Morley's trisector theorem

Remarks

Let $R$ be the circumradius and write $A$, $B$, $C$ also for the angles. The side of the Morley triangle is $8R \sin \frac{A}{3} \sin \frac{B}{3} \sin \frac{C}{3}$, which is symmetric in the angles; this length is one route to a trigonometric proof. A well-known proof runs backwards: start from an equilateral triangle, attach suitable triangles with angles built from $\frac{A}{3}$, $\frac{B}{3}$, $\frac{C}{3}$, and check that the result is similar to $ABC$.

Example: in an equilateral triangle all angles are $\pi/3$, and the figure is symmetric under rotation by $2\pi/3$ about the centre, so the three points form an equilateral triangle. The theorem says that this equilateral pattern survives for every triangle, with no symmetry.

Going deeper

For a student

The theorem is surprising because angle trisection is not a ruler-and-compass operation, see trisecting the angle, so classical synthetic geometry did not study trisectors. Proofs use trigonometry or a reverse construction, and Alain Connes gave an algebraic proof using the affine group of a field. It is a standard example of a deep-looking fact of triangle geometry with an elementary proof.