# Morley's Theorem

Hubert Shutrick

For any triangle ABC, pairs of trisectors of it's angles meet in points, A', B' and C', as illustrated, that are vertices of an equilateral triangle.

To prove it, we consider how compass and straight edge constructions can be used. Since angles cannot be trisected, we have to assume that one third of each angle of the triangle is given, say \alpha, \beta and \gamma adding up to 60°. This suggests that we start with an equilateral triangle A'B'C', as size doesn't matter, and try to construct the triangle ABC round it with \angle CAB = 3\alpha and similarly for the other angles.

The locus of points A such that \angle C'AB' = \alpha as required is the arc of a circle through B' and C'. Any other chord of the same length of this arc will subtend the same angle. This suggests constructing the point B_{a} such that B'B_{a} = B'C' and \angle B'B_{a}C' = \alpha. What is evident is that B_{a} must lie on the side AC of the triangle if it exists and so must B_{c} constructed similarly using A'B' and \gamma. In general, there will be three pairs B_{a},B_{c}, A_{b},A_{c} and C_{a},C_{b} and the lines through them should be AC, BC and AB. Since the triangle B_{a}B'B_{c} is isosceles, the segment joining B' to the midpoint B_{m} of B_{a}B_{c} should be orthogonal to AC. In special cases it can happen that B_{a} = B_{c}, but then it is natural to take B_{m} = B_{a} = B_{c} and again draw AC at right angles to B'B_{m}.

The proof will be complete if we can show that \angle CAB constructed in this way is 3\alpha since the other two angles follow similarly. The angle between the constructed CA and BA is the same as between their orthogonals B'B_{m} and C'C_{m}.

The angle C'B'B_{m} is 150° + \gamma - \alpha because it is the mean of \angle C'B'B_{a} = 180° -2\alpha and \angle C'B'B_{c} = 360° - 60° - (180° - 2\gamma) and similarly C'B'C_{m} is 150° + \beta - \alpha. The angle we want is the other exterior angle of the triangle B'C'X where X is the intersect of B'B_{m} and C'C_{m} and it is 360° - (150° + \gamma - \alpha) - (150° + \beta - \alpha) = 60° -\beta - \gamma + 2\alpha = 3\alpha .