CANDV: THE TRlSKCTION OF AN ANCJl.E. 



43 



Then PF=2PE^PH, since e=^2. 



. •. arc FP-=:Arc PH^ Arc AH=:^ Arc APF. 



Since PC also trisects <FCN, one side of the given angle could 

 have been taken for the directrix. 



Now the lines OP, OP' and OP" (fig. V) invert into FP,FP' and 

 FP" (fig. VIII) respectively, 



. •. P"F bisects <PFP'. and henceP, P', and P" trisect the circle. 



FOURTH METHOD. 



Let DFB (fig. IX) 

 be the given angle. 

 With F as focus and FD 

 as axis construct a hy- 

 perbola whose eccen- 

 tricity is 2 and inter- 

 secting FD in A and A'. 



Produce BF to C, 

 making FC=FA. 



Pass a circle through 

 A, F, and C, intersect- 

 ing the hyperbola in P 

 and P'. 



Then PF trisects <AFB and P'F trisects the reflex <AFB. 



Proof. This is the Inverse of the Second Method and is therefore 

 proved by Theorem (II) since <AFB^Arc APF. 



The direct proof is the same as in the preceding solution. 



INVERSION OF FIG. VIII FROM O, THE CENTER OF THE HVPERBOLA. 



Substituting formula (5) in {7). and reducing and drcjpping the 

 primes, we get 



(10) 



(x2 J-y2)2--,^- (3x2— y2), where a ' of 



a curve of the fourth degree, consisting of two ciual symmetrical 

 loops, forming approximately a figure 8, and having the origin for a 

 double point. 



The following (fig. X) is the inverse figure; the ])oints .\, C, F, P 

 and P' being ihe inverse of the corresponding jioints of fig. \TII. 



h 



