On a given finite straight line to construct an equilateral triangle. Let AB be the given finite straight line. Thus it is required to con 5 struct an equilateral triangle on the straight line AB. With centre A and distance AB let the circle BCD be described; B E [Post. 3] and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1] 15 Now, since the point A is the centre of the circle CDB, [Def. 15] 20 Again, since the point B is the centre of the circle CAE, BC is equal to BA. But CA was also proved equal to AB; [Def. 15] therefore each of the straight lines CA, CB is equal to AB. And things which are equal to the same thing are also equal to one another; therefore CA is also equal to CB. [C. N. 1] Therefore the three straight lines CA, AB, BC are 25 equal to one another. H. E. 16 Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB. (Being) what it was required to do. 1. On a given finite straight line. The Greek usage differs from ours in that the definite article is employed in such a phrase as this where we have the indefinite. ènì Tĥs δοθείσης εὐθείας πεπερασμένης, “on the given finite straight line,” i.e. the finite straight line which we choose to take. 3. Let AB be the given finite straight line. To be strictly literal we should have to translate in the reverse order "let the given finite straight line be the (straight line) AB"; but this order is inconvenient in other cases where there is more than one datum, e.g. in the setting-out of 1. 2, "let the given point be A, and the given straight line BC," the awkwardness arising from the omission of the verb in the second clause. Hence I have, for clearness' sake, adopted the other order throughout the book. 8. let the circle BCD be described. Two things are here to be noted, (1) the elegant and practically universal use of the perfect passive imperative in constructions, yeɣpápow meaning of course "let it have been described" or "suppose it described," (2) the impossibility of expressing shortly in a translation the force of the words in their original order. Kúkλos yeɣpáøow & BгA means literally "let a circle have been described, the (circle, namely, which I denote by) BCD." Similarly we have lower down "let straight lines, (namely) the (straight lines) CA, CB, be joined,” èπešeúxowσav evleîai ai гA, гB. There seems to be no practicable alternative, in English, but to translate as I have done in the text. 13. from the point C.... Euclid is careful to adhere to the phraseology of Postulate I except that he speaks of "joining” (érešeúx0wσav) instead of "drawing" (ypapew). He does not allow himself to use the shortened expression "let the straight line FC be joined" (without mention of the points F, C) until 1. 5. 20. each of the straight lines CA, CB, èkaтépа тŵν гA, гB and 24. the three straight lines CA, AB, BC, al тpeîs ai гA, AB, Bг. I have, here and in all similar expressions, inserted the words "straight lines" which are not in the Greek. The possession of the inflected definite article enables the Greek to omit the words, but this is not possible in English, and it would scarcely be English to write "each of CA, CB" or "the three CA, AB, BC." It is a commonplace that Euclid has no right to assume, without premising some postulate, that the two circles will meet in a point C. To supply what is wanted we must invoke the Principle of Continuity (see note thereon above, p. 235). It is sufficient for the purpose of this proposition and of 1. 22, where there is a similar tacit assumption, to use the form of postulate suggested by Killing. "If a line [in this case e.g. the circumference ACE] belongs entirely to a figure [in this case a plane] which is divided into two parts [namely the part enclosed within the circumference of the circle BCD and the part outside that circle], and if the line has at least one point common with each part, it must also meet the boundary between the parts [i.e. the circumference ACE must meet the circumference BCD].” Zeno's remark that the problem is not solved unless it is taken for granted that two straight lines cannot have a common segment has already been mentioned (note on Post. 2, p. 196). Thus, if AC, BC meet at F before reaching C, and have the part FC common, the triangle obtained, namely FAB, will not be equilateral, but FA, FB will each be less than AB. But Post. 2 has already laid it down that two straight lines cannot have a common segment. Proclus devotes considerable space to this part of Zeno's criticism, but satisfies himself with the bare mention of the other part, to the effect that it is also necessary to assume that two circumferences (with different centres) cannot have a common part. That is, for anything we know, there may be any number of points C common to the two circumferences ACE, BCD. It is not until III. 10 that it is proved that two circles cannot intersect in more points than two, so that we are not entitled to assume it here. The most we can say is that it is enough for the purpose of this proposition if one equilateral triangle can be found with the given base; that the construction only gives two such triangles has to be left over to be proved subsequently. And indeed we have not long to wait; for 1. 7 clearly shows that on either side of the base AB only one equilateral triangle can be described. Thus 1. 7 gives us the number of solutions of which the present problem is susceptible, and it supplies the same want in 1. 22 where a triangle has to be described with three sides of given length; that is, 1. 7 furnishes us, in both cases, with one of the essential parts of a complete Siopiapós, which includes not only the determination of the conditions of possibility but also the number of solutions (Toσaɣws ¿yxwpei, Proclus, p. 202, 5). This view of 1. 7 as supplying an equivalent for III. 10 absolutely needed in I. 1 and 1. 22 should serve to correct the idea so common among writers of text-books that 1. 7 is merely of use as a lemma to Euclid's proof of 1. 