Morley's Theorem

Roger Smyth

Morley's theorem asserts that in the diagram below the triangle XYZ is equilateral, whatever ΔABC.

Morley's trisector theorem

In the next diagram, the unit of measure is the perpendicular DX, and the lengths of BX and CX are s and s'. E and F are points on BC where ∠CEX = 60° + β and ∠XFB = 60° + γ. P and P' are where BP = s and CP' = s' and S is constructed so that BS = s and SBX = 120°. This makes the two marked angles (TSB and SPX) 60°. The rest of the construction is hopefully as self-explanatory.

Morley's trisector theorem. A direct proof by Roger Smyth

What can be said of the diagram?

  1. α + β + γ = 60° (so ∠EXF = α)
  2. ∠CYA = ∠CLP' = 120° + β
  3. ∠TSB = ∠SPX = 60°
  4. AY = (AC/s')LP' (for, ΔCP'L is similar to ΔCAY)
  5. LP' = PS/ST (for ∠P'LY = ∠SPT = 60° - β)
  6. XR = RV = 2/s (for triangles BDX, PRX, PVR are similar)
  7. SU = s/2 (for ∠BSU = 60°)

Now, by (vi) and (vii),

2 ST = 2 SU + 2 UT = s + 2(s - 2/s) = 3s - 4/s.

Also, since triangles BQV and BDX are similar, VQ = 1 - 4/s² so

PQ = PV + VQ = 3 - 4/s²,

leading to 2 ST = s PQ.

Next, by (iv) and (v),

AY = (AC / s')(PS / ST) = 2 AC·PS / (ss' PQ).

But XE = 2s/PS as ΔPBS is isosceles and ∠SBP = 2(60° + β) = 2∠DEX. It follows that

XE·AY = 4 AC / (s' PQ)

and, by symmetry,

(XE·AY) / (XF·AZ) = [4 AC / (s' PQ)] [(s P'Q') / (4 AB)] = (s AC·P'Q') / (s' AB·PQ) = 1

because PQ(AB/s) = P'Q'(AC/s') is the altitude of ΔABC. However, this means AZ / AY = XE / XF which together with ∠ZAY = α = ∠EXF implies that triangles AZY and XEF are similar. Hence ∠YZA = ∠FEX = 60° + β and ∠AYZ = ∠XFE = 60° + γ. Analogous arguments for ΔBXZ and ΔCYX give ∠ZXB = 60° + γ, ∠CXY = 60° + β and ∠BZX = ∠CYX = 60° + α. All the angles in the diagram may now be deduced in terms of α, β, γ and it transpires that every angle of ΔXYZ is 60°.

References

  1. M. R. F. Smyth, MacCool's proof of Morley's Miracle, Irish Math. Soc. Bulletin 63 (2009), 63-66

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