74 Trisection of an Angle. [Frs. 
them, thatis, is double of DGB; but the angle BAD is equal to the 
angle BDA, because BD is equal to BA (5.1), therefore BAD, that is 
BAG, is double of DGB, that is, AGB. Wherefore the base AG is drawn 
so that the angle BAG, one of the angles at the base of the triangle 
ABG, is double of BGA, the other angle at the base. Which was re- 
quired to be done. 
Proposition 3rd, Problem. 
To divide any given rectilineal angle into three equal angles. 
Let ABC be any given rectilineal angle, it is required to divide it 
into three equal angles. 
Consider whether the given angle is greater or less than three half 
right angles. First, let the angle ABC be less than three half right an- 
gles. Takeany point Din AB, and through D draw DE, parallel to BC 
(31.1); then the angle BDE is the angle to be placed as the vertical 
angle of the triangle BDL; which having obtained, draw the line BL 
in the same manner as was done in the diagram for the foregoing pro- 
position No. 2; and bisect the angle ABL by the straight line BO 
(9.1). The straight lines BL and BO divide the angle ABC into three 
equal angles. 
Because DE is parallel to BC, and LB falls upon them, the angle 
DLB is equal to the angle LBC (29.1); and because the angle DBL is 
double of the angle DLB, as demonstrated in the 2nd proposition above 
written; thereforethe angle DBI. is double of the angle LBC; also 
because the angle DBL is bisected by the straight line BO, the three 
angles DBO, OBL, LBC are equal to one another. 
Secondly. But if the given angle ABC be greater than three half 
right angles, bisect it by the straight line BN (9.1); and take any 
point D in AB, and through D draw DE, parallel to BN (81.1). Hay- 
ing thus got the vertical angle for the triangle BDL, viz. BDE, draw 
the line BL, as was done in the diagram for the foregoing proposition 
No. 2; and bisect the angle LBC by the straight line BO (9.1); the 
straight lines BL and BO divide the given angle ABC into three equal 
angles. 
As before, because the angle DBL is double of the angle LBN, it is 
two thirds of the angle DBN ; but because the angle DBN is one-half of 
the angle ABC, and that two-thirds of the half is one-third of any given 
whole, therefore DBL is one-third, and the remaining angle LBC is 
two-thirds of the whole angle ABC; and because the angle LBC is 
bisected by the straight line BO, the three angles ABL, LBO, OBC 
are equal to one another. Wherefore the given angle ABC is divided 
into three equal angles by the straight lines BL and BO. Which was 
required to be done. 
