## Morley's Theorem

### Proof

Construct the parallelogram $$[ABCD]$$, their angle trisectors and the intersection points of adjacent angle trisectors, $$E$$, $$F$$, $$G$$ and $$H$$. Join these points consecutively to form a quadrilateral. As $$A\hat{D}E=C\hat{B}G$$, $$D\hat{A}E=B\hat{C}G$$ and $$\overline{AD}=\overline{BC}$$, the triangles $$[ADE]$$ and $$[CBG]$$ are congruent, $$\overline{AE}=\overline{CG}$$ and $$\overline{DE}=\overline{BG}$$. Similarly, as $$A\hat{B}F=C\hat{D}H$$, $$B\hat{A}F=D\hat{C}H$$ and $$\overline{AB}=\overline{CD}$$, the triangles $$[ABF]$$ and $$[CDH]$$ also are congruent. Then $$\overline{AF}=\overline{CH}$$ and $$\overline{BF}=\overline{DH}$$. As $$E\hat{A}F=G\hat{C}H$$, $$\overline{AE}=\overline{CG}$$ and $$\overline{AF}=\overline{CH}$$, the triangles $$[EAF]$$ and $$[GCH]$$ are congruent, $$\overline{EF}=\overline{GH}$$. Similarly, as $$F\hat{B}G=H\hat{D}E$$, $$\overline{BF}=\overline{DH}$$ and $$\overline{BG}=\overline{DE}$$, we have that the triangles $$[ FBG]$$ and $$[ HDE]$$ also are congruent, and then $$\overline{FG}=\overline{HE}$$. Therefore, the quadrilateral $$[EFGH]$$ is a parallelogram.

If $$[ABCD]$$ is a rectangle, then all its interior angles are right angles, so the amplitude of the angles obtained by its angles trisectors is $$30^{\circ}$$. So, $$[ AED]$$, $$[ AFB]$$, $$[ BGC]$$ and $$[ CHD]$$ are isosceles triangles, so that $$\overline{DE}=\overline{AE}=\overline{BG}=\overline{CG}$$ (they are sides of $$2$$ congruent isosceles triangles) and $$\overline{AF}=\overline{BF}=\overline{CH}=\overline{DH}$$ (they are also sides of $$2$$ congruent isosceles triangles). So, as $$E\hat{A}F=F\hat{B}G$$, $$\overline{AE}=\overline{BG}$$ and $$\overline{AF}=\overline{BF}$$, we conclude that $$[EAF]$$ and $$[FBG]$$ are congruent, and so $$\overline{FE}=\overline{FG}$$. Since $$[EAF]$$ is congruent with $$[GCH]$$, it follows that $$\overline{FE}=\overline{HG}$$ and, as $$[FBG]$$ is congruent with $$[HDE]$$, it follows that $$\overline{FG}=\overline{HE}$$. Therefore, the sides of the parallelogram $$[EFGH]$$ are all equal, that is, $$[EFGH]$$ is a rhombus.

If $$[ABCD]$$ is a rhombus, then $$\overline{AD}=\overline{AB}$$ and, as $$D\hat{A}E=F\hat{A}B$$ and $$A\hat{D}E=A\hat{B}F$$, we conclude that $$[ADE]$$ and $$[ABF]$$ are congruent, and $$\overline{AE}=\overline{AF}$$. As $$[ADE]$$ is congruent with $$[CBG]$$, we have that $$[ABF]$$ and $$[CBG]$$ also are congruent, and then $$\overline{BF}=\overline{BG}$$. Therefore, the triangles $$[EAF]$$ and $$[FBG]$$ are isosceles and thus the triangles $$[ GCH]$$ and $$[ HDE]$$ are also isosceles, since they are congruent with the previous one. Then we have that $$A\hat{E}D=A\hat{F}B=B\hat{G}C=C\hat{H}D$$ (they are angles of $$4$$ congruent triangles), $$A\hat{E}F=A\hat{F}E=C\hat{G}H=C\hat{H}G$$ (they are angles of $$2$$ congruent isosceles triangles) and $$B\hat{F}G=B\hat{G}F=D\hat{H}E=D\hat{E}H$$ (they are also angles of $$2$$ congruent isosceles triangles). Therefore, the interior angles of the parallelogram $$[EFGH]$$ are equal, since their amplitudes can be obtained subtracting measures of equal angles to $$360^{\circ}$$ (for example, $\begin{array}{ccl} E\hat{F}G & = & 360^{\circ}-A\hat{F}E-A\hat{F}B-B\hat{F}G\\ & = & 360^{\circ}-C\hat{G}H-C\hat{G}B-B\hat{G}F\\ & = & F\hat{G}H). \end{array}$ Moreover, the sum of the interior angles of any quadrilateral is $$360^{\circ}$$, therefore all interior angles of the parallelogram $$[EFGH]$$ are right angles, that is, $$[EFGH]$$ is a rectangle.