L 
236 
THE TRISECTION OF ANY ANGLE 
Suppose E to he such a point that, if BE he joined meeting 
AG in D, the intercept BE between AG and AE is equal 
to 2 AB. 
Bisect BE at G, and join A G. 
Then BG = GE = AG = AB. 
Therefore Z ABG = Z A GB = 2 Z AEG 
= 2 Z BBG, since FE, BG are parallel. 
Hence ZBBG=\LABG, 
and the angle ABG is trisected by BE. 
Thus the problem is reduced to dr caving BE from B to cut 
AG and AE in such a way that the intercept BE — 2 AB. 
In the phraseology of the problems called revueis the 
problem is to insert a straight line EB pf given length 
2AB between AE and AG in such a way that EB verges 
towards B. 
Pappus shows how to solve this problem in a more general 
form. Given a parallelogram ABGB (which need not be 
rectangular, as Pappus makes it), to draw AEF to meet GB 
and BG produced in points E and F such that EF has a given 
length. 
Suppose the problem solved, EF being of the given length. 
Complete the parallelogram 
EBGF. 
Then, EF being given in length, 
BG is given in length. 
Therefore G lies on a circle with 
centre B and radius equal to the 
given length. 
Again, by the help of Eucl. I. 43 relating to the complements