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The Postulate of Parallels
Published online by Cambridge University Press: 03 November 2016
Extract
There are in the Mathematical Gazette for October 1924, three communications on the above subject.
It will make my remarks on these clearer if I refer to the contents of my paper on the Postulate of Parallels in the Gazette for December 1923, and make an addition thereto. In that paper the demonstrations of Euc. I. 27 and 29 are deduced from Wallis’s Postulate of Similarity, viz.: “To every figure there exists a similar figure of arbitrary magnitude” taken as a substitute for Euclid’s Postulate of Parallels. The demonstration of Euc. I. 27 in this manner was first given (I believe) by Professor Nunn in the Gazette of May 1922. He has not, however, published in full his demonstration of Euc. I. 29. He was so kind as to communicate it to me, but I hope he will some day publish it in full. It is entirely different from the demonstration of Euc. I. 29 which I published in the Gazette of December 1923, which involves three successive applications of the Postulate of Similarity and makes use of Euc. I. 27 as proved by Professor Nunn, but does not include the proof of any other proposition.
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- Copyright © Mathematical Association 1925
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page 271 note * Euclid’s proof involves the infinity of the straight line.
page 272 note * We may go further and show that what has been described above as the second part of the definition of similar figures can be deduced as a conclusion from the first part if we take account of the postulate of parallels (or similarity) and the concept.of ratio. For if A, B, C be the points of tbe first figure corresponding to A′, B′, C′ respectively of the second figure, then by the first part of the definition of similar figures the triangles ABC, A′B′C′ have the angles of the one respectively equal to those of the other. Then making use of the postulate of parallels and the concept of ratio, we may obtain the result given in Eue. VI. 4, that
(BC: B′C′) = (CA . C′A′)=(AB:A′B′).
Now let the points P, Q, R, S of the one figure correspond respectively to P′, Q′, R′, S′ of the other figure. To prove the second part of the definition of similar figures, it is enough to snow that
(PQ:P′Q′)=(RS:′S′).
From the similarity of the triangles PQR, P′Q′R′, we get
(PQ: P′Q′) =(QR: Q′R′) =(RP: R′P′).
From the similarity of the triangles RPS, R′P′S′, we get
(RP: R′P′) = (PS: P′S′) =(RS: R′S′).
Consequently (PQ: P′Q′) = (RS: R′S′).