proof of Morley’s theorem

The scheme of this proof, due to A. Letac, is to use the sines law to get formulas for the segments AR, AQ, BP, BR, CQ, and CP, and then to apply the cosines law to the triangles ARQ, BPR, and CQP, getting RQ, PR, and QP.

To simplify some formulas, let us denote the angle π/3, or 60 degrees, by σ. Denote the angles at A, B, and C by 3a, 3b, and 3c respectively, and let R be the circumradiusMathworldPlanetmath of ABC. We have BC=2Rsin(3a). Applying the sines law to the triangle BPC,

BP/sin(c) = BC/sin(π-b-c)=2Rsin(3a)/sin(b+c) (1)
= 2Rsin(3a)/sin(σ-a) (2)



Combining that with the identity


we get




Using the cosines law now,


But we have


whence the cosines law can be applied to those three angles, getting




Since this expression is symmetricMathworldPlanetmath in a, b, and c, we deduce


as claimed.

Remarks: It is not hard to show that the triangles RYP, PZQ, and QXR are isoscoles.

By the sines law we have

ARsinb=BRsina  BPsinc=CPsinb  CQsina=AQsinc



This implies that if we identify the various vertices with complex numbersMathworldPlanetmathPlanetmath, then


provided that the triangle ABC has positive orientation, i.e.


I found Letac’s proof at, with the reference Sphinx, 9 (1939) 46. Several shorter and prettier proofs of Morley’s theorem can also be seen at cut-the-knot.

Title proof of Morley’s theorem
Canonical name ProofOfMorleysTheorem
Date of creation 2013-03-22 13:45:44
Last modified on 2013-03-22 13:45:44
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 6
Author mathcam (2727)
Entry type Proof
Classification msc 51M04