Morley's Theorem

Morley's theorem asserts that in the diagram below $\Delta XYZ$ is equilateral, whatever $\Delta ABC.$

Morley's trisector theorem

What follows has been communicated to me in private correspondence by Roger Smyth. Roger has based his new direct proof on the following


Assume point $G$ in the interior of $\Delta DEF,$ with $\angle FED = 3\alpha,$ satisfies $\angle FEG=\alpha$ and $DE=EG.$

Lemma for a proof of Morley's trisector theorem

Then $DG=FG$ iff $\angle DFE$ equals $60^{\circ}-\alpha.$


Assume $DG=FG$ and let $H$ be the foot of the perpendicular from $G$ to $EF.$

Proof of Lemma towards Morley's trisector theorem

Since $GF = GD = 2GH$ it follows that $\angle GFE = 30^{\circ}.$ Thus, for the reflex $\angle DGF,$

$\begin{align} \angle DGF &= \angle DGE +\angle EGH + \angle FGH \\ &= 2(90^{\circ}-\alpha) + 60^{\circ} \\ &= 240^{\circ} -2\alpha. \end{align}$

Since we assumed $\Delta DGF$ isosceles, $\angle DFG =\angle FDG=30^{\circ}-\alpha;$ therefore, $\angle DFE=60^{\circ}-\alpha,$ as required.

Proof of Morley's Theorem

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

Above (as usual) $X$ is the Morley vertex adjacent to $BC$ and $P$ and $Q$ are the points on $AB$ and $AC$ such that $BP = BX$ and $CQ = CX.$ The circle through $A,$ $P,$ $Q$ is drawn and the angle trisectors from $A$ cut it at $Z$ and $Y.$ Thus the arcs $PZ,$ $ZY,$ $YQ$ are equal. Let the centre of the circle be $O.$ Since $\Delta XPQ$ is isosceles, the perpendicular bisector of $YZ$ goes through both $O$ and $X.$ Moreover, $\angle XOQ = 3\alpha$ and $\angle XOY = \alpha$ and $OY = OQ,$ so the conditions of the lemma are met in $\Delta QOX.$ $(Q \leftrightarrow D,$ $O \leftrightarrow E,$ $X \leftrightarrow F,$ $Y \leftrightarrow G).$

Now comes the crunch. The Morley point next to $A$C lies at the intersection of two trisectors, one of which is $AY$ and the other is also the perpendicular bisector of $QX.$ But $\angle QXO = \frac{1}{2}\angle QXP = 60^{\circ}-\alpha$ and, according to the lemma, this means that $QY=XY;$ thus the perpendicular bisector of $QX$ actually goes through $Y.$ So our search for the Morley point is over almost before it has begun! By symmetry $Z$ is the Morley vertex adjacent to $AB$ and $\Delta XYZ$ is equilateral because $PZ = ZX = XY = YQ = YZ.$

Morley's Miracle

On Morley and his theorem

  1. Doodling and Miracles
  2. Morley's Pursuit of Incidence
  3. Lines, Circles and Beyond
  4. On Motivation and Understanding
  5. Of Looking and Seeing

Backward proofs

  1. J.Conway's proof
  2. D. J. Newman's proof
  3. B. Bollobás' proof
  4. G. Zsolt Kiss' proof
  5. Backward Proof by B. Stonebridge
  6. Morley's Equilaterals, Spiridon A. Kuruklis' proof
  7. J. Arioni's Proof of Morley's Theorem

Trigonometric proofs

  1. Bankoff's proof
  2. B. Bollobás' trigonometric proof
  3. Proof by R. J. Webster
  4. A Vector-based Proof of Morley's Trisector Theorem
  5. L. Giugiuc's Proof of Morley's Theorem
  6. Dijkstra's Proof of Morley's Theorem

Synthetic proofs

  1. Another proof
  2. Nikos Dergiades' proof
  3. M. T. Naraniengar's proof
  4. An Unexpected Variant
  5. Proof by B. Stonebridge and B. Millar
  6. Proof by B. Stonebridge
  7. Proof by Roger Smyth
  8. Proof by H. D. Grossman
  9. Proof by H. Shutrick
  10. Original Taylor and Marr's Proof of Morley's Theorem
  11. Taylor and Marr's Proof - R. A. Johnson's Version
  12. Morley's Theorem: Second Proof by Roger Smyth
  13. Proof by A. Robson

Algebraic proofs

  1. Morley's Redux and More, Alain Connes' proof

Invalid proofs

  1. Bankoff's conundrum
  2. Proof by Nolan L Aljaddou
  3. Morley's Theorem: A Proof That Needs Fixing

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny