Thai Mathematical Olympiad

Since $D$ is the midpoint of the arc $B'C'$ and $AC$ touches the incircle at $B'$, we see that
$$\therefore \angle AB'D = \angle BA'D = \angle DA'C' = \angle DB'C'$$
Thus, $B'D$ bisects $\angle AB'C'$ and similarly, $DC'$ bisects $\angle B'C'A$. Hence $D$ is the incenter of $\triangle AB'C'$. By the same arguments, we can conclude that the incenters of $A'CB'$ and $A'BC'$, denoted by $O_1$ and $O_2$, are the midpoints of the arcs $A'B'$ and $A'C'$, respectively.
Since $O_1, O_2, D$ are the midpoints of the arcs $A'B', A'C', B'C'$, then $C'O_1, A'D$, and $B'O_2$ intersect at the incenter of $\triangle A'B'C'$.
$$\therefore \angle A'DO_1 + \angle DO_1C' + \angle C_1O_1O_2 = \frac{1}{2}(\angle A'C'B' + \angle B'A'C' + \angle A'B'C') = \frac{\pi}{2}$$
Thus, $A'D \perp O_1O_2$. Since $A'$ lies on $BC$, the common tangent of the circles $O_1$ and $O_2$, the reflection of $A'$ with respect to the line $O_1O_2$ is $E$. By the reflection, $\angle EO_2O_1 = \angle A'O_2O_1 = \angle O_1O_2B'$. Since $E$ and $B'$ lie on the same side with respect to $O_1O_2$, $E$ must lie on $B'O_2$. Therefore, $E$ is the incenter of $\triangle A'B'C'$.
Now, it is left to show that the common tangent is parallel to $B'C'$; this will implies that $\angle FEC' = \angle B'C'E = \angle FC'E$. This follows from the reflection of the tangent $BC$ with respect to $O_1O_2$ that,
$$\angle FEA' = \angle EA'B = \angle A'B'D = \angle A'B'C' + \angle C'B'D = \angle A'DC' + \angle BC'D = \angle EGC'$$
where $G$ is the intersection of $A'D$ and $B'C'$.