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A Mathematical History of the Golden Number

Dover Publications, Inc.

THE EUCLIDEAN TEXT

Section 1. The Text

As I indicated in the Introduction, I felt that my study of DEMR must start off with a detailed analysis of the Euclidean text and that for various reasons it was appropriate to gather together all the relevant Theorems here in Section 1. My reasons for the method of presenting the statements, proofs or sketches of proofs, and notations are explained in the Introduction and in the list of symbols which precedes it. I have omitted most of the congruence and similarity Theorems as well as some other results which do not play an important conceptual role in the discussions.

The diagrams follow those of the critical edition of Euclid-Heiberg, except that the Greek letters have been replaced by their romanized versions as in Eu-clid-Heath. This has been done for ease of comparison even though great clarity could sometimes have been achieved (for example, in XIII,1,2) by relabelling as in the Euclid-Frajese edition. I have, however, followed the lead of this edition by identifying quadrilaterals by four letters instead of two, although sometimes regions are simply identified by roman numerals.

The results are followed by a partial list of those Theorems of the Elements in which they are used. Sometimes just a reference to a book is given; a series of three dots indicates that the listing is not complete. In some cases I have also given a partial list of the principal results used in the proof. For all these lists I have followed, with some corrections and additions, the lists found in Euclid-Frajese. I have not methodically compared them with other lists and indications (such as those given in Euclid-Peyrard; Euclid-Heiberg; Euclid-Heath; Neuenschwander [1972; 1973]; Mueller [1981], although I have used these for various results. The interconnection between Theorems will play an important role in the discussions of Section 2. In some cases I have added a few explanatory notes, none of which is related to DEMR. Where applicable I have indicated where in this book the Theorem is discussed in relationship to DEMR.

BOOK I. This book contains some basic results concerning lines, triangles, and parallelograms.

Theorem I,32. In any triangle an exterior angle is the sum of the two opposite interior angles. The sum of the interior angles is equal to two right angles.

Used in:IV,10; XIII,8,9,10,11,18 lemma...

Theorem I,42 (a transformation of areas result). To construct a parallelogram one of whose angles is given and whose area is equal to the area of a given triangle.

Proof: See Section 4.

Used in:I,44,45.

Discussed in:Sections 4; 5,H.

Theorem I,43 (complements). In the parallelogram ABCD the so-called complements—with respect to the diagonal—I and II have the same area.

Proof: See Section 3.

Used in:I,44; II,4(?),5,6,7,8; VI,27,28,29; X; XIII,1,2,3,4,5(?) ...

Discussed in:Sections 2.D; 3; 4.

Note: See II,def.2 (gnomon); also Section 3.

Theorem I,44 (an application of areas result). To construct (to apply) on a given line a parallelogram under the constraints that one angle is given and that the area is equal to the area of a given triangle.

Proof: See Section 4.

Used in:I,45; VI,25; X.

Discussed in:Sections 4; 5,C.

Theorem I,45 (a transformation of areas result). To construct a parallelogram, one of whose angles is given, and whose area is equal to the area of a given rectilineal figure.

Proof: See Section 4.

Used in:II,14; VI,25; X (27 times).

Discussed in:Section 4.

Theorem I,47 (Pythagorean Theorem). In a right-angle triangle with hypotenuse BC and legs AB and AC,S (BC) = S (AB) + S (AC).

Used in:II,9,10,14; III,36; XIII,18 ...

BOOK II. The origin, context, and meaning of the results of this book have been the subject of much controversy; I shall discuss these questions in detail in Section 5. In order to avoid any prejudgments, it is important to read these results in the first instance as simply dealing with the "geometry of areas," to use the terminology of Gow [1884, 190] and Szabo [1968, 226], that is, geometric statements concerning the areas of squares and rectangles.

Theorem II,def.2. With respect to the parallelogram of Figure 1 both of the regions I + III + II and I + IV + II (i.e., the two complements with respect to the diagonal together with one of the parallelograms III or IV) are called a gnomon.

Used in:II,5,6; VI,27,28,29; XIII,1,2,3,4 ...

Discussed in:Section 3.

Theorem II,1. In the diagram let D and E be arbitrary division points of the line BC. Then R (BG,BC) = R (BG,BD) + R (BG,DE) + R (BG,EC) = I + II + III.

Used in: Never used.

Discussed in:Section 5,A,G.

Theorem II,2. Let AB be any line and on it draw a square. If C is an arbitrary division point of AB, then S(AB) = R (AB,BC) + R (AB,AC).

Used in:XIII,10; implicitly in I,47.

Note: The proof does not use II,1.

