243 
THE TRISECTION OF ANY ANGLE 
Draw BD perpendicular to AG, and cut off BE along DA 
equal to DC. Join BE. 
Then, since BE = BC, 
¿BEG = BCE. 
But ¿BEG=ABAE+AEBA, 
and, by hypothesis, 
A F EGD C ' i. 
LBGA = 2 ZBAE. 
Therefore ¿BAE+l EBA = 2 ¿BAE; 
therefore Z BAE — Z ABE, 
or AE — BE. 
Divide AG at G so that AG = 2 GC, or GG = ^AG. 
Also let FE be made equal to ED, so that GD = i GF. 
It follows that GD = |-(AG—GF) = ^AF. 
Now BD 2 = BE 2 —ED 2 
= BE 2 ~EF 2 . 
Also DA. AF= AE 2 -EF 2 (Eucl. II. 6) 
= BE 2 —EF 2 . 
Therefore BD 2 = DA . AF 
= 3 AD. DG, from above, 
so that BD 2 : AD .DG =3:1 
= 3 AG 2 : AG 2 . 
Hence D lies on a hyperbola with AG as transverse axis 
and with conjugate axis equal to Vs . AG. 
Now suppose we are required 
to trisect an arc AB of a circle 
with centre 0. 
Draw the chord AB, divide it 
at C so that AG — 2 GB, and 
construct the hyperbola which 
has AG for transverse axis and 
a straight line equal to V3 . AG for conjugate axis. 
Let the hyperbola meet the circular arc in P. Join FA, 
DO,- PB.