I found how to determine the proportion of angles, and thus solve the trisection angles
Given the angle CAB
Divider AD (Point D is the free choice of the branch AB), from point A, creates a circular arc ED
Bisection circular arc ED obtained item H
Divider AD, from point D, obtained point L
Divider AD, from the point of L, we get the point F
Divider AF, from point A, creates a circular arc FG
Divider DH, from the point F, intersects a circular arc FG, obtained point I
Divider DH, referred to in Clause, cuts a circular arc FG, obtained point J
Divider FJ, from point J, cuts a circular arc FG, obtained point K
[tex] \angle GAK = \angle KAJ = \angle JAF = {\angle GAF \over 3} [/tex]
Next - my character
- Solution of the construction of a regular n (n> 2) of the polygon