148
PYTHAGOREAN GEOMETRY
it independently of I, 47 by means of proportions. This
seems to suggest that he proved I. 47 by the methods of
Book I instead of by proportions in order to get the proposi
tion into Book I instead of Book YI, to which it must have
been relegated if the proof by proportions had been used.
If, on the other hand, Pythagoras had proved it by means
of the methods of Books I and II, it would hardly have been
necessary for Euclid to devise a new proof of I, 47. Hence
it would appear most probable that Pythagoras would prove
the proposition by means of his (imperfect) theory of pro
portions. The proof may have taken one of three different
shapes.
(1) If ABC is a triangle right-
angled at A, and AD is perpen
dicular to BC, the triangles DBA,
DAG are both similar to the tri
angle ABC.
It follows from the theorems of
Eucl. VI. 4 and 17 that
BA 2 = BD.BG,
AC 2 = CD. BG,
whence, by addition, BA 2 + AC 2 = BG 2 .
It will be observed that this proof is in substance identical
with that of Eucl. I. 47, the difference being that the latter
uses the relations between parallelograms and triangles on
the same base and between the same parallels instead of
proportions. The probability is that it was this particular
proof by proportions which suggested to Euclid the method
of I. 47 ; but the transformation of the proof depending on
proportions into one based on Book I only (which was abso
lutely required under Euclid’s arrangement of the Elements)
was a stroke of genius.
(2) It would be observed that, in the similar triangles
DBA, DAG, ABC, the corresponding sides opposite to the
right angle in each case are BA, AG, BG.
The triangles therefore are in the duplicate ratios of these
sides, and so are the squares on the latter.
But of the triangles two, namely DBA, DAG, make up the
third, ABC.