Commentaar van Proclus op Propositie I-47 (Euclides: Elementen)

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  1. Proclus' commentaar
  2. Propositie VI-31 (Uitgebreide Stelling van Pythagoras)
  3. Bewijs van VI-31

Zie ook de pagina "Boek VI"

1. Proclus' commentaar terug
Vertaling uit het Grieks:
Glen R. Morrow in Proclus: A Commentary on the First Book of Euclid's Elements (Princeton University Press, 1970; herdruk 1992)
[Zie verder ook History of Mathematics: Proclus (411-485)]

    If we listen to those who like to record antiquities, we shall find them attributing this theorem to Pythagoras and saying that he sacrificed an ox on its discovery. For my part, though I marvel at those who first noticed the truth of this theorem, I admire more the author of the Elements, not only for his lucid proof by which he made it fast, but also because in the sixth book he laid hold of a theorem even more general then this and secured it by irrefutable scientific arguments. For in that book he proofs generally that in right angled triangles the figure on the side that subtends the right angle is equal to the similar and similarly drawn figures that contain the right angle. Every square is of course similar to every other square, but not all similar rectilinear figures are squares, for there is similarity in triangles and other polygonal figures. Hence the argument establishing that the figure on the side subtending the right angle, whether it be a square or any other kind of figure, is equal to similar and similarly drawn figures on the sides about the right angle, proves something more general and scientific then that which shows only that the square is equal to the squares. For there the cause of the more general proposition that is proved becomes clear: it is the rightness of the angle that makes the figure on the subtending side equal to the similar and similarly drawn figures on the containing side, just as the obtuseness of the angle is the cause of its being greater an the acuteness of the angle is the cause of its being less.
    How he proves the theorem in the sixth book will be evident there. But now let us consider how he shows the theorem before us to be true, remarking only that he does not prove the universal proposition here, since he has not yet explained similarity in rectilinear figures, nor proved anything in general about proportion. Hence many of the things here proved in a partial fashion are proved in that book more generally through the use of the above method. In the present proposition the author of the Elements proves his conclusion by means of the ordinary theory of parallelograms.
    There are two sorts of right-angled triangles, isosceles and scalene. In isosceles triangles you cannot find numbers that fit the sides; for there is no square number that is de double of a square number, if you ignore approximations, such as the square of seven which lacks one of being double the square of five. But in scalene triangles it is possible to find such numbers, and it has been clearly shown that the square on the side subtending the right angle may be equal to the squares on the sides containing it. Such is the triangle in the republic, in which sides of three and four contain the right angle and five subtends it, so that the square on five is equal to the squares on those sides. For this is twenty-five, and of those the square of three is nine and that of four sixteen. The statement, then, is clear for numbers.

[] Certain methods have been handed down for finding such triangles, one of them attributed to Plato, the other to Pythagoras. The method of Pythagoras begins with odd numbers, positing a given odd number as being the lesser of the two sides containing the angle, taking its square, subtracting one from it, and positing the half of the remainder as the greater of the sides about the right angle; then adding one to it this, it gets the remaining side, the one subtending the angle. For example, it takes three, squares it, subtract one from nine, takes the half of eight, namely, four, then adds one to this and gets five; and thus is found the right-angled triangle with sides of three, four, and five. The Platonic method proceeds from even numbers. It takes a given even number as one of the sides about the right angle, divides it into two and squares the half, then by adding one to the square gets the subtending side, and by subtracting one from the square gets the other side about the right angle. For example, it takes four, halves it and squares the half, namely two, getting four; then subtracting one it gets three and adding one gets five, and thus it has constructed the same triangle that was reached by the other method. [] For the square of this number is equal to the square of three and the square of four taken together.
    These remarks are somewhat outside our subject. But since the proof given by the author of the Elements is clear, I do not think I should add anything superfluous but should be content with what he has written, especially since those who have made additions, such as the disciples of Heron and Pappus, have been obliged to assume something proved in the sixth book, and for no material purpose.


Het gedeelte van de tekst hierboven tussen [] staat in vertaling van E.J. Dijksterhuis op de pagina over de Stelling van Pythagoras.
[einde Opmerking]

2. Propositie VI-31 (Uitgebreide Stelling van Pythagoras) terug
Bedoelde propositie luidt (in vertaling van E.J. Dijksterhuis):

Propositie VI-31
In rechthoekige driehoeken is een figuur, beschreven op de den rechten hoek onderspannende zijde, gelijk aan de op de den rechten hoek omvattende zijden op gelijke wijze beschreven gelijkvormige figuren.


In het bovenstaande commentaar wordt Propositie VI-31 door Proclus dus uitdrukkelijk als eigen vondst van Euclides aangemerkt.
Op de pagina "Boek VI" wordt het bewijs van VI-31 gegeven in een "moderne" notatie.
We laten hieronder het bewijs volgen zoals Euclides dat gegeven heeft (naar een vertaling van C. Thaer: Die Elemente).
[einde Opmerkingen]

3. Bewijs van Propositie VI-31 terug

propVI-7.gif (1458 bytes) ABC zij een rechthoekige driehoek met rechte hoek BAC. Ik beweer, dat een figuur op BC aan gelijkvormige [figuren], op BA, AC gelijkgetekende figuren, gelijk is.
Men trekke de loodlijn AD.
Daar men in de rechthoekige driehoek ABC uit de rechte hoek bij A op de basis BC de loodlijn AD getrokken heeft, zijn de driehoeken op de loodlijn, namelijk ABD, ADC gelijkvormig met zowel de gehele driehoek ABC als met elkaar. Daar ABC, ABD gelijkvormigzijn, is CB : BA = AB : BD. Daar hier de lijnstukken in verhouding tot elkaar staan, verhoudt zich de figuur op de eerste tot de op de tweede gelijkvormige en gelijkgetekende, als het eerste lijnstuk tot het derde [lijnstuk].
Zo verhoudt zich, als CB : BD, aldus de figuur op CB tot de op BA gelijkvormige en gelijkgetekende, en op dezelfde grond ook, als BC : CD, aldus de figuur op BC tot die op CA. Vervolgens verhoudt zich ook, als BC : (BD+DC), aldus de figuur op BC tot de gelijkvormige, op BA, AC gelijkgetekende samen. Nu is BC = BD + DC. Dus is ook de figuur op BC gelijkvormig met de gelijkgetekende figuren op BA, AC samen.

Op de webpagina "Boek VI" staat ook een CabriJavapplet bij Prop. VI-31.
[einde Opmerking]

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