$54 On the Curves of Trisection. 
~The angle ACG, or HCF, is one third of the angle ACP, 
or HCB. ; 
Produce GC to L, and the opposite angles GCE and 
LCP are equal. Draw GN parallel with DB, and produce 
AG to M; and the angle ECG is equal to CGN, and MGN 
equal to GEC. 
‘The angle FGA being an exterior angle of the triangle 
AGC is equal to the angles ACG and GAC, and these are 
equal to each other, because GA is equal to GC. But the 
angle MGC is equal to FGA; the angle ACG is therefore 
one half of the angle MGC. ” 
Again, the angle GEC at the circumference is one half of 
the angle GCP, or ECL, at the centre; the angle MG 
therefore, which is equal to GEC, is one half of the angle 
ECL. 
The angle CGN being equal to LCP and also equal to 
CG, it is one half of these two angles together. 
Therefore the whole MGC, (composed of the angles 
MGN and NGC) is one half of the whole composed of 
ECL, LCP, and ECG. But this whole is the same as 
the two angles ACP and ACG; wherefore the angle MGC 
—— to one half of the whole composed of ACP and 
_ But ACG has been proved to be equal to one half of the 
angle MGC. It is therefore one quarter of the whole com- 
posed of ACP and ACG. Abstract this quarter, ACG, 
and three parts are lefteach equal to ACG ; that is ACGis 
one third of the given angle ACP, or HCB. Set off then 
the arc HF twice from H towards B, and through the points 
thus found draw straight lines from the centre, and the angle 
HCB is trisected. 
given angle. The proof is as follows. 
~Join AF. The exterior angle AGF is equal to the two 
are together equal to one right angle. FAC, composed of 
