## Morley's Theorem

### Proof

Let us consider two distinct cases.

1st Case

In this case, the area of Morley’s parallelogram can be obtained by subtracting the areas of the yellow and red triangles from the area of the initial parallelogram. Assuming that the measures of the sides of the initial parallelogram are $$a=\overline{AD}$$ and $$b=\overline{AB}$$ and form an angle with amplitude $$\alpha = D\hat{A}B$$, its area is equal to $$ab\sin \alpha$$. The area of the triangles is half of the product of the length of its two sides by the sine of the angle adjacent to these sides.

For example, to calculate the area of the triangle $$[ADE]$$ we need to know the length of the sides $$[AE]$$ and $$[DE]$$ and the amplitude of the angle $$\measuredangle AED$$. Using the law of sines we get: $D\hat{A}E=\frac{D\hat{A}B}{3}=\frac{\alpha}{3}$ $A\hat{D}E=\frac{A\hat{D}C}{3}=\frac{\pi-\alpha}{3}$ $A\hat{E}D=\pi-D\hat{A}E-A\hat{D}E=\pi-\frac{\alpha}{3}-\frac{\pi-\alpha}{3}=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$ $\begin{array}{lll} \frac{\overline{AD}}{\sin A\hat{E}D}=\frac{\overline{AE}}{\sin A\hat{D}E}\Rightarrow\overline{AE} & = & \overline{AD}\centerdot\frac{\sin A\hat{D}E}{\sin A\hat{E}D}\\ & = & a\frac{\sin\frac{\pi-\alpha}{3}}{\sin\frac{2\pi}{3}}\\ & = & \frac{2a}{\sqrt{3}}\sin\frac{\pi-\alpha}{3} \end{array}$ $\begin{array}{lll} \frac{\overline{AD}}{\sin A\hat{E}D}=\frac{\overline{DE}}{\sin D\hat{A}E}\Rightarrow\overline{DE} & = & \overline{AD}\centerdot\frac{\sin D\hat{A}E}{\sin A\hat{E}D}\\ & = & a\frac{\sin\frac{\alpha}{3}}{\sin\frac{2\pi}{3}}\\ & = & \frac{2a}{\sqrt{3}}\sin\frac{\alpha}{3} \end{array}$

$\begin{array}{lll} Area\left(\left[ADE\right]\right) & = & \frac{1}{2}\overline{AE}\centerdot\overline{DE}\sin A\hat{E}D\\ & = & \frac{1}{2}\left(\frac{2a}{\sqrt{3}}\sin\frac{\pi-\alpha}{3}\right)\left(\frac{2a}{\sqrt{3}}\sin\frac{\alpha}{3}\right)\sin\frac{2\pi}{3}\\ & = & \frac{2a^{2}}{3}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\frac{\sqrt{3}}{2}\\ & = & \frac{a^{2}}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3} \end{array}$

Similarly, we have: $Area \left(\left[ADE\right]\right)=\frac{b^{2}}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}$

Notice that the triangles $$[ADE]$$ and $$[CBG]$$ are congruent, as well as the triangles $$[ABF]$$ and $$[CDH]$$. Therefore, the sum of the areas of the yellow triangles is: $\begin{array}{l} Area\left(\left[ADE\right]\right)+Area \left(\left[ABF\right]\right)+Area\left(\left[CBG\right]\right)+Area\left(\left[CDH\right]\right)=\\ \;=2\centerdot\frac{a^{2}}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}+2\centerdot\frac{b^{2}}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\\ \;=\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3} \end{array}$

Considering now the red triangles and using the fact that the triangles $$[EAF]$$ and $$[GCH]$$ are congruent, as are also the triangles $$[FBG]$$ and $$[HDE]$$, we get: $\begin{array}{lll} Area\left(\left[GCH\right]\right) & = & Area\left(\left[EAF\right]\right)\\ & = & \frac{1}{2}\overline{AE}\cdot\overline{AF}\cdot\sin E\hat{A}F\\ & = & \frac{1}{2}\left(\frac{2a}{\sqrt{3}}\sin\frac{\pi-\alpha}{3}\right)\left(\frac{2b}{\sqrt{3}}\sin\frac{\pi-\alpha}{3}\right)\sin\frac{\alpha}{3}\\ & = & \frac{2}{3}ab\sin^{2}\frac{\pi-\alpha}{3}\sin\frac{\alpha}{3} \end{array}$ $\begin{array}{lll} Area\left(\left[HDE\right]\right) & = & Area\left(\left[FBG\right]\right)\\ & = & \frac{1}{2}\overline{BF}\cdot\overline{BG}\cdot\sin F\hat{B}G\\ & = & \frac{1}{2}\left(\frac{2b}{\sqrt{3}}\sin\frac{\alpha}{3}\right)\left(\frac{2a}{\sqrt{3}}\sin\frac{\alpha}{3}\right)\sin\frac{\pi-\alpha}{3}\\ & = & \frac{2}{3}ab\sin^{2}\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3} \end{array}$

