The official solution uses complex number Isn't there any synthetic proof??? I am really interested to see a very elegant one...

Denote $ \mathcal N_{PQR}$ the 9-point circle of a $ \triangle PQR.$ Consider quadrilaterals $ ABCE, ABDF.$ The 9-point circles $ \mathcal N_{ABC}, \mathcal N_{BCE}, \mathcal N_{CEA}, \mathcal N_{EAB}$ concur at a point $ X$ and the 9-point circles $ \mathcal N_{ABD}, \mathcal N_{BDF}, \mathcal N_{DFA}, \mathcal N_{FAB}$ concur at a point $ Y$ [1]. $ ABCE, ABDF$ are cyclic with a common circumcircle $ \mathcal O \Longrightarrow$ all 9-point circles are half the size of $ \mathcal O$ and congruent. Moreover, $ X, Y$ are anticenters of $ ABCE, ABDF,$ respectively, i.e., reflections of the circumcenter $ O$ in their respective centroids $ G, H.$ In addition, Simson lines $ l(E, ABC), l(A, BCE), l(B, CEA), l(C, EAB)$ concur at $ X$ and Simson lines $ l(F, ABD), l(A, BDF), l(B, DFA), l(D, FAB)$ concur at $ Y$ [2]. Let $ M, K, L$ be midpoints of $ AB, CE, DF.$ The centroids $ G, H$ are midpoints of $ MK, ML.$ The anticenters $ X, Y$ are identical $ \Longleftrightarrow$ the centroids $ G, H$ are identical $ \Longleftrightarrow$ the midpoints $ K, L$ of the chords $ CE, DF$ are identical $ \Longleftrightarrow$ the chords $ CE, DF$ are identical, in which case $ CDEF$ is a rectangle degenerated into a line segment, or $ CE, DF$ are both diameters of $ \mathcal O,$ in which case $ CDEF$ is a rectangle. In the second case, $ X \equiv M \equiv Y.$ References: [1] http://www.mathlinks.ro/viewtopic.php?t=85989 [2] http://www.mathlinks.ro/viewtopic.php?t=179089 (and elsewhere)

Perhaps I am missing something, but it seems to me that if $AD$ and $BE$ are diameters, then $l(A,BDF)=l(D,ABF)=BF$, and $l(B,ACE)=l(E,ABC)=AC$ and there is a point of intersection of all four lines, without $CDEF$ needing to be a rectangle. Anyway, assume that neither $AD$ nor $BE$ are diameters. Let $X$ be intersection of the four Simpson lines. Let $A_1, A_2$ be the altitudes from $A$ to $BD,DF$ and let $D_1, D_2$ be the altitudes from $D$ to $AB,AF$. Then $A,A_1,A_2,D,D_1,D_2$ all lie on the circle with diameter $AD$. So by Pascal's theorem applied to $A_1A_2AD_1D_2D$, we have that $X,B,H_{AFD}$ are collinear, where $H_{AFD}$ is the orthocenter of $AFD$. Similarly we have that $X,F,H_{ABD}$ are collinear. Let $a,b,c,d,e,f,x$ denote the vectors from the center of the circle to $A,B,C,D,E,F,X$. Then $x=r(a+f+d)+(1-r)b=s(a+b+d)+(1-s)f$, for some $0 \le r,s \le 1$. We, (or at least I) have assumed that $AD$ is not a diameter, so $l(A,BDF)$ and $l(D,ABF)$ are distinct lines with a unique point of intersection, so we have that $x=(a+b+f+d)/2$. Similarly, working on the other two Simpson lines yields that $x=(a+b+c+e)/2$, which in turn means that $CE$ and $FD$ have the same midpoint. Each non-diameter chord of the circle has a unique midpoint, which means that $CE$ and $FD$ are diameters, which in turn means that $CDEF$ is a rectangle.