THE‘THEOREM OF PYTHAGORAS’ 
149 
The same must therefore be the case with the squares, or 
BA 2 + AG 2 = BG 2 . 
(3) The method of VI. 31 might have been followed 
exactly, with squares taking the place of any similar recti 
lineal figures. Since the triangles DBA, ABC are similar, 
BD:AB — AB: BG, 
or BD, AB,.BG are three proportionals, whence 
AB 2 : BG' 1 = BD 2 : AB 2 = BD : BG. 
Similarly, A C 2 : BG 2 = CD : BG. 
Therefore (BA 2 + AG 2 ) : BG 2 = (BD + DG) : BG. [V. 24] 
If, on the other hand, the proposition was originally proved 
by the methods of Euclid, Books I, II alone (which, as I have 
said, seems the less probable supposition), the suggestion of 
a b a b 
Bretschneider and Hankel seems to be the best. According 
to this we are to suppose, first, a figure like that of Eucl. 
II. 4, representing a larger square, of side («■ + &), divided 
into two smaller squares of sides a, h respectively, and 
two complements, being two equal rectangles with a, h as 
sides. 
Then, dividing each complementaiy rectangle into two 
equal triangles, we dispose the four triangles round another 
square of side a + h in the manner shown in the second figure. 
Deducting the four triangles from the original square in 
each case we get, in the first figure, two squares a 2 and h 2 
and, in the second figure, one square on c, the diagonal of the 
rectangle (a, h) or the hypotenuse of the right-angled triangle 
in which a, h are the sides about the right angle. It follows 
that à 2 + b 2 = c 2 .