Using the formula $$\sin x = 4 \sin \frac{x}{3} \sin \frac{x+\pi}{3} \sin \frac{x+2\pi}{3}$$ deduced in the trigonometric proof of Morley's theorem, we conclude that the sum of the areas of the red triangles is: $\begin{array}{l} \;Area\left(\left[EAF\right]\right)+Area\left(\left[FBG\right]\right)+Area\left(\left[GCH\right]\right)+Area\left(\left[HDE\right]\right)=\\ =2\cdot\frac{2}{3}ab\sin^{2}\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}+2\cdot\frac{2}{3}ab\sin^{2}\frac{\pi-\alpha}{3}\sin\frac{\alpha}{3}\\ =\frac{4}{3}ab\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\left(\sin\frac{\alpha}{3}+\sin\frac{\pi-\alpha}{3}\right)\\ =\frac{4}{3}ab\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\cdot2\sin\frac{\frac{\alpha}{3}+\frac{\pi-\alpha}{3}}{2}\cdot\cos\frac{\frac{\alpha}{3}.\frac{\pi-\alpha}{3}}{2}\\ =\frac{8}{3}ab\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\sin\frac{\pi}{6}\cdot\cos\frac{2\alpha-\pi}{6}\\ =\frac{8}{3}ab\sin\frac{\alpha}{3}\sin\left(\pi-\frac{\pi-\alpha}{3}\right)\cdot\frac{1}{2}\cdot\sin\left(\frac{\pi}{2}+\frac{2\alpha-\pi}{6}\right)\\ =\frac{4}{3}ab\sin\frac{\alpha}{3}\sin\frac{\alpha+2\pi}{3}\sin\frac{\alpha+\pi}{3}\\ =\frac{1}{3}ab\cdot4\sin\frac{\alpha}{3}\sin\frac{\alpha+\pi}{3}\sin\frac{\alpha+2\pi}{3}\\ =\frac{1}{3}ab\sin\alpha \end{array}$

(that is, the sum of the areas of the red triangles is exactly one-third of the area of the parallelogram $$[ABCD]$$).

We can now calculate the area of the parallelogram $$[EFGH]$$: $\begin{array}{lll} Area\left(\left[EFGH\right]\right) & = & ab\sin\alpha-\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}-\frac{1}{3}ab\sin\alpha\\ & = & \frac{2}{3}ab\sin\alpha-\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3} \end{array}$

2nd Case

In this case, the sum of the areas of the triangles $$[ADE]$$, $$[CBG]$$, $$[ABF]$$, $$[CDH]$$, $$[EAF]$$, $$[GCH]$$, $$[FBG]$$ and $$[HDE]$$ exceeds the area of the parallelogram $$[ABCD]$$ and the excess is exactly equal to the area of the parallelogram $$[EFGH]$$. Furthermore, it is easy to see that the values obtained in the previous case for the areas of the referred triangles remain valid. Hence: $\begin{array}{lll} Area\left(\left[EFGH\right]\right) & = & \frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}+\frac{1}{3}ab\sin\alpha-ab\sin\alpha\\ & = & \frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}-\frac{2}{3}ab\sin\alpha \end{array}$

In both cases we have: $\begin{array}{lll} Area\left(\left[EFGH\right]\right) & = & \left|\frac{2}{3}ab\sin\alpha-\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\right|\\ & = & \left|\frac{2}{3}ab\cdot4\sin\frac{\alpha}{3}\sin\frac{\alpha+\pi}{3}\sin\frac{\alpha+2\pi}{3}-\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\right|\\ & = & \left|\frac{8}{3}ab\sin\frac{\alpha}{3}\sin\frac{\alpha+\pi}{3}\sin\frac{\pi-\alpha}{3}-\frac{2}{\sqrt{3}}\left(a^{2}+b^{2}\right)\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\right|\\ & = & \frac{2}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right| \end{array}$