8, and therefore may be left out if an alternative proof of that proposition is adopted. Agreeably to his notion that it is from 1. I that we must satisfy ourselves that isosceles and scalene triangles actually exist, as well as equilateral triangles, Proclus shows how to draw, first a particular isosceles triangle, and then a scalene triangle, by means of the figure of the proposition. To make an isosceles triangle he produces AB in both directions to meet the respective circles in D, E, and then describes circles with A, B as centres and AE, BD as radii respectively. The result is an isosceles triangle with each of two sides double of the third side. To make an isosceles triangle in which the equal sides are not so related to the third side but have any given length would require the use of 1. 3; and there is no object in treating the question at all in advance of I. 22. An easier way of satisfying ourselves of the existence of some isosceles triangles would surely be to conceive any two radii of a circle drawn and their extremities joined. E B There is more point in Proclus' construction of a scalene triangle. Suppose AC to be a radius of one of the two circles, and D a point on AC lying in that portion of the circle with centre A which is outside the circle with centre B. Then, joining BD, as in the figure, we have a triangle which obviously has all its ⚫ sides unequal, that is, a scalene triangle. The above two constructions appear in an-Nairīzī's commentary under the name of Heron; Proclus does not mention his source. E B In addition to the above construction for a scalene triangle (producing a triangle in which the "given" side is greater than one and less than the other of the two remaining sides), Heron has two others showing the other two possible cases, in which the "given" side is (1) less than, (2) greater than, either of the other two sides. PROPOSITION 2. To place at a given point (as an extremity) a straight line equal to a given straight line. Let A be the given point, and BC the given straight line. Thus it is required to place at the point A (as an extremity) 5 a straight line equal to the given straight line BC. From the point A to the point B let the straight line AB be joined; K H 20 25 30 and again, with centre D and distance DG let the circle GKL be described. [Post. 3] Then, since the point B is the centre of the circle CGH, BC is equal to BG. Again, since the point D is the centre of the circle GKL, And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. But BC was also proved equal to BG; [C.N. 3] therefore each of the straight lines AL, BC is equal to BG. And things which are equal to the same thing are also equal to one another; [C.N. 1] therefore AL is also equal to BC. Therefore at the given point A the straight line AL is placed equal to the given straight line BC. (Being) what it was required to do. 1. (as an extremity). I have inserted these words because "to place a straight line at a given point” (πpòs tŵ dobévti onμely) is not quite clear enough, at least in English. 10. Let the straight lines AE, BF be produced........ It will be observed that in this first application of Postulate 2, and again in 1. 5, Euclid speaks of the continuation of the straight line as that which is produced in such cases, ἐκβεβλήσθωσαν and προσεκβεβλήσθωσαν meaning little more than drawing straight lines "in a straight line with" the given straight lines. The first place in which Euclid uses phraseology exactly corresponding to ours when speaking of a straight line being produced is in 1. 16: "let one side of it, BC, be produced to D” (προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ). 22. the remainder AL...the remainder BG. The Greek expressions are our AA and λo Tŷ BH, and the literal translation would be "AL (or BG) remaining," but the shade of meaning conveyed by the position of the definite article can hardly be expressed in English. This proposition gives Proclus an opportunity, such as the Greek commentators revelled in, of distinguishing a multitude of cases. After explaining that those theorems and problems are said to have cases which have the same force, though admitting of a number of different figures, and preserve the same method of demonstration while admitting variations of position, and that cases reveal themselves in the construction, he proceeds to distinguish the cases in this problem arising from the different positions which the given point may occupy relatively to the given straight line. It may be (he says) either (1) outside the line or (2) on the line, and, if (1), it may be either (a) on the line produced or (b) situated obliquely with regard to it; if (2), it may be either (a) one of the extremities of the line or (b) an intermediate point on it. It will be seen that Proclus' anxiety to subdivide leads him to give a case," (2) (a), which is useless, since in that "case" we are given what we are required to find, and there is really no problem to solve. As Savile says, "qui quaerit ad ß punctum ponere rectam aequalem τô ßy rectae, quaerit quod datum est, quod nemo faceret nisi forte insaniat." Proclus gives the construction for (2) (6) following Euclid's way of taking G as the point in which the circle with centre B intersects DB produced, and then proceeds to "cases," of which there are still more, which result from the different ways of drawing the equilateral triangle and of producing its sides. This last class of cases "he subdivides into three according as AB is (1) equal to, (2) greater than or (3) less than BC. Here again "case" (1) serves no purpose, since, if AB is equal to BC, the problem is already solved. But Proclus' figures for the other two cases are worth giving, because in one of them the point G is on BD produced beyond D, and in the other it lies on BD itself and there is no need to produce any side of the equilateral triangle. B K K H B A glance at these figures will show that, if they were used in the proposition, each of them would require a slight modification in the wording (1) of the construction, since BD is in one case produced beyond D instead of B and in the other case not produced at all, (2) of the proof, since BG, instead of being the difference between DG and DB, is in one case the sum of DG and DB and in the other the difference between DB and DG. |