1833.] Trisection of an Angle. 73 
base of the triangle ABG, is so drawn that BAG one of the angles at 
the base is double of BGA the other angle at the base. 
AD and DG are in the same straight line ; for through D draw DK, 
making the angles KDA, KDG equal to one another (9.1); take any 
point L in the line KD, and from the centre L at the distance LD 
describe the circle DNMK ; if necessary, produce DB to meet the circle 
DNMK in M; join MK; and make the angle MND in the segment 
DNM of the circle DNMK. Now the angles KDG, KDA are either 
together, equal to, or greater, or less than two right angles ; if greater, 
then KDG is greater than a right angle, and GD being produced in 
the direction of D, will fall within the circle DNMK on the opposite 
side of KD from DG (Cor. 16.3), which it does not, therefore KDG is 
not greater than a right angle: neither is it less than a right angle, for 
then DG would fall within the circle DNMK (16.3), which it does not; 
therefore KDG must be a right angle: and because at the point D the 
extremity of the diameter DK, DG makes a right angle with DK, 
therefore GD touches the circle DNMK (Cor. 16.3), and because DM, 
drawn from the point of contact D, cuts the circle DNMK, the angle 
MDG is equal to the angle DNM in the alternate segment MND 
(32.3). Again, because the angle KDA is equal to the angle KDG, it 
is: a right angle, and also touches the circle KMND (Cor. 16.3); and 
because DM, drawn from the point of contact D, cuts the circle DNMK, 
the angle MDA is equal to the angle MKD (82.3); and because 
KDNM is a quadrilateral figure, described in a circle, the opposite 
angles MKD, DNM are equal to two right angles (22.3), therefore 
the angles MDG, MDA, being equal to the angles DNM, MKD are 
also equal to two right angles ; and since at the point D, in the straight 
line MD or BD, the two straight lines DA, DG, upon the opposite sides 
of MD, make the adjacent angles MDA, MDG, equal to two right 
angles, AD is in the same straight line with DG (14.1) : and the figure 
AGB being contained by three straight lines, is therefore a rectilineal 
triangle (16 def.) Now BD is equal to BA, being radii of the same 
circle ADE (11 def.), and because BH is a radius of the circle of which 
FH is an arch, it is equal to BF and double of BA or BD, and bisected 
in D by the circle ADE; also because BGH is a right angle, subtended 
by BH, it is anangle in half the circle, having BH for its diameter and 
DB for its radius (31.3); and because GD joins the vertex of the 
right angle BGH and D, the point of bisection of the diameter, it is 
equal to DB (11 def.) Now, because DG is equal to DB, the angle 
DBG is equal to the angle DGB (5.1), and they are the two interior 
and opposite angles to BDA, the exterior angle of the triangle BDG, 
therefore BDA is equal to them both (32.1), and double of either of 
L 
