[f.point (OP)]

FIRST PART
AB divider, circuit 1 (c pictured), AG (Analytical Geometry)
Ruler, the line referred to in Section A and B ( pictured) , AG provided the point D (intersection circle 1 and line )
Divider AB from point D cuts the line and afforded the point E
Bisection angle DAB, a point C
Ruler, the line referred to in Section B and C (b pictured), AG [latex]-x-y=-\sqrt{d}[/latex]
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[evan_au]

Trisection of an angle with straightedge & compass was proved impossible by Pierre Wantzel in 1837.

The reason is that these instruments cannot construct a cube root (on the other hand, today we can easily calculate cube roots on a calculator, but that is not permitted by the rules defined by the ancient Greek mathematicians).

See: https://en.wikipedia.org/wiki/Angle_trisection

I think I'll stick with Pierre....

[f.point (OP)]

Pierre Wantzel

a man he did not know well mathematics,

SECOND PART
Divider AB, at point A (arm angle rotates around point C), from point B to create a circle 2 (g pictured), AG
Divider BC, at point B, cut the circle 2, we get the point F
Ruler, line through the points A and F, AG
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[Bored chemist]

a man he did not know well mathematics,

It looks like some man doesn't understand mathematics.

You seem to be trying to draw a sphere with a compass.

[wolfekeeper]

There are many, many ways to trisect an angle, but you can't do it on a plane with a compass and straight edge.

[f.point (OP)]

PART THREE
Divider BF from point B, cuts line e is afforded point J
point G on circuit 1 (free choice),
ruler connect points A and G, we get along AG, we get the angle BAG
Divider GB, from the point B, we cut a circle 1, we get the point I
Divider GB, from the point I, we cut a circle1, we get the point H
Ruler join the dots G and J, we get along GJ
Ruler join the dots H and J, we get along JH, we get the angle GJH
angle GAB = angle GJH
Ruler merge point B and J, JB get along, we get the angle GJB
Ruler merge points I and J, we get along IJ, BJI get the angle, we get the angle IJH

ladies and gentlemen looking for a mistake ...
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[chiralSPO]

As far as I can tell, this is not trisecting anything. It is easy to add 3 equal angles to get one angle that is three times the size. This is very different from taking one angle, and (using only a compass and straight edge) cut it into three equal parts.

[f.point (OP)]

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



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Next - my character
- Solution of the construction of a regular n (n> 2) of the polygon

[f.point (OP)]

Divider DH, referred to in Clause, cuts a circular arc FG, obtained point J

should - Divider DH, from the point I, intersects a circular arc FG, obtained point J

[Bored chemist]

I'm not certain, but I think you have accidentally applied the small angle approximation.
In any even, your construction is wrong because it is known that no correct construction exists.

I would suggest making it clearer- but there's no point. The fact that I'm not sure what you have done doesn't matter because you have not trisected the angle.

Do you understand that this has been proven to be impossible?