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.