Morley's Theorem

Trigonometric proof

First, we are going to deduce a trigonometric formula that will be very useful. Given any angle $$x$$, we have: $\begin{array}{lll} \sin x & = & \sin\left(\frac{2x}{3}+\frac{x}{3}\right)\\ & = & \sin\frac{2x}{3}\cos\frac{x}{3}+\sin\frac{x}{3}\cos\frac{2x}{3}\\ & = & \left(2\sin\frac{x}{3}\cos\frac{x}{3}\right)\cos\frac{x}{3}+\sin\frac{x}{3}\left(\cos^{2}\frac{x}{3}-\sin^{2}\frac{x}{3}\right)\\ & = & \sin\frac{x}{3}\left(2\cos^{2}\frac{x}{3}+\cos^{2}\frac{x}{3}-\sin^{2}\frac{x}{3}\right)\\ & = & \sin\frac{x}{3}\left(4\cos^{2}\frac{x}{3}-1\right)\\ & = & \sin\frac{x}{3}\left(4\cdot\frac{1+\cos\frac{2x}{3}}{2}-1\right)\\ & = & \sin\frac{x}{3}\left(2\cos\frac{2x}{3}+1\right)\\ & = & 2\sin\frac{x}{3}\left(\cos\frac{2x}{3}-\cos\frac{2\pi}{3}\right)\\ & = & 2\sin\frac{x}{3}\cdot2\sin\left(\frac{\pi}{3}+\frac{x}{3}\right)\sin\left(\frac{\pi}{3}-\frac{x}{3}\right)\\ & = & 4\sin\frac{x}{3}\sin\frac{\pi+x}{3}\sin\frac{\pi-x}{3}\\ & = & 4\sin\frac{x}{3}\sin\frac{\pi+x}{3}\sin\left(\pi-\frac{\pi-x}{3}\right)\\ & = & 4\sin\frac{x}{3}\sin\frac{x+\pi}{3}\sin\frac{x+2\pi}{3} \end{array}$

Let $$a=C\hat{A}B$$, $$b=A\hat{B}C$$, $$c=B\hat{C}A$$ and $$R$$ be the circumradius of $$[ABC]$$. Then, applying the law of sines to the triangle $$[ABC]$$, we have: $\overline{BC}=2R\sin a = 8R \sin \frac{a}{3} \sin \frac{a+\pi}{3} \sin \frac{a+2\pi}{3}$

Now consider the triangle $$[DBC]$$. The measures of their angles are given by: $D\hat{B}C=\frac{b}{3},$ $D\hat{C}B=\frac{c}{3}$ and $\begin{array}{lll} B\hat{D}C & = & \pi-\frac{b}{3}-\frac{c}{3}\\ & = & \frac{3\pi-b-c}{3}\\ & = & \frac{\left(\pi-b-c\right)+2\pi}{3}\\ & = & \frac{a+2\pi}{3} \end{array}$

Applying the law of sines to the triangle $$[DBC]$$, we have: $\begin{array}{lll} \frac{\overline{DC}}{\sin\frac{b}{3}}=\frac{\overline{BC}}{\sin\frac{a+2\pi}{3}}\Rightarrow\overline{DC} & = & \frac{\overline{BC}}{\sin\frac{a+2\pi}{3}}\cdot\sin\frac{b}{3}\\ & = & \frac{8R\sin\frac{a}{3}\sin\frac{a+\pi}{3}\sin\frac{a+2\pi}{3}}{\sin\frac{a+2\pi}{3}}\sin\frac{b}{3}\\ & = & 8R\sin\frac{a}{3}\sin\frac{b}{3}\sin\frac{a+\pi}{3} \end{array}$

Similarly, considering the triangle $$[AEC]$$, we have: $\overline{EC}= 8R \sin \frac{a}{3} \sin \frac{b}{3} \sin \frac{b+\pi}{3}$

Note that $\begin{array}{lll} \frac{a+\pi}{3}+\frac{b+\pi}{3}+\frac{c}{3} & = & \frac{a+\pi+b+\pi+c}{3}\\ & = & \frac{\left(a+b+c\right)+2\pi}{3}\\ & = & \frac{\pi+2\pi}{3}\\ & = & \pi \end{array}$ therefore it is possible to build a triangle $$[E'D'C']$$ with angle amplitudes $$\frac {a + \pi}{3}$$, $$\frac {b + \pi}{3}$$ and $$\frac {c}{3}$$. Further, assuming that its circumradius is given by $$4R \sin \frac{a}{3} \sin \frac{b}{3}$$, by the law of sines the adjacent sides of the angle with amplitude $$\frac{c}{3}$$ have the same lengths obtained for $$\overline{DC}$$ and $$\overline{EC}$$.

This triangle is congruent with the triangle $$[EDC]$$ because they have an angle and the corresponding adjacent sides equal, which is one of the congruence criteria for triangles. Hence, we have: $\overline{ED}=\overline{E'D'}=8R\sin \frac{a}{3} \sin \frac{b}{3} \sin \frac{c}{3}$