On the Curves af Trisection. 351 
Through H, the point where Em intersects the interior 
semi-circle, draw CG. Because E is an exterior an- 
gle to the triangle HmC, it is equal to the two angles HmC 
and HCm. And these two angles are equal to each other 
because by the construction Hm is equal toHC. Therefore 
the angle HCA or GCA, is one half of the angle EHC. 
Because HCF is an angle at the centre, it is double the 
angle HEC at the circumference. But the angle HEC is 
equal to the angle EHC, because by the construction HC is 
equal to EC. Therefore the angle EHC is one half of the 
angle HCF. And as the angle GCA has been proved to be 
one half of the angle EHC, it consequently is one quarter of 
the angle HCF, of which the angle EHC is one half; or 
the angle GCA is one quarter of the angle GCB, which is 
the same with the angle HCF. ‘The angle GCA is there- 
fore one third of the angle ACB, for it being one quarter of 
the whole angle GCB, if it be abstracted from this whole, 
there will be left three parts, each equal to this part abstract- 
ed. Make AI then equal to AG, and draw CI, and the an- 
gle ACT is one third of the given angle ACB. 
his demonstration applies to every angle less than angle 
of 135°. The angle of 135° is known by the construction; 
and that the supplement is 45°, or a third of 135° anda 
quarter of 180° needs no proof. 
The curve being described and ACB being the angle to 
be trisected ; take the radius of the interior circle in the 
compasses, and from m, the point of intersection of the side 
A and the curve, intersect the arc EF. The point of in- 
tersection of radius CH and radius mH, each equal to the 
other, being H, through this point draw CG. The demon- 
an angle in a semi-circle, which is a right angle. Gm is 
therefore perpendicular to mC or x 
