I want to preface this by saying I have been trying for a while to prove myself wrong because my results appear to contradict the work of some previous work by people who have studied much more than me. Anyway. I have what I believe to be trisection of a line which is easily extended to trisection of an arbitrary angle. I have even backed up my work algebraically. Please tell me whether I am right or wrong as it would take a heavy pressure off of myself of being conflicted on my correctness. Thank you in advance.
Here is the paper that contains my work.
Immediate Edit: If I missed something small here please don't downvote, just tell me and I'll delete. I really am driving myself crazy here through.
One can trisect a given segment with a compass and straightedge (one method is to create a triangle with that segment as the median, as quasi suggested, and use the fact that a triangle with median $AM$ and centroid $G$ has $\frac{GM}{AM} = \frac{1}{3}$). However, this does not imply that one can trisect an angle.
For a simple example, consider a 45-45-90 triangle:
where we have trisected the hypotenuse. The areas of the three smaller red triangles we have must all be equal to each other. However, in a triangle with two sides $b$,$c$ and angle $A$ between them, the area is
$$\frac{bc\sin(A)}{2}.$$
Consider the leftmost red triangle. If this length trisection indeed gives an angle trisection, we must have that the product of its non-horizontal sides equals the product of those for the middle red triangle, which would imply that the length of the blue segments equal the length of the green segments. However, this is false by the Pythagorean theorem (if we drop an altitude from the right-angled vertex, we see that one leg of the triangle with blue hypotenuse is the same as the corresponding leg of the triangle with green hypotenuse, but the other legs are of different lengths). Thus, this length trisection doesn't necessarily give an angle trisection, even though it may look that way.