]—is four times the (square) on NB [Def. 11.3] [Prop. 1.47].
Thus, VB is double BN.
And BC (is) also double BN.
Thus, BV is equal to BC.
And since the two (straight-lines) BU and UV are equal to the two (straight-lines) BW and WC (respectively), and the base BV (is) equal to the base BC, angle BUV is thus equal to angle BWC [Prop. 1.8].
So, similarly, we can show that angle UVC is equal to angle BWC.
Thus, the three angles BWC, BUV, and UVC are equal to one another.
And if three angles of an equilateral pentagon are equal to one another then the pentagon is equiangular [Prop. 13.7].
Thus, pentagon BUVCW is equiangular.
And it was also shown (to be) equilateral.
Thus, pentagon BUVCW is equilateral and equiangular, and it is on one of the sides, BC, of the cube.
Thus, if we make the same construction on each of the twelve sides of the cube then some solid figure contained by twelve equilateral and equiangular pentagons will have been constructed, which is called a dodecahedron.
So, it is necessary to enclose it in the given sphere, and to show that the side of the dodecahedron is that irrational (straight-line) called an apotome.
For let XP have been produced, and let (the produced straight-line) be XZ.
Thus, PZ meets the diameter of the cube, and they cut one another in half.
For, this has been proved in the penultimate theorem of the eleventh book [Prop. 11.38].
Let them cut (one another) at Z.
Thus, Z is the center of the sphere enclosing the cube, and ZP (is) half the side of the cube.
So, let UZ have been joined.
And since the straight-line NS has been cut in extreme and mean ratio at P, and its greater piece is NP, the (sum of the squares) on NS and SP is thus three times the (square) on NP [Prop. 13.4].
And NS (is) equal to XZ, inasmuch as NP is also equal to PZ, and XP to PS.
But, indeed, PS (is) also (equal) to XU, since (it is) also (equal) to RP.
Thus, the (sum of the squares) on ZX and XU is three times the (square) on NP.
And the (square) on UZ is equal to the (sum of the squares) on ZX and XU [Prop. 1.47].
Thus, the (square) on UZ is three times the (square) on NP.
And the square on the radius of the sphere enclosing the cube is also three times the (square) on half the side of the cube.
For it has previously been demonstrated (how to) construct the cube, and to enclose (it) in a sphere, and to show that the square on the diameter of the sphere is three times the (square) on the side of the cube [Prop. 13.15].
And if the (square on the) whole (is three times) the (square on the) whole, then the (square on the) half (is) also (three times) the (square on the) half.
And NP is half of the side of the cube.
Thus, UZ is equal to the radius of the sphere enclosing the cube.
And Z is the center of the sphere enclosing the cube.
Thus, point U is on the surface of the sphere.
So, similarly, we can show that each of the remaining angles of the dodecahedron is also on the surface of the sphere.
Thus, the dodecahedron has been enclosed by the given sphere.
So, I say that the side of the dodecahedron is that irrational straight-line called an apotome.
For since RP is the greater piece of NP, which has been cut in extreme and mean ratio, and PS is the greater piece of PO, which has been cut in extreme and mean ratio, RS is thus the greater piece of the whole of NO, which has been cut in extreme and mean ratio.
[Thus, since as NP is to PR, (so) PR (is) to RN, and (the same is also true) of the doubles.
For parts have the same ratio as similar multiples (taken in corresponding order) [Prop. 5.15].
Thus, as NO (is) to RS, so RS (is) to the sum of NR and SO.
And NO (is) greater than RS.
Thus, RS (is) also greater than the sum of NR and SO [Prop. 5.14].
Thus, NO has been cut in extreme and mean ratio, and RS is its greater piece.]
And RS (is) equal to UV.
Thus, UV is the greater piece of NO, which has been cut in extreme and mean ratio.
And since the diameter of the sphere is rational, and the square on it is three times the (square) on the side of the cube, NO, which is the side of the cube, is thus rational.
And if a rational (straight)-line is cut in extreme and mean ratio then each of the pieces is the irrational (straight-line called) an apotome.
Thus, UV, which is the side of the dodecahedron, is the irrational (straight-line called) an apotome [Prop. 13.6].
Corollary
So, (it is) clear, from this, that the side of the dodecahedron is the greater piece of the side of the cube, when is cut in extreme and mean ratio.
(Which is) the very thing it was required to show.
Proposition 18
To set out the sides of the five (aforementioned) figures, and to compare (them) with one another.
Let the diameter, AB, of the given sphere be laid out.
And let it have been cut at C, such that AC is equal to CB, and at D, such that AD is double DB.