Theorem II,4. In the diagram let C be an arbitrary division point, thenS (AB) = S (AC) + S (CB) + 2.R (AC,CB) = I + I + 2 III.

Uses:I,43 (complements)(?).

Used in: II,12; IX; X; XIII,2 lemma (possibly not genuine).

Discussed in:Section 5,B.

Note: One usually presumes (e.g., Euclid-Frajese and Neuenschwander [1972]) that I,43 (complements) was what the mathematician had in mind when stating the equality of the rectangles marked III; however, the word "complements" is not used. Since it is shown earlier in the proof that CB = BK, it is possible that the author had an extended version of I,35 in mind.

Theorem II,5. Let AB be a line segment divided into two unequal segments AD and DB, and let C be the midpoint of AB. ThenR (AD,DB) + S (CD) = S (CB).

Proof: Construct square CBFE and draw line BE; this determines points H and K. We have that R (AD,DB) [equivalent] ρεctangle ADHK [equivalent] I + II. But I = II + III, since C is the midpoint. Also by I,43, II = IV. Therefore R (AD,DB) = (II + III) + (IV). This latter L-shaped region, the gnomon of II,def.2, is simply the difference of the two squaresS (CB) and S (CD). Hence, by adding S (CD) to both R (AD,DB) and the gnomon, the result is proved.

Uses:I,43 (complements).

Used in:II,14; III,35; X,17,41 lemma, 59 lemma.

Discussed in: Introduction; Sections 2; 5,B,E,J.

Theorem II,6. Let AB be a straight line and C ITS midpoint. Add another straight line BD to AB. Then R (AD,DB) + S (CB) = S (CD).

Proof: Construct square CDFE and draw line DE. This determines points H,K, and M. We have that R (AD,DB) = rectangle ADMK = I + II + S (BD). But I = II, since C is the midpoint. Also, by I,43, II = III. Therefore R (AD,DB) = II + S (BD) + III. This L-shaped region, the gnomon (II,def.2), is simply the difference of the two squares S(CD) and S(CB). Hence, by adding S (CB) to both R (AD,BD) and the gnomon, the result is proved.

Uses:I,43 (complements).

Used in:II,11; III,36; X,28 lemmas 1,2.

Discussed in:Sections 2; 5,B,E,G.

THEOREM II,9. Let AB be a line with C as its midpoint. Let D be an arbitrary point of AB differing from C. Then S(AD) + S (DB) = 2[S (AC) + S (CD)].

Used: Never used.

Discussed in:Section 5,F.

Theorem II,10. Let AB be a line with C as its midpoint. Let the line segment BD be added to AB. Then S (AD) + S (DB) = 2[S (AC) +S (CZ))].

Used in: Never used.

Discussed in:Section 5,F.

THEOREM II,11 (the area formulation of DEMR). To divide a line AB into two segments, a larger one AH and a smaller one HB so thatS (AH) = R (AB,BH).

Note: While the picture conjured up by the statement is that of Figure I-9a, the figure of the text is that of Figure I-9b.

Proof (Fig. I-9b): Construct square ABDC and let E be the midpoint ofAC. Obtain point F on the extension of EC so that FE = EB and then construct square AFGH.

Since E is the midpoint of AC and AF is added to AC,II,6 tells us that R (CF,FA) + S (AE) = S (EF) = S (BE). By the Pythagorean Theorem (I,47), S (BE) = S (AE) + S (AB) so that when we subtractS (AE) we obtain

(*) R(CF,FA) = S(AB).

In terms of the labelled areas, the equation (*) can be written I + III = II + III so that by subtracting III = R (AH,HK) we have S(AH) =R (BD,HB) = R (AB,HB). This shows that H is the desired division point.

Uses:I,47; II,6.

Used in:IV,10.

Discussed in:Sections 2,B,ii; 5,K,L ...

Theorem II,14 (a transformation of areas result). To construct a square whose area is the same as that of a given rectilineal figure.

Uses:I,45,47.

Used in:Book X (5 times).

Discussed in:Sections 4; 5,F; 22.

BOOK III. This book deals with the properties of circles.

Theorem III,20 (Fig. I-10). Let B and C be two points on the circumference of a circle. Let [??] BOC be the central angle and let BAC, where A is any point on the circumference, be an inscribed angle. Then [??] BOC = 2 · [??] BAC.

Used in:III,27; VI,33 ...

Theorem III,26 (Fig. I-10). Let BC and EF be arcs of equal circles. Suppose that central angles BOC and EOF are equal, then arcs BC and EF are equal. The same is true if the inscribed angles BAC andEDF are equal.

