On the Curves of Trisection. 353 
been proved under figure 3, the angle HCm, or HCA, is 
one quarter of HCB. But ACL is obviously equal to 
HCA;; it is therefore one third of ACB. The rule CKL 
therefore cuts off one third of the given angle. 
If this curve, which terminates at 0, was continued to E, 
any angle could be trisected by it though larger than 135° 
without the necessity of bisection. 
By the following methods the curve may be continued 
to 
In figure 6 let Bo be the part of the curve of sines already 
described. By the construction Eo is equal to the radius 
. The semi-circlés on the opposite side of the diame- 
ter being formed, let Eo be supposed to move upon the fix- 
ed point E at the extremity of the radius CE, the extremity 
E of Eo moving in the are EGI, until TE be equal to Eo. 
The point o will describe the curve oAE, which continues 
the curve Bo from o to E. 
Or with the radius in the compasses, if it be set off from 
various points of the arc EGI towards 0, so that a straight 
line from each point shall pass through E, the points between 
o and E, thus found, will be points of the curve ; and 
through these points with a steady hand the curve oAE may 
be drawn. 
Or this curve may be drawn mechanically, by a continued 
motion, as follows. Let CE, in figure 7, be a straight rule, 
fastened bya pin so as to be moveable about the centre C. 
Let Eo be another rule of the same length, fastened by a pin 
pan the extremity of the rule CE, so as to be moveable 
ut E 
Through this rule Jet there be a slit, so as to allow the 
to move upon a fixed pin at E. Leta pencil pass 
through a hole of this rule ato. By P pe the pencil 
towards E the rule will move on the pin E and being fasten- 
ed to the radius CE will always, at its extremity E, be on 
= circumference EJ. The pencil will describe the curve 
r) 
The curve being thus formed, let HCB, (in figure 6,) av 
angle. greater than 135°, be the angleto be trisected. From 
A, the point of intersection of the side CH and the curve, 
draw through E the straight line AG, terminating at G on 
the circumference ‘of the inner circle. By the construction 
AG is equal to Eo, or to the radius. Through G draw CF. 
