Morley's Miracle
A Vector-based Proof

This proof by Cesare Donolato has appeared in Forum Geometricorum, 13 (2013) 233-235.

As in several other proofs, let the angles of $\Delta ABC$ be $3\alpha,$ $3\beta,$ $3\gamma$ so that $\alpha +\beta +\gamma =60^{\circ}.$ The adjacent angle trisectors meet to form Morley's triangle $PQR;$ $BR$ and $CQ$ extended meet at $D.$ Point $P$ is evidently the incenter of $\Delta BDC,$ implying that $DP$ is the bisector of the angle at D.

Lemma

The line $DP$ is perpendicular to the line $RQ.$

Proof of Lemma

Let $\mathbf{e}$ be the unit vector along $DP$ and $\mathbf{s}_1$ the vector representing $QR.$ We need to show that the scalar product $\mathbf{e}\cdot \mathbf{s}_1 =0.$ With a reference to the above diagram, this is equivalent to $(\mathbf{v}_{3}-\mathbf{v}_{2})\cdot\mathbf{e}=0.$

A Vector-based Proof of Morley's Trisector Theorem

In $\Delta BDC,$

$\begin{align} \angle D &= 180^{\circ} - 2\beta -2\gamma\\ &=180^{\circ} - 2(60^{\circ} -\alpha)\\ &=60^{\circ} +2\alpha . \end{align}$

Since $DP$ is the bisector of this angle, $\angle QDP=\angle RDP=30^{\circ}+\alpha.$ By the Exterior Angle Theorem, $\angle AQD=\alpha +\gamma$ and $\angle ARD=\alpha +\beta.$ The angle between vectors $\mathbf{v}_{3}$ and $\mathbf{e},$ being the difference between angles $RDP$ and $ARD,$ is $30^{\circ}-\beta.$ Similarly, the angle between vectors $\mathbf{v}_{2}$ and $\mathbf{e}$ is $30^{\circ}-\gamma.$ If so,

(1)

$\begin{align} (\mathbf{v}_{3}-\mathbf{v}_{2})\cdot\mathbf{e} &= \mathbf{v}_{3}\cdot\mathbf{e}-\mathbf{v}_{2}\cdot\mathbf{e}\\ &=v_{3}\cos (30^{\circ} - \beta)-v_{2}\cos (30^{\circ} - \gamma)\\ &=v_{3}\sin (60^{\circ} + \beta)-v_{2}\sin (60^{\circ} + \gamma). \end{align}$

where $v_{2}$ and $v_{3}$ are the magnitudes of vectors $(\mathbf{v}_{2}$ and $(\mathbf{v}_{3},$ respectively, and can be found with the Law of Sines in triangles $AQC$ and $ARB:$

$\begin{align}\displaystyle v_{2} &= \frac{c\sin\beta}{\sin(\alpha +\beta)}=\frac{c\sin\beta}{\sin(60^{\circ}-\gamma)},\\ v_{3} &= \frac{b\sin\gamma}{\sin(\alpha +\gamma)}=\frac{b\sin\gamma}{\sin(60^{\circ}-\beta)}. \end{align}$

Substituting these into (1) yields

$\begin{align}\displaystyle (\mathbf{v}_{3}-\mathbf{v}_{2})\cdot\mathbf{e} &= \frac{c\sin\beta\sin(60^{\circ}+\beta)}{\sin(60^{\circ}-\gamma)}-\frac{b\sin\gamma\sin(60^{\circ}+\gamma)}{\sin(60^{\circ}-\beta)}\\ &=\frac{c\sin\beta\sin(60^{\circ}+\beta)\sin(60^{\circ}-\beta)-b\sin\gamma\sin(60^{\circ}+\gamma)\sin(60^{\circ}-\gamma)}{\sin(60^{\circ}-\gamma)\sin(60^{\circ}-\beta)}\\ &=\frac{1}{4}\frac{c\sin 3\beta-b\sin 3\gamma}{\sin(60^{\circ}-\gamma)\sin(60^{\circ}-\beta)}, \end{align}$

where we used the identity $\displaystyle\sin x\sin(60^{\circ}-x)\sin(60^{\circ}+x)=\frac{1}{4}\sin 3x.$

But from the Law of sines in $\Delta ABC$ $c\sin 3\beta -b\sin 3\gamma=0,$ $(\mathbf{v}_{3}-\mathbf{v}_{2})\cdot\mathbf{e}=0,$ proving Lemma.

Proof of Morley's Theorem

$DP$ being perpendicular to $RQ,$ it divides the latter in half and $\Delta DQR$ into two congruent right triangles, implying $DQ=DR.$ It follows that triangles $DPQ$ and $DPR$ are also congruent (by SAS), and $s_{2}=s_{3}.$ Similarly, $s_{1}=s_{2},$ and, by transitivity, all three sides of $\Delta PQR$ are equal, completing the proof of Morley's theorem


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

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