With respect to the area of the parallelogram $$[IJKL]$$, we note first that the triangles $$[EAF]$$ and $$[LAK]$$ are similar. In fact, we have: $E\hat{A}F=L\hat{A}K$ $\overline{AE}=\frac{2a}{\sqrt{3}}\sin\frac{\pi-\alpha}{3}$ $\overline{AF}=\frac{2b}{\sqrt{3}}\sin\frac{\pi-\alpha}{3}$ $\overline{AL}=\frac{2b}{\sqrt{3}}\sin\frac{2\left(\pi-\alpha\right)}{3}$ $\overline{AK}=\frac{2a}{\sqrt{3}}\sin\frac{2\left(\pi-\alpha\right)}{3}$ $\begin{array}{lll} \frac{\overline{AK}}{\overline{AE}} & = & \frac{\overline{AL}}{\overline{AF}}\\ & = & \frac{\sin\frac{2\left(\pi-\alpha\right)}{3}}{\sin\frac{\pi-\alpha}{3}}\\ & = & \frac{2\sin\frac{\pi-\alpha}{3}\cos\frac{\pi-\alpha}{3}}{\sin\frac{\pi-\alpha}{3}}\\ & = & 2\cos\frac{\pi-\alpha}{3} \end{array}$

Similarly, the triangles $$[EDH]$$ and $$[JDK]$$ are also similar, since we have: $E\hat{D}H=J\hat{D}K$ $\overline{DE}=\frac{2a}{\sqrt{3}}\sin\frac{\alpha}{3}$ $\overline{DH}=\frac{2b}{\sqrt{3}}\sin\frac{\alpha}{3}$ $\overline{DJ}=\frac{2b}{\sqrt{3}}\sin\frac{2\alpha}{3}$ $\overline{DK}=\frac{2a}{\sqrt{3}}\sin\frac{2\alpha}{3}$ $\begin{array}{lll} \frac{\overline{DK}}{\overline{DE}} & = & \frac{\overline{DJ}}{\overline{DH}}\\ & = & \frac{\sin\frac{2\alpha}{3}}{\sin\frac{\alpha}{3}}\\ & = & \frac{2\sin\frac{\alpha}{3}\cos\frac{\alpha}{3}}{\sin\frac{\alpha}{3}}\\ & = & 2\cos\frac{\alpha}{3} \end{array}$

Using the congruence between the triangles$$[BGF]$$ and $$[DEH]$$ and the similarity of triangles $$[EAF]$$ and $$[LAK]$$ and of triangles $$[EDH]$$ and $$[JDK]$$, we have: $I\hat{F}G=G\hat{B}F+B\hat{G}F$ $E\hat{F}K=E\hat{A}F+A\hat{E}F$ $\begin{array}{lll} A\hat{F}B & = & \pi-B\hat{A}F-A\hat{B}F\\ & = & \pi-\frac{1}{3}\left(B\hat{A}B+A\hat{B}C\right)\\ & = & \pi-\frac{\pi}{3}\\ & = & \frac{2\pi}{3} \end{array}$ $G\hat{B}F + E\hat{A}F = \frac{1}{3} \left( C\hat{B}A + D\hat{A}B \right) = \frac{\pi}{3}$ $B\hat{G}F=D\hat{E}H=D\hat{K}J$ $A\hat{E}F=A\hat{K}L$ $\begin{array}{ccl} E\hat{F}G & = & I\hat{F}G+E\hat{F}K-I\hat{F}K\\ & = & \left(G\hat{B}F+B\hat{G}F\right)+\left(E\hat{A}F+A\hat{E}F\right)-A\hat{F}B\\ & = & \left(G\hat{B}F+E\hat{A}F\right)+\left(B\hat{G}F+A\hat{E}F\right)-\frac{2\pi}{3}\\ & = & \frac{\pi}{3}+\left(D\hat{K}J+A\hat{K}L\right)-\frac{2\pi}{3}\\ & = & \left(D\hat{K}J+A\hat{K}L\right)-\frac{\pi}{3}\\ & = & A\hat{K}L+D\hat{K}J-A\hat{K}D\\ & = & J\hat{K}L \end{array}$ $\begin{array}{ccc} \overline{KL} & = & \frac{\overline{AK}}{\overline{AE}}\cdot\overline{FE}\\ & = & 2\cos\frac{\pi-\alpha}{3}\cdot\overline{FE} \end{array}$ $\begin{array}{ccc} \overline{KJ} & = & \frac{\overline{DK}}{\overline{DE}}\cdot\overline{HE}\\ & = & 2\cos\frac{\pi-\alpha}{3}\cdot\overline{HE}\\ & = & 2\cos\frac{\pi-\alpha}{3}\cdot\overline{FG} \end{array}$

