Manchester Memoirs, Vol. //It'. (191 5), No. 13. 15 

 the branch of the hyperbola which passes through F. 



A \ M 



Fig. 10. Methods of Pappus and Clairaut. 



Draw PQ parallel to BH cutting ME at N. Join BP, 

 QH and SP. 



Then, as before, the arcs BP, PQ and QH are equal. 

 Therefore 



L PBH= I L PSH= L BSP= ~ L BSH~ It ABC. 



2 3 3 



Method of Clair ant. This method is the most direct of 

 the three. In Fig. 10 let BSHbe the angle to be trisected. 

 With S as centre describe a circle cutting off SB and SH. 

 Join BH. Trisect BH at F and K. Bisect the angle BSH 

 by SM, cutting BH at E. Construct a hyperbola of 

 eccentricity =2, with B as focus and 5 M as directrix, and 

 let the branch which passes through .Fcut the circle at P. 

 Join PS. Then the angle BSP is one-third of the angle 

 BSH y as previously proved. 



It can easily be shown that the circle with centre 5 

 cuts the hyperbola at H, P, and at two other points P u P s , 

 such that PPyP, is an equilateral triangle. 



