208 THE ELEMENTS DOWN TO PLATO’S TIME 
Cut off CD from CA equal to CB, and draw DE at right 
angles to CA. Then DE = EB. 
Now AD = 75 — 2, and by similar triangles 
DE = 2 AD = 2 (>/5-2). 
Cut off from EA the portion EF equal to 
ED, and draw FG at right angles to AE. 
Then AF — AB — BF = AB—2DE 
= 1-4(75-2) 
= ( 75 — 2) 2 . 
Therefore ABC, ADE, AFG are diminishing 
similar triangles such that 
AB:AD.AF = 1: (75-2): (75- 2) 2 , 
and so on. 
Also AB > FBf i.e. 2DE or 4 AD. 
Therefore the side of each triangle in the series is less than 
\ of the corresponding side of the preceding triangle. 
In the case of 7 3 the process of finding the G. C. M. of 
7 3 and 1 gives 
1 ) 73 (1 
73-1) '1 (1 
73-1 
1(73 
1)2) 73_1 (2 
(73-1) 2 
*(</3-1 ) 3 
the ratio of -|(73 — l) 2 to |(73 —l) 3 being the same as that 
1). 
of 1 to ( 7 3 
This case 
is more difficult to show in geometrical form 
because we have to make one more 
division before recurrence takes place. 
The cases 710 and 717 are exactly 
similar to that of 75. 
The irrationality of 7 2 can, of course, 
be proved by the same method. If A BCD 
is a square, we mark off along the diagonal 
AC a length AE equal to A Sand draw 
EF at right angles to AG. The same 
thing is then done with the triangle CEF