Morley's Miracle
Leo Giugiuc's Proof

The three points of intersection of the adjacent trisectors of the angles of any triangle form an equilateral triangle.

Morley's theorem - statement

In $\Delta QAC,$ we have $\displaystyle\frac{AQ}{\sin\frac{C}{3}}=\frac{AC}{\sin\frac{A+C}{3}};$ but $AC=2R\cdot\sin B,$ where $R$ is the circumradius of $\Delta ABC.$ It follows that

$\displaystyle\frac{AQ}{\sin\frac{C}{3}}=\frac{2R\cdot\sin B}{\sin\frac{A+C}{3}}=8R\cdot\sin\frac{B}{3}\sin\frac{A+C}{3}\sin\frac{\pi +B}{3},.$

implying $\displaystyle AQ=8R\cdot\sin\frac{B}{3}\sin\frac{C}{3}\sin\frac{\pi +C}{3}.$ But the numbers $\displaystyle\frac{A}{3},$ $\displaystyle\frac{\pi +B}{3},$ and $\displaystyle\frac{\pi +C}{3}$ are strictly positive and their sum is $\pi,$ so there is a triangle $XYZ,$ with $\displaystyle X=\frac{A}{3},$ $\displaystyle Y=\frac{\pi +B}{3},$ and $\displaystyle Z=\frac{\pi +C}{3}$ and the circumradius of $\displaystyle\frac{1}{2}.$

In $\Delta XYZ,$ $\displaystyle YZ=\sin\frac{A}{3},$ $\displaystyle XZ=\sin\frac{\pi +B}{3},$ and $\displaystyle XY=\sin\frac{\pi +C}{3};$ but $\displaystyle\frac{AQ}{XZ}=\frac{AR}{XY}=8R\cdot\sin\frac{B}{3}\cdot\sin\frac{C}{3}$ and $\displaystyle\angle QAR=X=\frac{A}{3},$ so that $\Delta AQR$ is similar to $\Delta XZY$ and, therefore, $\displaystyle\frac{QR}{YZ}=8R\cdot\sin\frac{B}{3}\cdot\sin\frac{C}{3}$ which finally gives $\displaystyle QR=8R\cdot\sin\frac{A}{3}\cdot\sin\frac{B}{3}\cdot\sin\frac{C}{3},$ a symmetric function of the angles of $\Delta ABC.$ This leads to $QR=PQ=PR.$


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