Hint(I will provide the whole solution later): Take P a point on the circumcenter such that $\triangle PAB$ is equilateral, easy to see that $B'-T-A'$ are collinear

Let $C'$ is the intersection of $BA'$ and $AB'$ and $I$ be the incenter of $\triangle ABC$. So, $\angle ACB =\angle AC'B = 60^\circ$ and $\angle AIB =120^\circ \implies \ AC'BI$ is a cyclic quadrilateral . Now, $$\angle AA'C' =\angle CAA' =\angle IAB =\angle IC'B $$$$\implies IC'=IA' \ and \ A'B =AB$$As the same, $AB'=AB =A'B$ and $IB'=IC'\implies \ I$ is the circumcenter of $\triangle A'B'C'$ Now, Let $X$ be the intersection of the circumcircle of $IBA'$ with $A'B'$. It is easy to show that $IAB'X$ is also cyclic. So, $$\angle B'XA =\angle BXA' = 60^\circ $$$$\implies \angle XBA' =\angle XB'A; \angle XA'B =\angle XAB'$$As, $AB'=BA'$, $\triangle AB'X \cong \triangle A'BX$ and both are similar to $\triangle A'B'C'$ $\implies \frac {A'C'}{B'C'} =\frac{A'X}{B'X}$ So, $C'X$ is the angle bisector of $\angle B'C'A'$ Now , $\angle AXB' =\angle A'XB =60^\circ \implies \angle AXB =60^\circ$ So, X is one of the intersection of $( \triangle ABC)$ with $A'B'$. Let $$X=E$$Now, let $A_1$ and $B_1$ be the reflection of $A,B$ over the angle bisector $C'E$. As, $\angle A'C'B' =60^\circ$ it is easy to show that, $A_1,B_1$ lies on the circumcircle of $\triangle ABC$. So the reflection of the circumcircle of $\triangle ABC$ over $C'E$ sends it to itself. Which implies that $C'E$ goes through the center of $(\triangle ABC)$. Now, let $D'$ be a point on ($\triangle ABC$) such that, $ABD'$ is equilateral. And it is easy to show that $B',A',D'$ are co linear because $BD'=AD'=AB=A'B=AB'$. So, $$D'=D$$And $C'D$ is the symmedian of triangle $ABC'$ [$DA,DB$ are tangent to $(\triangle ABC')$] As, $C'C$ is the median of the triangle [$ACBC'$ is a parallelogram] $C'D$ is the reflection of $C'C$ over the bisector of $\angle B'C'A'$ so $DE=EC$ and we are done.

A very nice problem. I'll present a solution that was more intuitive for me: Construct the circumcenter $I$ as in the previous solution. This is the one step of the solution which is not so obvious to see, since it's not clear at first that we should think of intersecting $A'B$ and $B'A$. At least, drawing an accurate diagram can give some clues, and this construction could easily be found after some tinkering. Now drop altitudes from $A$ to $BB'$ and $B$ to $AA'$, and let them intersect at $D'$. From a quick angle chase, we see that $\angle BD'A = 60^\circ \implies D' \in (ABC)$ and $\angle A'D'B' = 180^\circ \implies D' \in A'B'$, so $D'$ is one of our desired intersection points. WLOG $D = D'$. This step is pretty clear just from drawing the diagram and noticing the perpendicularity. Now draw $E' \in A'B'$ such that $\angle E'AB = 60^\circ$. From another quick angle chase, we see $\angle B'E'A = \angle E'B'A$, so $EA=AB'$. But since we previously found that $AB = AB'$ (see above solution for details, just easy angle chase), $E'AB$ is an isosceles triangle with vertex angle $60^\circ$, so it is obviously equilateral. Thus $\angle BE'A = 60^\circ$, so $E' \in (ABC)$ and $E = E'$. This step is also pretty clear from the diagram. Therefore, we have characterized the location of the two intersection points $D, E$, and almost all the angles in the diagram are now available for us to use. Now it suffices to show that arcs $CD$ and $DE$ are equal, which is obvious after a quick angle chase (in particular, it suffices to compute and equate the angles formed by $BD, AC$ and $BE, AD$). Nice problem!