### 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.