And let the semi-circle AEB have been drawn on AB.
And let CE and DF have been drawn from C and D (respectively), at right-angles to AB.
And let AF, FB, and EB have been joined.
And since AD is double DB, AB is thus triple BD.
Thus, via conversion, BA is one and a half times AD.
And as BA (is) to AD, so the (square) on BA (is) to the (square) on AF [Def. 5.9].
For triangle AFB is equiangular to triangle AFD [Prop. 6.8].
Thus, the (square) on BA is one and a half times the (square) on AF.
And the square on the diameter of the sphere is also one and a half times the (square) on the side of the pyramid [Prop. 13.13].
And AB is the diameter of the sphere.
Thus, AF is equal to the side of the pyramid.
Again, since AD is double DB, AB is thus triple BD.
And as AB (is) to BD, so the (square) on AB (is) to the (square) on BF [Prop. 6.8] [Def. 5.9].
Thus, the (square) on AB is three times the (square) on BF.
And the square on the diameter of the sphere is also three times the (square) on the side of the cube [Prop. 13.15].
And AB is the diameter of the sphere.
Thus, BF is the side of the cube.
And since AC is equal to CB, AB is thus double BC.
And as AB (is) to BC, so the (square) on AB (is) to the (square) on BE [Prop. 6.8] [Def. 5.9].
Thus, the (square) on AB is double the (square) on BE.
And the square on the diameter of the sphere is also double the (square) on the side of the octahedron [Prop. 13.14].
And AB is the diameter of the given sphere.
Thus, BE is the side of the octahedron.
So let AG have been drawn from point A at right-angles to the straight-line AB.
And let AG be made equal to AB.
And let GC have been joined.
And let HK have been drawn from H, perpendicular to AB.
And since GA is double AC.
For GA (is) equal to AB.
And as GA (is) to AC, so HK (is) to KC [Prop. 6.4].
HK (is) thus also double KC.
Thus, the (square) on HK is four times the (square) on KC.
Thus, the (sum of the squares) on HK and KC, which is the (square) on HC [Prop. 1.47], is five times the (square) on KC.
And HC (is) equal to CB.
Thus, the (square) on BC (is) five times the (square) on CK.
And since AB is double CB, of which AD is double DB, the remainder BD is thus double the remainder DC.
BC (is) thus triple CD.
The (square) on BC (is) thus nine times the (square) on CD.
And the (square) on BC (is) five times the (square) on CK.
Thus, the (square) on CK (is) greater than the (square) on CD.
CK is thus greater than CD.
Let CL be made equal to CK.
And let LM have been drawn from L at right-angles to AB.
And let MB have been joined.
And since the (square) on BC is five times the (square) on CK, and AB is double BC, and KL double CK, the (square) on AB is thus five times the (square) on KL.
And 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 [Prop. 13.16 corr.].
And AB is the diameter of the sphere.
Thus, KL is the radius of the circle from which the icosahedron has been described.
Thus, KL is (the side) of the hexagon (inscribed) in the aforementioned circle [Prop. 4.15 corr.].
And since the diameter of the sphere is composed of (the side) of the hexagon, and two of (the sides) of the decagon, inscribed in the aforementioned circle, and AB is the diameter of the sphere, and KL the side of the hexagon, and AK (is) equal to LB, thus AK and LB are each sides of the decagon inscribed in the circle from which the icosahedron has been described.
And since LB is (the side) of the decagon.
And ML (is the side) of the hexagon—for (it is) equal to KL, since (it is) also (equal) to HK, for they are equally far from the center.
And HK and KL are each double KC.
MB is thus (the side) of the pentagon (inscribed in the circle) [Prop. 13.10] [Prop. 1.47].
And (the side) of the pentagon is (the side) of the icosahedron [Prop. 13.16].
Thus, MB is (the side) of the icosahedron.
And since FB is the side of the cube, let it have been cut in extreme and mean ratio at N, and let NB be the greater piece.
Thus, NB is the side of the dodecahedron [Prop. 13.17 corr.].
And since the (square) on the diameter of the sphere was shown (to be) one and a half times the square on the side, AF, of the pyramid, and twice the square on (the side), BE, of the octahedron, and three times the square on (the side), FB, of the cube, thus, of whatever (parts) the (square) on the diameter of the sphere (makes) six, of such (parts) the (square) on (the side) of the pyramid (makes) four, and (the square) on (the side) of the octahedron three, and (the square) on (the side) of the cube two.
Thus, the (square) on the side of the pyramid is one and a third times the square on the side of the octahedron, and double 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.
