Let $A_i$, $i = 1, 2,\ldots, 6$, be six points on plane. Taking subscripts modulo $6$, we denote, for $i = 1, 2,\ldots, 6$, the intersection of the lines $A_iA_{i+1}$ and $A_{i+2}A_{i+3}$ by $B_{i+3}$, and the second intersection of the circumcircles of triangles $A_iA_{i+1}B_{i+2}$ and $A_{i+1}A_{i+2}B_{i+3}$ by $C_{i+1}$, and the circumcenter of the triangle $C_iB_{i+1}B_{i+2}$ by $D_i$.
Please provide the lines $D_1D_4$, $D_2D_5$, and $D_3D_6$ are concurrent.
Here is my solution and some remarks about this problem:
Denote $(D_i) = C_iB_{i+1}B_{i+2}$, with $i = 1, 2,\ldots, 6$ and taking subscripts modulo $6$.
The main idea is to prove that there exists the radical center of $(D_i), (D_{i+1}), (D_{i+3}), (D_{i+4})$ for all $i = 1, 2,\ldots, 6$.
We consider the case $(D_2), (D_3), (D_5), (D_6)$ in the figure below:
Let $T_1$ be the second intersection of $(D_5),(D_6)$.
We have $\angle B_2T_1B_6 = 360^{\circ} - \angle B_2T_1B_1-\angle B_1T_1B_6 = 360^{\circ} - \angle B_2C_6B_1 - \angle B_1C_5B_6 = 360^{\circ} - \angle B_2A_1B_1 - \angle B_1A_5B_2 - \angle B_1A_6B_6 - \angle B_1A_4B_6 = 360^{\circ} - ( \angle B_2A_1A_6 + \angle B_2A_6A_1) - (\angle A_4A_5B_6 + \angle A_4B_6A_5) = 180^{\circ} - \angle B_2B_4B_6 $, hence $T_1 \in (B_2B_4B_6)$. Let $R$ be the second intersection of $B_2B_4$ and $(D_6)$, then $\angle RB_1C_6 = \angle C_6B_2B_4 = \angle C_6A_5B_1$, so $RB_1$ is tangent to $(B_1A_5A_6)$ or $ \angle RB_1O_1 = 90^{\circ}$, with $O_1$ is the circumcenter of $\bigtriangleup B_1A_5A_6$. Let $B_1T_1 \cap (B_2B_4B_6) = X_1$, by Reim we have $B_4X_1\perp B_1O_1$. Define analogously $T_3,T_4,X_3,O_3$. By the same proof, we have $T_4\in (B_1B_3B_5)$ and $T_3\in (B_2B_4B_6)$. Note that $\angle B_1T_4B_4 = \angle B_3T_4B_4 - \angle B_3T_4B_1 = \angle B_4T_3X_3 - \angle B_1B_5B_3$.
We will prove that $\angle B_4T_1X_1 = \angle B_4T_3X_3 - \angle B_1B_5B_3$ $(*)$.
It's equilvalent to $ \angle B_1B_5B_3= \angle B_4T_1X_3 - \angle B_4T_1X_1 = \angle X_3T_1X_1 $. Let $U_2,U_4,U_6\in (B_2B_4B_6)$ such that $B_2U_2\| B_1B_3, B_4U_4\|B_3B_5,B_6U_6\|B_1B_5$. By the properties of antiparallel lines, we have $U_2U_4\|X_3B_6, U_2U_6\|X_1B_4$. So $ \angle X_3T_1X_1 = \angle X_3B_4X_1= 180^{\circ} - \angle X_3U_6U_2 - \angle U_6X_3B_4 = \angle U_6B_6B_4 - \angle B_6B_4U_4 = \angle B_6A_4B_5 - \angle A_4A_3B_5 = \angle B_1B_5B_3$, so $(*)$ is true and from $(*)$ we have $B_1,T_1,B_4,T_4$ lie on a circle. Thus the intersection $M_0$ of $B_1T_1$ and $B_4T_4$ is the radical center of $(D_2),(D_3),(D_5),(D_6)$. By this proof, we can construct the radical center $N_0$ of $(D_3),(D_4),(D_1),(D_3)$ and the radical center $P_0$ of $(D_4),(D_5),(D_1),(D_2)$.
Now, just note that if $S$ is the radical center of $(D_1), (D_3), (D_5)$, then $SN_0\perp D_1D_3$, $SP_0\perp D_5D_1$ and $SM_0\perp D_3D_5$, or $\bigtriangleup D_1D_3D_5$ and $\bigtriangleup M_0P_0N_0$ are orthologic, but it's easy to see that $D_1D_4 \perp N_0P_0$, $D_2D_5 \perp M_0P_0$ and $D_3D_6 \perp N_0M_0$, so $D_1D_4, D_2D_5, D_3D_6$ are concurrent ( $Q.E.D$ ).
Remarks: When working on this problem, I've found some interesting ideas ( I checked it by computer and it's true), but it's still unsolved:
$i)$ Let the point of concurrency $V$ in the problem is the "center" of six points $(B_1, B_2, B_3, B_4, B_5, B_6)$, then take a fixed point $H$, take a homothetic transformation with center $H$, it transforms $\bigtriangleup B_1B_3B_5$ to $\bigtriangleup B_1'B_3'B_5'$. Let $D_1', D_4'$ (wrt six points $(B_1', B_2, B_3', B_4, B_5', B_6)$ are the points that have the same definition with $D_1, D_4$ wrt six points $(B_1, B_2, B_3, B_4, B_5, B_6)$ and $V'$ is the "center" of six points $(B_1, B_2, B_3, B_4, B_5, B_6)$, then $\dfrac {\overline {VD_1}}{\overline{VD_4}}= \dfrac {\overline {V'D_1'}}{\overline{V'D_4'}}. $
$ii)$ By applying $i)$ and $E.R.I.Q$ lemma, I've found a nice result: With $B_2, B_4, B_6, H$ fixed and the ratio of the homothetic transformation with center $H$ varies, then the "center" of six points $(B_1, B_2, B_3, B_4, B_5, B_6)$ lies on a fixed line.
Sincerely,
XH
(Figure for this solution at: Concurrent lines with six any points)