But HE is equal to CE, and DH is equal to HC; therefore DH is also equal to HE. Therefore the whole DK is equal to the whole AE. for AD is equal to DL; and AE is the square on AB; therefore the rectangle BD, DA is equal to the squar on AB. Therefore, as DB is to BA, so is BA to AD. And DB is greater than BA; therefore BA is also greater than AD. [VI. I [v. I Therefore DB has been cut in extreme and mean ratio a A, and AB is the greater segment. We have Q. E. D. (sq. DH) (sq. HC) = = (rect. CE), by hypothesis, or Add to each side the rectangle AK, and = (rect. DK) (sq. AE), (rect. BD, DA) = (sq. on AB). The result is of course obvious from II. II. There is an alternative proof given in P after XIII. 6, which depends Book v. If a rational straight line be cut in extreme and mean rati each of the segments is the irrational straight line call apotome. H. E. III. 29 straight line called apotome. For let BA be produced, and let AD be made hal Since then the straight line AB has been cut in and mean ratio, and to the greater segment AC is added AD whic of AB, therefore the square on CD is five times the square c Therefore the square on CD has to the square on ratio which a number has to a number; therefore the square on CD is commensurable with th on DA. But the square on DA is rational, for DA is rational, being half of AB which is rationa therefore the square on CD is also rational; therefore CD is also rational. And, since the square on CD has not to the s DA the ratio which a square number has to a square therefore CD is incommensurable in length with DA therefore CD, DA are rational straight lines comme in square only; therefore AC is an apotome. Again, since AB has been cut in extreme and me and AC is the greater segment, therefore the rectangle AB, BC is equal to the squar [vi. De Therefore the square on the apotome AC, if a the rational straight line AB, produces BC as breadt But the square on an apotome, if applied to a straight line, produces as breadth a first apotome ; therefore CB is a first apotome. In the first place, there is a scholium to XIII. 17 in P itself which proves 1 same thing as XIII. 6, and which would therefore have been useless if XIII had preceded. Hence, when the scholium was written, this proposition h not yet been interpolated. Secondly, P has it before the alternative proof XIII. 5; this proof is considered, on general grounds, to be interpolated, a it would appear that it must have been a later interpolation (x111. 6) whi divorced it from the proposition to which it belonged. Thirdly, there is ca for suspicion in the proposition itself, for, while the enunciation states t each segment of the straight line is an apotome, the proposition adds that t lesser segment is a first apotome. The scholium in P referred to has not t blot. What is actually wanted in XIII. 17 is the fact that the greater segme is an apotome. It is probable that Euclid assumed this fact as evident enou from XIII. I without further proof, and that he neither wrote XIII. 6 nor t quotation of its enunciation in XIII. 17. PROPOSITION 7. If three angles of an equilateral pentagon, taken either order or not in order, be equal, the pentagon will be equiangula For in the equilateral pentagon ABCDE let, first, thr angles taken in order, those at A, B, C, be equal to one another; I say that the pentagon ABCDE is equiangular. For let AC, BE, FD be joined. Now, since the two sides CB, BA are equal to the two sides BA, AE respectively, and the angle CBA is equal to the angle BAE, B therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, F A and the remaining angles will be equal to the remaining angl namely those which the equal sides subtend, [1. that is, the angle BCA to the angle BEA, and the ang ABE to the angle CAB; hence the side AF is also equal to the side BF. and the base FD is common to them; therefore the angle FCD is equal to the angle FED. But the angle BCA was also proved equal to the АЕВ; therefore the whole angle BCD is also equal to the angle AED. But, by hypothesis, the angle BCD is equal to the at A, B; therefore the angle AED is also equal to the angles a Similarly we can prove that the angle CDE is als to the angles at A, B, C ; therefore the pentagon ABCDE is equiangular. Next, let the given equal angles not be angles t order, but let the angles at the points A, C, D be equ I say that in this case too the pentagon ABCDE is equi For let BD be joined. Then, since the two sides BA, AE are equal to 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 namely those which the equal sides subtend; therefore the angle AEB is equal to the angle CDB. But the angle BED is also equal to the angle B since the side BE is also equal to the side BD. Therefore the whole angle AED is equal to t angle CDE. But the angle CDE is, by hypothesis, equal to th at A, C ; therefore the angle AED is also equal to the angles thus I. Suppose that the angles at A, B, C are all equal. Then the isosceles triangles BAE, ABC are equal in all respects; BE = AC, L BCA = L BEA, By the last equality, so that, since BE = AC, L CAB FA = FB, The As FED, FCD are now equal in all respects, and But whence, by addition, LEBA. [1. LFCD=L FED. LACB=LAEB, from above, LBCD = LAED. Similarly it may be proved that CDE is also equal to any one of angles at A, B, C. II. Suppose the angles at A, C, D to be equal. Then the isosceles triangles ABE, CBD are equal in all respects, hence BE = BD (so that BDE = L BED), and LCDB= = LAEB. By addition of the equal angles, LCDEL DEA. Similarly it may be proved that ABC is also equal to each of the a at A, C, D. PROPOSITION 8. If in an equilateral and equiangular pentagon stra lines subtend two angles taken in order, they cut one ano in extreme and mean ratio, and their greater segments are e to the side of the pentagon. A For in the equilateral and equiangular pentagon ABC. let the straight lines AC, BE, cutting one another at the point H, subtend two angles taken in order, the angles at A, B; I say that each of them has been cut in extreme and mean ratio at the point H, and their greater segments are equal to the side of the pentagon. For let the circle ABCDE be circumscribed about the pentagon ABCDE. IV |