Therefore, the aforementioned sides of the three figures—I mean, of the pyramid, and of the octahedron, and of the cube— are in rational ratios to one another.
And (the sides of) the remaining two (figures)—I mean, of the icosahedron, and of the dodecahedron—are neither in rational ratios to one another, nor to the (sides) of the aforementioned (three figures).
For they are irrational (straight lines): (namely), a minor [Prop. 13.16], and an apotome [Prop. 13.17].
(And), we can show that the side, MB, of the icosahedron is greater that the (side), NB, or the dodecahedron, as follows.
For, since triangle FDB is equiangular to triangle FAB [Prop. 6.8], proportionally, as DB is to BF, so BF (is) to BA [Prop. 6.4].
And since three straight-lines are (continually) proportional, as the first (is) to the third, so the (square) on the first (is) to the (square) on the second [Def. 5.9] [Prop. 6.20 corr.].
Thus, as DB is to BA, so the (square) on DB (is) to the (square) on BF.
Thus, inversely, as AB (is) to BD, so the (square) on FB (is) to the (square) on BD.
And AB (is) triple BD.
Thus, the (square) on FB (is) three times the (square) on BD.
And the (square) on AD is also four times the (square) on DB.
For AD (is) double DB.
Thus, the (square) on AD (is) greater than the (square) on FB.
Thus, AD (is) greater than FB.
Thus, AL is much greater than FB.
And KL is the greater piece of AL, which is cut in extreme and mean ratio—inasmuch as LK is (the side) of the hexagon, and KA (the side) of the decagon [Prop. 13.9].
And NB is the greater piece of FB, which is cut in extreme and mean ratio.
Thus, KL (is) greater than NB.
And KL (is) equal to LM.
Thus, LM (is) greater than NB [and MB is greater than LM ].
Thus, MB, which is (the side) of the icosahedron, is much greater than NB, which is (the side) of the dodecahedron.
(Which is) the very thing it was required to show.
So, I say that, beside the five aforementioned figures, no other (solid) figure can be constructed (which is) contained by equilateral and equiangular (planes), equal to one another.
For a solid angle cannot be constructed from two triangles, or indeed (two) planes (of any sort) [Def. 11.11].
And (the solid angle) of the pyramid (is constructed) from three (equiangular) triangles, and (that) of the octahedron from four (triangles), and (that) of the icosahedron from (five) triangles.
And a solid angle cannot be (made) from six equilateral and equiangular triangles set up together at one point.
For, since the angles of a equilateral triangle are (each) two-thirds of a right-angle, the (sum of the) six (plane) angles (containing the solid angle) will be four right-angles.
The very thing (is) impossible.
For every solid angle is contained by (plane angles whose sum is) less than four right-angles [Prop. 11.21].
So, for the same (reasons), a solid angle cannot be constructed from more than six plane angles (equal to two-thirds of a right-angle) either.
And the (solid) angle of a cube is contained by three squares.
And (a solid angle contained) by four (squares is) impossible.
For, again, the (sum of the plane angles containing the solid angle) will be four right-angles.
And (the solid angle) of a dodecahedron (is contained) by three equilateral and equiangular pentagons.
And (a solid angle contained) by four (equiangular pentagons is) impossible.
For, the angle of an equilateral pentagon being one and one-fifth of right-angle, four (such) angles will be greater (in sum) than four right-angles.
The very thing (is) impossible.
And, on account of the same absurdity, a solid angle cannot be constructed from any other (equiangular) polygonal figures either.
Thus, beside the five aforementioned figures, no other solid figure can be constructed (which is) contained by equilateral and equiangular (planes).
(Which is) the very thing it was required to show.
Lemma
It can be shown that the angle of an equilateral and equiangular pentagon is one and one-fifth of a right-angle, as follows.
For let ABCDE be an equilateral and equiangular pentagon, and let the circle ABCDE have been circumscribed about it [Prop. 4.14].
And let its center, F, have been found [Prop. 3.1].
And let FA, FB, FC, FD, and FE have been joined.
Thus, they cut the angles of the pentagon in half at (points) A, B, C, D, and E [Prop. 1.4].
And since the five angles at F are equal (in sum) to four right-angles, and are also equal (to one another), (any) one of them, like AFB, is thus one less a fifth of a right-angle.
Thus, the (sum of the) remaining (angles in triangle ABF), FAB and ABF, is one plus a fifth of a right-angle [Prop. 1.32].
And FAB (is) equal to FBC.
Thus, the whole angle, ABC, of the pentagon is also one and one-fifth of a right-angle.
(Which is) the very thing it was required to show.