On the Curves of T'risection. 347 
Let DG be a third rule, moveable about D, where it is 
fastened by a pin, having a slit through it of the length of 
DB ; and the distance CD being equal to CG. Let the 
pin G in CG pass through this slit of DG, so as to move 
in it. 
The instrument being thus constructed, let a single pencil 
pass through each of the slits of the three rules at the com- 
mon point of intersection, 0. By pushing the pencil either 
way the rules will move one upon another, and a portion of 
the curve of trisection will be described. IfCG be brought 
to the position CB; then the pencii will begin the curve at 
F, and will have described it and reached D, when CG 
shall have reached the position Cg, cutting off two thirds of 
the arc BAD. 
When CG is in the position CB, oG will coincide with 
FB, and oC€ with FC, that is, will each be equal to the radi- 
us of the interior circle ; but as the curve is described, they 
continually increase, until at the completion of the curve 
they each are double to the radius of the exterior circle, 
or equal to the radius of the exterior circle, that is, oC coin- 
cides with DC, and oG with Dg. 
From the construction of the instrument, HK being per- 
pendicular to CG, and H equidistant from C and G, it is 
evident, that Co, and oG are always equal. Supposing the 
curve to be described ;—then whatever line is drawn from 
the centre to the circumference between B and D, as CA, 
in fig. 1, a straight line from D passing through the point of 
intersection of the line CA and the curve, and proceeding 
to the circumference, as at G, will give that part of the line 
DG, which is without the curve of trisection, as oG, equal 
to that part of the line from the centre, as CA, which is 
within the curve,as oC. Thatis, oG and Co are always equal. 
Suppose then, the curve has been described, as in figure 
1, and the semi-circles drawn. Let ACB be the angle to 
be trisected. Through the point 0, where CA. intersects 
the curve, draw, from D, a straight line DG to the exterior _ 
circumference. ‘Then oG is equal to oC. From the cen- 
tre draw CG. And because 0G is equal to oC, the angle 
oGC is equal to the angle oCG ; and AoG being an exterior 
angle to the triangle oCG, is equal to the two angles oCG 
and oGC, that is, is equal to ACG and DGC, and these are 
equal to each other. Because GCB is an angle at the cen- 
