I've been doing something wrong for quite some time, but the problem is quite simple. Note that $\angle AB'C = \angle ADC = 120^{\circ} = \angle ABC \Rightarrow A, B, B', C$ are concyclic. From symmetry: $DB = BA' = BC' $ $\Rightarrow B$ is circumcentre of triangle $A'C'D$ and $BA, BC$ are perpendicular bisectors to its sides $\Rightarrow \angle A'BC' = 360^{\circ} - 2\angle ABC = 120^{\circ} \Rightarrow \angle A'DC' = 60^{\circ}$. Note that $CD = CB' = CA'; AD = AB' = AC'$ $\Rightarrow$ $\angle A'B'C' = 360^{\circ} - 120^{\circ} - \angle CB'A' - \angle AB'C'$ $= 240^{\circ} - 180^{\circ} + \angle A'DC'$ $= 120^{\circ} \Rightarrow A', B, B', C'$ are concyclic. Easy see that $\angle AC'A' + \angle CA'C' = \angle DC'A' + \angle CA'C' + \angle ADC' + \angle CDA' =$ $120^{\circ} + 120^{\circ} - 60^{\circ} = 180^{\circ} \Rightarrow AC' \parallel A'C$. Let $AA', CC'$ interesect at point $X$; $AA'$ intersect $(ABC)$ again at point $P$; $CC'$ intersect $(A'B'C')$ again at point $Q$. Note that $\angle CQA' = 60^{\circ} = \angle CPA' \Rightarrow A', C, P, Q$ are concyclic $\Rightarrow \angle C'QP = \angle AA'C = \angle C'AP \Rightarrow A, C', P, Q$ are concircle $\Rightarrow QX \cdot XC' = PX \cdot XA$. It means that $X$ lie on rad. axe of $(ABC)$ and $(A'B'C') \Rightarrow AA', BB', CC'$ are concurrent.

A idea to another synthetic solution: Let $O$ is circumcenter $(A'B'C')$. It is obvious that $O$ is the midpoint of the lesser arc $A'C'$ of $(DA'C')$ (the trillium lemma because $\angle A'DC' = 60^{\circ}$). We need proof that $AA', CC', DO$ are concurrent. Then note that $B, B'$ are $A'C'D$ and $ACO$ triangles orthology centers and by Sondat's theorem the original condition will follow. But i dont see good explanation why $AA', CC', DO$ are concurrent. But it's easy do using trigonometry.

This problem is clearly related to the conditions of degeneracy of the Neuberg cubic. In this case, it decomposes into a line (bisector of ABC) and a circle (ACD).