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. a 
By the following methods the curve may be continued _ 
to E. : , 
In figure 6 let Bo be the part of the curve of sines already 
described. By the construction Eo is equal to the radius 
EC. » The semi-circles 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 arc 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 o, 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 oAR 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 by a pin'so as to be moveable about the centre C.- 
Let Eo be another rule of the same length, fastened by a pin - 
at E, the extremity of the rule CE, so as to be moveable is 
about E. 
Through this rule let there be a ahi so as to allow ee 
rule to move upon a fixed pin at E. J.eta pencil pass. 
through a hole of this rule ato. By pushing the pencik 
towards EK the rule will move on the pin E and being fasten- 
ed to the radius CE will always, at its extremity E, be on- 
the circumference EI. The pencil will describe the curve 
oie! Bars i 
‘The curve being thus formed, let HCB, (in figure 6,) an. 
angle greater than 135°, be the angle to 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. 