Used in:III,27,28; IV,11; XIII,10 ...

Note: The statement says that the arcs will be equal if either the central or the inscribed angles are equal. In IV,11 it is equality of the inscribed angles that is used, whereas in XIII,10 it is equality of the central angles that is used. However, for some reason the proof of III,26 assumes that both the central and the inscribed angles are equal. This does not seem to have been observed in the various commentaries. See also the proof of XIII,8' in Section 2,C.

Theorem III,27 (Fig. I-10). With the notation of III,26, if the arcs are equal then the angles are equal.

Used in:III,29; IV,11; VI,33 ...

Discussed in:Section 2.

Note: Compare with VI,33.

Theorem III,28 (Fig. I-10). In two equal circles if chords CB and EF are equal then arc CB is equal to arc EF.

Used in:XIII,8 ...

Theorem III,29 (Fig. I-10). In two equal circles if arcs CB and EF are equal, then chords CB and EF are equal. Used in:IV,11; XIII,10 ...

Theorem III,32 (Fig. I-11). Let BC be the arc of a circle and BF the tangent at point B. Let D be a point on that arc of the two arcs determined by B and C which is not included by [??] CBF, then [??] BDC and [??] FBC are equal.

Sketch of Proof: Draw the diameter AB so that [??] ABF = 90° = [??] AOB. Further, [??] BAD and [??] BCD add to 180° since their associated arcs correspond to the entire circumference. Combining these results—via subtractions and additions—with the Theorems on the sum of the angles of a triangle and the angles along a line, the Theorem is immediate.

Used in:IV,10 ...

Theorem III,35. Let AC and BC be two chords of a circle that meet atE, then R (AE,EC) = R (DE,EB).

Sketch of Proof: A perpendicular is dropped from the centre F and this divides AC into two parts. This brings II,5 into play which gives an expression for R (AE,EC). This expression is related to FE and the radius FC via the Pythagorean Theorem (I,47). By working in the same way with BD,R (DE,EB) is related to FE and FB. But FE is common and the radii FB and FC are equal, giving the desired equality.

Uses:I,47; II,5.

Used in: Never used.

Theorem III,36. If from a point D outside a circle, a tangent line DB and a secant line DA are drawn, and if DA also cuts the circle at C, then R (DA,DC) = S (DB).

Note: Euclid considers two cases: Case 1—when AC is a diameter; and Case 2—when AC is a chord. Case 2 does not use Case 1. In both, which are treated independently, there are two essential points to the proof: first, F is the midpoint of AC and so II,6 gives us a value for R (AD,DC); and second, the right angle allows use of the Pythagorean Theorem.

Proof of Case 1: By II,6, R (AD,DC) + S (FC) = S (FD). Applying the Pythagorean Theorem to the term S (FD) and substituting FB for FC results in R (AD,DC) + S (FB) = S (FD) = S (FB) + S (BD). Since S (FB) is common to both sides, we have R (AD,DC) = S (BD).

Proof of Case 2: By II,6R (AD,DC) + S (FC) = S (FD). The perpendicular EF is common to the two right triangles EFD and EFC; and so if we add S (EF) to both sides of the equality and use the Pythagorean Theorem, we obtain R (AC,DC) + S (EC) - S (ED). Now substituting EB for EC and applying the Pythagorean Theorem once more, but this time to the right triangle EDB, we have R (AD,DC) +S (EB) = S (DB) + S (EB). Since S (EB) is common to both sides, we have R (AD,DC) = S (BD).

Uses:I,47; II,6.

Used in:III,37.

Discussed in:Section 2.

Theorem III,37 (the converse of III,36). Let D be a point outside a circle with centre F and let DA be a secant line which cuts the circle at C. Now suppose that B is a point on the circle such that the relationship R (AD,DC) = S (DB) holds, then the line DB is in fact tangent to the circle at B.

Proof: Let DE be tangent to the circle then, by III,36 and the hypothesis, S (DE) = R (AD,DC) = S(DB) so that DE = DB. Since FE = BF and DF is common, we have that triangles DEF and DFB are congruent. This means that DBF is also a right angle and thus DB is tangent.

Uses:III,36.

Used in:IV,10.

Discussed in:Section 2.

BOOK IV. This book gives the construction of various regular polygons.

Theorem IV,2. To inscribe an arbitrary triangle in a circle.

Theorem IV,6. To inscribe a square in a circle.

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Excerpted from A Mathematical History of the Golden Number by Roger Herz-Fischler. Copyright © 1987 Wilfrid Laurier University Press. Excerpted by permission of Dover Publications, Inc..
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