Thai Mathematical Olympiad

Since $D$ is the midpoint of the arc $B^{\prime }{C}^{\prime }$ and $AC$ touches the incircle at $B^{\prime }$, we see that $$\therefore \angle A{B}^{\prime }D=\angle B{A}^{\prime }D=\angle D{A}^{\prime }{C}^{\prime }=\angle D{B}^{\prime }{C}^{\prime }$$ Thus, $B^{\prime }D$ bisects $\angle A{B}^{\prime }{C}^{\prime }$ and similarly, $D{C}^{\prime }$ bisects $\angle B^{\prime }{C}^{\prime }A$. Hence $D$ is the incenter of $△A{B}^{\prime }{C}^{\prime }$. By the same arguments, we can conclude that the incenters of $A^{\prime }C{B}^{\prime }$ and $A^{\prime }B{C}^{\prime }$, denoted by $O_1$ and $O_2$, are the midpoints of the arcs $A^{\prime }{B}^{\prime }$ and $A^{\prime }{C}^{\prime }$, respectively. Since $O_1,O_2,D$ are the midpoints of the arcs $A^{\prime }{B}^{\prime },A^{\prime }{C}^{\prime },B^{\prime }{C}^{\prime }$, then $C^{\prime }{O}_1,A^{\prime }D$, and $B^{\prime }{O}_2$ intersect at the incenter of $△A^{\prime }{B}^{\prime }{C}^{\prime }$. $$\therefore \angle A^{\prime }D{O}_1+\angle D{O}_1{C}^{\prime }+\angle C_1{O}_1{O}_2=\frac{1}{2}(\angle A^{\prime }{C}^{\prime }{B}^{\prime }+\angle B^{\prime }{A}^{\prime }{C}^{\prime }+\angle A^{\prime }{B}^{\prime }{C}^{\prime })=\frac{\pi }{2}$$ Thus, $A^{\prime }D\perp O_1{O}_2$. Since $A^{\prime }$ lies on $BC$, the common tangent of the circles $O_1$ and $O_2$, the reflection of $A^{\prime }$ with respect to the line $O_1{O}_2$ is $E$. By the reflection, $\angle E{O}_2{O}_1=\angle A^{\prime }{O}_2{O}_1=\angle O_1{O}_2{B}^{\prime }$. Since $E$ and $B^{\prime }$ lie on the same side with respect to $O_1{O}_2$, $E$ must lie on $B^{\prime }{O}_2$. Therefore, $E$ is the incenter of $△A^{\prime }{B}^{\prime }{C}^{\prime }$. Now, it is left to show that the common tangent is parallel to $B^{\prime }{C}^{\prime }$; this will implies that $\angle FE{C}^{\prime }=\angle B^{\prime }{C}^{\prime }E=\angle F{C}^{\prime }E$. This follows from the reflection of the tangent $BC$ with respect to $O_1{O}_2$ that, $$\angle FE{A}^{\prime }=\angle E{A}^{\prime }B=\angle A^{\prime }{B}^{\prime }D=\angle A^{\prime }{B}^{\prime }{C}^{\prime }+\angle C^{\prime }{B}^{\prime }D=\angle A^{\prime }D{C}^{\prime }+\angle B{C}^{\prime }D=\angle EG{C}^{\prime }$$ where $G$ is the intersection of $A^{\prime }D$ and $B^{\prime }{C}^{\prime }$.