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.