# Morley's Theorem, a Proof

### Brian Stonebridge 2009

In the diagram below, the near trisectors of the internal angles at the vertices $A,$ $B,$ and $C$ of a triangle meet in $X,$ $Y,$ and $Z.$ Morley's theorem states that the triangle $XYZ$ is equilateral.

Using the notation in the diagram, since, in $\Delta ABC, $3\alpha + 3\beta + 3\gamma = \pi ,$

$\alpha + \beta + \gamma = \pi /3.$

Start with an arbitrary equilateral triangle $XYZ.$.

Let $P,$ $Q,$ $R$ be point on the altitudes $\Delta XYZ$ (produced) such that

$\displaystyle\angle XPY=\angle XPZ=\alpha+\frac{\pi}{6},\\ \angle YQZ=\angle YQX=\beta+\frac{\pi}{6},\\ \angle ZRX=\angle ZRY=\gamma+\frac{\pi}{6}. $

Define $A$ to be the intersection of $QZ$ and $RY,$ $B$ the intersection of $RX$ and $PZ,$ and $C$ that of $PY$ and $QX.$ Then in quadrilateral $XRAQ,$ $\displaystyle\angle AQX=2\beta +\frac{\pi}{3},$ $\displaystyle\angle ARX=2\gamma +\frac{\pi}{3},$ and $\displaystyle\angle QXR=2\alpha +\beta +\gamma+\frac{\pi}{3}.$ It follows that $\angle ZAY=\alpha.$ Similarly, $\angle XBZ =\beta$ and $\angle YCX=\gamma.$

Draw circle with center $X$ touching $PB$ and, since $PX$ bisects $\angle BPC,$ it also touches $PC.$ Next, draw tangents $BT$ and $CU$, set $V$ as the intersection of the two lines. Then,

$\angle XBT=\angle XBZ=\beta$ and $\angle XCU =\angle XCY=\gamma.$

Now, the sum of angles $P,$ $B,$ and $C$ in quadrilateral $PBVC$ equals

$\displaystyle\angle QXR=2\alpha+\frac{\pi}{3} +2\beta +2\gamma=\pi,$

implying that $\angle TVU=0.$ In other words, $BTVUC$ is a straight line so that $T$ and $U$ coincide in $V.$ It follows that $\angle XBC=\beta$ and $\angle XCB=\gamma.$ In the same manner the angles of triangles $YCA$ and $ZAB$ are determined, letting one to conclude that triangle $ABC$ has angles $3\alpha,$ $3\beta,$ and $3\gamma.$ It may be scaled (if need be) to coincide with the original triangle.

- B. Stonebridge, A Simple Geometric Proof of Morley's Trisector Theorem, Applied Probability Trust, 2009

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