Many proofs of Morley's Theorem start with an equilateral triangle and then proceed to show that given arbitrary positive angles that sum to 180 degrees, a unique triangle can be constructed such that the original given triangle is the intersection of the angle trisectors. However, direct proofs can also be given using trigonometry.

It can be shown that using only a straightedge and a compass not every angle can be trisected. For example a 60 degree triangle cannot be trisected with only a straightedge and a compass. Thus it is not possible to construct an arbitrary Morley equilateral triangle using a straightedge and a compass. However, if two marks are placed on the straightedge, then every Morley equilateral triangle can be constructed.

The theorem can be generalized by using angle trisectors of the interior and exterior angles. This can be done in two ways. The following way yields three equilateral triangles one lying off of each side. To obtain the three new equilateral triangles we trisect two exterior angles and the other interior angle.

The following way gives one additional central equilateral triangle. To obtain the additional new central equilateral triangle, we trisect all three exterior angles.

We now show all five equilateral triangles.

Since Morley's Theorem is true for a triangle, one might wonder whether it is true for a general quadrilateral. That is perhaps we always get a rhombus for instance. The answer is no. Even if the vertices of the quadrilateral lie on a circle, the answer is still no.