Thank you all for your contributions. Special thanks to Ray Roussel and John Conway.
The theorem has to be amended to read;
"The angular trisectors of a triangle cut the circumcircle at 18 points which form 3 sets of 9 parallel lines centered on the circumcenter and parallel to the Morley system."
This array is rich in equilateral triangles. Much order and symmetry can be seen in the trisector indices. The intersection points of the Morley system correspond to secant lines and chords in this system. When we draw lines connecting various cuts, interesting things happen. I especially like the triangle formed by connecting the auxillary trisector cuts, a3a6-b3b6-c3c6. This forms a triangle similar to the original. Likewise, for a1a4-b1b4-c1c4. The system is centered on the circumcenter. The central triangle is b1c2-b3c4-b5c6. Curious that there is no third vertex involved for what should be a critical triangle.
The 54 equilateral triangles among the 108 intersection points, as charted by the Morley team at Abracadabri http://www-cabri.imag.fr/abracadabri/GeoPlane/Geo2DGene.html (lots of good stuff here), are all accounted for in this configuration. All equilateral triangles in the Morley system appear as equilateral triangles in this system, yet there are some equilateral triangles in this system that are not equilateral triangles in the Morley system.
There are 3 pairs of equilateral triangles on the circumcircle. These 3 pairs are special because when the circle is brought into play we only need 1 node (i.e., the origin of a 60 degree axis, e.g., (a1a3a5)) to produce an equilateral triangle on the circle. The chord connecting this vertex to the corresponding vertex of its companion triangle (b2b4b6), another 1 node creation, forms a line parallel to a Morley edge. These 9 chords in the 3 pairs of triangles form a miniature Morley grid of 3 sets of 3 parallel lines intersecting in sixes at 27 points. These triangles might be considered as pure lighthouse triangles or as "les etoiles pure", from theorems by Richard Guy and Andre Viricel.
For Divine reasons, trisections produce equilateral triangles. Sometimes the converse is also true. Notice that:
"In a circle, if an equilateral triangle is erected on the middle third of a trisected arc, then the other arc is also trisected."
This can be seen by extending the sides of the triangle to the circle to form a pair of equilateral triangles.
A quick 3 stroke trisection results from this statement.
Stroke 1: Draw a circle of radius R.
Stroke 2: With center on the circle and radius <sqr(3)R, draw an arc cutting the circle at A and B.
Stroke 3: With B as center and the same radius, draw an arc cutting the circle at C and D and cutting the arc AB at E.
The angle ACD is trisected by the angle ECD. The supplement of the trisected arc shows itself as the diagonal angle in the pure isoceles trapezoid (3 equal sides). This is what we have trisected. Of course you can NOT trisect a given angle.
This demonstrates that we can form pairs of equilateral triangles with any two points on a circle as base, as long as the base is < sqr(3)R. Our 18 cutting points provide us with many pairs of equilateral triangles. The pairs formed on the chords (a1a2-a4a5), (b1b2-b4b5), (c1c2-c4c5) have these chords parallel to the sides of the original triangle. As already mentioned, the triad a3a6, b3b6, c3c6 is parallel to the sides a,b,c of the original triangle and therefore, similar.
We may now have enough triangles to properly welcome in the centennary year of Frank Morley's observation. It is probably best to stop here, remarking that the original question is still open.