304 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 hd 
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 cries 
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 MGN 
therefore, which is equal to GEC, is one half of “the angle 
ECL. 
The angle CGN being equal to LCP and also equal’ to 
ECG, 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 
is ig to one half of the whole composed of ACP and 
ACG 
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 left each equal to ACG ; that is ACG is 
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. 
The same demonstration will apply to any other angle. 
But it yet remains to prove, that in this part of the curve 
a perpendicular to AC will be the Sine of one third of the 
given angle. The proof is as follows. 
Join AF. The exterior angle AGF is equal to the two 
angles ACG and GAC, and these are equal to each other. 
The exterior angle AGC ‘is also equal to GAF and GFA, 
and these are equal toeach other. But AGC and AGF are 
together equal to two right angles ; therefore GAC and GAF 
are together equal to one right angle. FAC, composed of 
