Cool problem! Here, quite briefly, is my solution Firstly, it is easy to show using that the angle of a regular nine-sided triangle is $140^{\circ}$, that the angle $\angle XBY = 40^{\circ}$, then it follows that $AX\parallel CY$ and all the similar. Next, assume that $YB\cap AX = K$ and $YB\cap ZC = R\not =K$. The angles $AKCB, ARCB$ are inscribed, but then the straight line $BK$ intersects $(ABC)$ at three points! Then $XA,YB,ZC,TD$ and $XB, YC,ZD,TE$ are competitive and intersect at points $K,L$, respectively, and $K,L$ lie on the circumscribed circle of the nine-sided Now it remains to note, again by counting angles, that $KXYL,KZTL,KYZL$ are isosceles trapezoids, from where all 6 points $X,Y,Z,T,K,L$ lie on the same circle