Dividing one angle into three equal angles seems a trivial problem. That is probably why it irked the Greeks so. Instead of being a simple problem, it is a complex, non-planar problem, as the Greeks soon discovered.
The trisection problem can probably credit its origin to the construction of regular polygons. The discover of the construction of a perfect pentagon(see The Golden Section) by the Pythagoreans would have roused great interest in the construction of other odd sided figures like the heptagon(which Archimedes succeeded at) and the 9-gon. If the mathematicians had the ability to divide any angle into any number of equal parts, all these figures would then become constructable. Thus, the problem began to attract a lot of attention. In the construction of a nine-gon, it is necessary to trisect a 120 deg.(2pi/3) angle. But, since a 90 degree(pi/2) angle can be trisected with the use of an equilateral triangle, it is only necessary to be able to trisect an acute angle of 30 degrees(pi/6). Okay, try it. No success? Well, surviving texts tell use that those great minds of the past did not have much success, either. Why? Because the problem was not one of the plane, but the solid.(Now there is a confusing statement. It may mean that the solution cannot be found using only a compass and straightedge, but I don't know for sure.) Thus, just as they were forced to turn to complex methods like curves for the squaring of the circle and doubling of the cube, so were they for trisecting an angle.
One of the earliest ways discovered was that of Hippias of Elis(circa 425 BC). Hippias used a curve he had invented, called the quadratrix. With this curve, the problem of trisecting an angle could be reduced to the trisection of a line segment. The following picture is one construction of such segment trisect. The great benefit of this method was that it could be generalized to divide any angle into any number of parts.
I don't really like this next solution, but maybe you will. This second method, perhaps the most well known of all, can be credited to Nicomedes(circa 180 BC). Nicomedes created a special device to use in his construction. As the upper part slide back and forth in its groove, the angle of the pointer changed so as to describe a curve known as a conchoid(as a function, y=K(x^2 + C)^(-1/2) is the simplest form).
In the above pictures, FH is made equal to 2AD and the conchoid represents all the segments that FH could be. What is desired is to have H on line ED, as in the pictures. Then, since FDH is a right triangle, all the red segments are equally(all radius's of the same circle are equal, see Euclid). Then, by another theorem of Euclid regarding inscribed angles, angle DKF and angle DAK(since ADK is an isosceles triangle) are twice as big as angle DHK and angle HAB(alternate interior angle theorem). Thus, angle HAB is one third angle DAB. Not the simplest solution(some not so intuitive logic), but it works.
Archimedes proposed method based on the following general theorem relating arcs and angles. Given an angle that has both sides intersecting a single circle at least once, the measure of the angle is 1/2 of the difference between the measures of the smaller central angle and the bigger central angle.
Thus, angle CAB is the one to be trisected. An arbitrary line is draw through C to intersect the circle at D. Then a circle sharing radius AD is constructed with center D. Line CD intersects this new circle at point E. Then, D is dragged around circle A, allowing E to trace the gray path. When E is on line AB, angle CEB is 1/3 of angle CAB, and all that remains is to construct line AJ parallel to line EC. Why is angle CEB 1/3 angle CAB? By the above theorem, angle CEB=1/2(angle CAB - angle DAE), and accordingly, 2CEB + DAE = CAB. But, since triangle ADE is an isosceles triangle, angle DAE= angle CEB. Thus, 3(angle CEB)= angle CAB.
The final way is one which no book used in this research mentions, but whose simplicity mandates acknowledgment. For a given angle, draw a circle centered at the vertex of that angle. Then, unroll the part of the circle which is on the inside of that angle.(Choose your own method. I prefer to paste mine to old mill stones and, using chalk to make the beginning and end, roll it down the sidewalk.) Trisect the unrolled segment, put it back in place(Once again, use your own method.), draw the appropriate rays to complete the task, and vale, trisected angle. The process may be able to be generalized through the use of Archimedes's spiral, but I don't know.
