io Gee AND ADAMSON, Trisecting an Angle. 



OA for a diameter and passing through the point C of 

 intersection of OB with the tangent AC to the circle at 

 A, will pass through one of the two points of trisection of 

 the arc. To draw this hyperbola : Bisect AC at D and 

 OA at L. Join LD and produce to F, making DF and 

 DE each equal to AD. Draw the rectangle AFCE. The 

 asymptotes XX } YY\ of the hyperbola will be parallel 

 to AF and AE and intersect at L. Construct the hyper- 

 bola intersecting the circle at P as well as at A and at 

 two other points, P v P 2 , not shown in Fig. 7. Join OP, 

 then the angle POB will be ^ of the angle ^40/?. 



Proof. — Construct the rectangle PGOH on OP as 

 diagonal with sides parallel to the asymptotes, and draw 

 the diagonal GKH. Then, as in the construction of 

 Pappus {Fig. 6) : 



_ 0LX= L 0LK+ LXLK 

 = L0XL+ _P0H 



= $LPOH. 



But 



L C0H= L FAC= L EAL = L GOL. 

 Therefore 



3 L POB = 31 POB - 3 L COH 



= 0LX-iLC0H 



- 90 + _ COH- 3 _ COH 



= 9 o°-2_COH 



= LAOB. 



Similar geometrical proofs could be given to show 



that the angles P\0B and P,0B are respectively equal to 



one-third of the angles A0B+2tt and A OB + 4tt, and 



that consequently P lt P 2 and P are the angular points of 



an equilateral triangle. These additional facts may also 



be proved analytically as follows : 



Let 



_A0B = «, 0A = 2a. 