Therefore we obtain: $\begin{array}{lll} Area \left(\left[IJKL\right]\right) & = & \overline{KL}\cdot\overline{KJ}\cdot\sin J\hat{K}L\\ & = & 2\cos\frac{\pi-\alpha}{3}\cdot\overline{FE}\cdot2\cos\frac{\pi-\alpha}{3}\cdot\overline{FG}\cdot\sin E\hat{F}G\\ & = & 4\cos\frac{\alpha}{3}\cos\frac{\pi-\alpha}{3}\cdot\left(\overline{FE}\cdot\overline{FG}\cdot\sin E\hat{F}G\right)\\ & = & 4\cos\frac{\alpha}{3}\cos\frac{\pi-\alpha}{3}\cdot Area\left(\left[EFGH\right]\right)\\ & = & 4\cos\frac{\alpha}{3}\cos\frac{\pi-\alpha}{3}\cdot\frac{2}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right|\\ & = & \frac{2}{\sqrt{3}}\left(2\sin\frac{\alpha}{3}\cos\frac{\alpha}{3}\right)\left(2\sin\frac{\pi-\alpha}{3}\cos\frac{\pi-\alpha}{3}\right)\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right|\\ & = & \frac{2}{\sqrt{3}}\sin\frac{2\alpha}{3}\sin\frac{2\left(\pi-\alpha\right)}{3}\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right| \end{array}$

Conclusion:

$Area\left(\left[EFGH\right]\right)=\frac{2}{\sqrt{3}}\sin\frac{\alpha}{3}\sin\frac{\pi-\alpha}{3}\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right|$ $Area\left(\left[IJKL\right]\right)=\frac{2}{\sqrt{3}}\sin\frac{2\alpha}{3}\sin\frac{2\left(\pi-\alpha\right)}{3}\left|\frac{4}{\sqrt{3}}ab\sin\frac{\alpha+\pi}{3}-\left(a^{2}+b^{2}\right)\right|$

Note that the ratio of the areas (assuming both nonzero) is: $\begin{array}{lll} r & = & \frac{Area \left(\left[IJKL\right]\right)}{Area\left(\left[EFGH\right]\right)}\\ & = & 4\cos\frac{\alpha}{3}\cos\frac{\pi-\alpha}{3}\\ & = & 4\cdot\frac{1}{2}\left(\cos\frac{\alpha-\left(\pi-\alpha\right)}{3}+\cos\frac{\alpha+\left(\pi-\alpha\right)}{3}\right)\\ & = & 2\left(\frac{1}{2}+\cos\frac{\pi-2\alpha}{3}\right)\\ & = & 1+2\cos\frac{\pi-2\alpha}{3},\mbox{ with }\alpha\in]0,\pi[. \end{array}$

Therefore, we have $$2<r \leq 3$$, that is, the parallelogram $$[IJKL]$$ has an area which is always greater than twice the area of the parallelogram $$[EFGH]$$ and never higher than its triple, which is reached when $$\alpha=\frac{\pi}{2}$$. We note also that, as $$J\hat{K}L=E\hat{F}G$$, $$\frac{\overline{KL}}{\overline{FE}}=2\cos \frac{\alpha}{3}$$ and $$\frac{\overline{KJ}}{\overline{FG}}=2\cos \frac{\pi-\alpha}{3}$$, the parallelograms $$[IJKL]$$ and $$[EFGH]$$ are similar when $$2\cos \frac{\alpha}{3} = 2\cos \frac{\pi - \alpha}{3}$$, that is, when $$\alpha = \frac{\pi}{2}$$. When this happens, the similarity ratio is $$2\cos \frac{\pi}{6}=\sqrt{3}$$, whereby the ratio between the areas is $$(\sqrt{3})^{2}=3$$, as indicated.