Official solution: Let $I$ be the incenter of triangle. $AB'C'I$ is cyclic. Fact 5 gives $IB'=IC'$. Consider circle $\omega$ centered at $I$ with radius $IB'$. We need to show that $A$ lies outside $\omega$. Drop feet from $I$ to $BC$ and $CA$. Angle chasing finishes this now. My solution which I found in exam: Let $\angle AA'B'=\alpha, \angle AA'C'=\beta$. Law of sines in $\triangle ABA'$, gives \[\frac{\sin (30+\alpha)}{\sin \alpha}=\frac{AA'}{AB'}\implies \frac{\cot \alpha + \sqrt{3}}{2}=\frac{AA'}{AB'}.\] Law of sines in $\triangle AA'C$, $AA'=2A'C\sin C.$ Thus \[\cot \alpha = \frac{\sqrt{3}(2a+c-b)}{b+c},\]similarly $\cot \beta = \frac{\sqrt{3}(2a+b-c)}{b+c}$. We need to show, $\cot (\alpha +\beta) \geq \frac{1}{\sqrt{3}}$. Now substituting $\cot \alpha, \cot \beta$ we need to show, \[3(2a+b-c)(2a+c-b)-(b+c)^2\geq 4a(b+c).\]Thus we need to show $3a^2\geq b^2+c^2-bc+ab+ac$. Cosine rule in triangle $ABC$ gives $a^2=b^2+c^2-bc$. Thus we need to show $2a\geq b+c$ or \[4b^2+4c^2-4bc=4a^2\geq b^2+c^2+2bc\Leftrightarrow b^2+c^2\geq 2bc.\]Proved.

Here's a sketch of a simpler proof. Use notations from here. Now it suffices to show that $IA' \ge IX$. For this wlog let $AB < AC$ (equality being a trivial case). Then we only need that $\angle XIA'$ is positive, that is, $\angle AIX < 180^\circ$. This is true since $\angle AIX = 120^\circ + \angle AB'C'$. Since $AB' > AC'$, we have $\angle B' < \angle C'$, and thus $\angle B' < \frac{\angle C' + \angle B'}{2} = 60^\circ$, and so we are done. Note that we also need that $\angle DIA'$ is positive, but it's easy to see that $\angle DIX$ is positive.

Nice problem! 8.7 wrote: Let $ABC$ be a triangle with $\angle A = 60^\circ$, and $AA', BB', CC'$ be its internal angle bisectors. Prove that $\angle B'A'C' \le 60^\circ$. WLOG $AB \le AC$. Clearly, $\angle A \ge \angle C$. Let $P \ne I$ be defined as $P \overset{\text{def}}{:=} \odot(BIC') \cap \odot(CIB')$. By Miquel's Theorem we know $P$ lies on $\overline{BC}$ since $AB'IC'$ is cyclic. Note that $IB'=IC'=IP$. Now $$\angle IPA'-\angle IA'P=\pi-\left(\angle B+\tfrac{1}{2}\angle C\right)-\left(\tfrac{1}{2}\angle A+\angle C\right)=\tfrac{1}{2}\left(\angle A-\angle C\right) \ge 0$$so $IA' \ge IP$. Consequently, $A'$ lies outside $\odot(I, IP)$ hence $\angle B'A'C' \le 60^{\circ}$.

TheDarkPrince wrote: Official solution: Let $I$ be the incenter of triangle. $AB'C'I$ is cyclic. Fact 5 gives $IB'=IC'$. Consider circle $\omega$ centered at $I$ with radius $IB'$. We need to show that $A$ lies outside $\omega$. Here's an alternative finish from here. Recall by the angle bisector theorem that $$\frac{AI}{IA'}=\frac{BA}{BA'} \text{ and } BA'=\frac{ac}{b+c}$$Hence, $$\frac{AI}{IA'}=\frac{b+c}{a}=\frac{\sin B+\sin C}{\sin A}=\frac{2}{\sqrt 3}(\sin C+\sin B).$$Recalling $\angle B+\angle C=120$, we have by Jensen's inequality, that $\sin C+\sin B\le 2\sin 60= 4/\sqrt{3}$. Hence, it follows $$2\ge AI/IA'=\frac{AI}{AA'-AI}\implies AA'\ge \frac{3}{2}AI.$$Let $X$ be the point on ray $IA'$ such that $IX=IB'$. Note that $X$ is the point of intersection between the circle with center $I$ and radius $IB'$ with $IA'$. Thus to show that $A'$ lies outside of the circle, it is enough to show that $AI+IX\le AA'$. But since $IX=AI/2$, this is obvious by the above inequality.