Morley's theorem asserts that in the diagram below the triangle XYZ is equilateral, whatever ΔABC.
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
What can be said of the diagram?
- α + β + γ = 60° (so ∠EXF = α)
- ∠CYA = ∠CLP' = 120° + β
- ∠TSB = ∠SPX = 60°
- AY = (AC/s')LP' (for, ΔCP'L is similar to ΔCAY)
- LP' = PS/ST (for ∠P'LY = ∠SPT = 60° - β)
- XR = RV = 2/s (for triangles BDX, PRX, PVR are similar)
- 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,
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
- M. R. F. Smyth, MacCool's proof of Morley's Miracle, Irish Math. Soc. Bulletin 63 (2009), 63-66
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