By the Incenter-Excenter Lemma on $BAA'$, $ABA'B'$ is cyclic, so $\angle AB'I=C+\frac A2=90-\angle IAO$, which implies that since $OI\perp AI$, $AB'IO$ is cyclic. Similarly, $AC'IO$ is cyclic, so $AB'C'IO$ is cyclic with diameter $AO$. Since $\angle IAB'=\frac B2$, $X$ is the midpoint of arc $ACB$, and $B'$ is the midpoint of $AX$. Therefore, $XA'\perp AI$, so $\angle ICX=\angle IA'X=90$ implies $IA'CX$ cyclic. Therefore, $\angle AXX'=180-B-\angle IXC=A/2=90-\angle X'AX$, so $XX'\perp AX'$. Therefore, the circle with diameter $AX$ passes through $X'$ and $A'$. Then, $\angle X'A'B=C+A/2-\angle X'A'I=C+A/2-\angle AXX'=C$ and $\angle X'IB=180-\angle BIX=90-A/2-\angle XA'C=90-A/2-B-A/2+90=C$, so $A'BIX'$ is cyclic. Therefore, $\angle XX'A'=\angle A'BI=\frac12\angle ABC$.

My problem! Claim 1: $IA'=M_{A}A'$ Note that by Ptolemy's theorem on $ABM_{A}C$, we get the following equation $$AB*M_{A}C+AC*BM_{A}=AM_{A}*BC$$Using the given condition $2BC=AB+AC$ and the fact that $BM_{A}=IM_{A}=CM_{A}$, it is equivalent to $$IM_{A}*2BC=AM_{A}*BC$$ Therefore, $AM_{A}=2IM_{A}$ Now note that by sine rule, $$\frac{IA'}{A'M_{A}}=\frac{\sin{\angle{IBA'}}}{\sin{\angle{A'BM_{A}}}}\frac{BI}{BM_{A}}=\frac{\sin{\angle{IBC}}}{\sin{\angle{BAI}}}\frac{BI}{AI}=1$$ Hence, our claim is true Claim 2: $B',C'$ are the midpoints of $M_{C}I,M_{B}I$ Since $M_{C}A=M_{C}I$ , $M_{B}A=M_{B}I$ , $M_{C}M_{B}$ is the perpendicular bisector of $AI$ By definition, $B'C'$ is the perpendicular bisector of $AA'$ From Claim 1 $AA'=\frac{3}{2}AI$ ,so $$B'C'=\frac{1}{2}M_{B}M{C}$$ Together with the fact that $B'C' \parallel M_{B}M{C}$, the claim is true Claim 3: $AC'A'C$ cyclic From POP on $AM_{C}M_{A}C$ , $$AI*IM_{A}=IM_{C}*IC$$ Using Claim 2, this is equivalent to $$AI*2IA'=2IC'*IC$$ Which is equivalent to our claim Claim 4: $AM_{C}YI$ is a parallelogram Extend $AC'$ and let it intersect $(ABC)$ again at $Y$ From Claim 2, we have $M_{C}C'=C'I$, therefore, we just have to show that $AM_{C} \parallel IY$ Note $$\angle{M_{C}YA}=\angle{M_{C}CA}=\angle{ICA'}=\angle{IAC'}=\angle{IAY}$$ Hence, our claim is true Claim 5: $I$ is the orthocenter of $AXY$ From claim 4, we get that $B',C'$ are the midpoints of $AX,AY$ Hence, $$2B'C'=M_{B}M_{C}=XY , XY \parallel B'C' \parallel M_{B}M_{C}$$ This implies that $A'$ lies on $XY$ and that $AA'$ is the altitude of $\triangle{AXY}$ Since $IA'=A'M_{A}$, it must mean that $I$ is the orthocenter The orthocenter condition implies that $$\angle{XX'A}=\angle{A'AX}=\angle{A'AB'}=\angle{B'BA'}=\angle{IBC}=\frac{1}{2}\angle{ABC}$$ Q.E.D

First we show that $BC'B'C$, $AB'A'B$ and $AC'A'C$ are cyclic. Let $M_a$ be the midpoint of minor arc $BC.$ Note that \[\angle B'C'I = 90 - \angle AIC' = 90 - \angle CIM_a = \frac{B}{2} = \angle B'BC.\] Hence, $BC'B'C$ is cyclic. Similarly letting $M_b$ and $M_c$ be the midpoint of arcs $AC$ and $AB$ respectively, we can analogously show that other two quadrilaterals are cyclic. Now we claim that $I$ is the midpoint of $AM_a$. Notice that Ptolemy's Theorem on quadrilateral $ABM_aC$ implies that \[AB \cdot M_aC + AC \cdot M_aB = BC \cdot AM_a \implies M_aB = 2AM_a \implies AI = 2AM_a\] as claimed. Now we show that $ADC'B'$ is cyclic. Indeed as \[\angle IC'B' = \angle CC'B' = \angle CBB' = \angle A'BB' = \angle A'AB' = \angle IAB'.\] We claim that $B'$ is the midpoint of $AX.$ This is equivalent to showing $B'I \parallel M_aX.$ Indeed, as \[\angle AB'I = 180 - \angle AC'I = 180 - \angle AC'C = 180 - \angle AA'C = 180 - (B+ A/2),\] and \[\angle AXM_a = 180 - (\angle ABM_a) = 180 - (B+ A/2).\] So $B'I \parallel M_aX \implies B'$ is the midpoint of $AX$ as needed. As $AB' = B'A' = B'X$ we see that $\angle AA'X = 90$. Now, \[\angle ICX = \angle M_cCX = 180 - \angle M_cAX = 180 - C/2 - A/2 - \angle A'AB' = 90 + B/2 + 90 - \frac{\angle AB'A'}{2} = 90.\] So, $IA'CX$ is cyclic with diameter $IX.$ Now, note that \[\angle X'AA' = \angle C'AA' = \angle C'CA' = \angle X'XA'\] so $X'AXA'$ are cyclic. Finally, \[\angle IX'A' = \angle XX'A' = \angle XAA' = \angle B'AI = \angle B'C'I = \frac{\angle ABC}{2}\] as desired.

Nice problem! The key is $\boxed{3\times 4=2\times 6}$ It is clear that $B’$ lies on $(ABA’)$. Similarly, $C’$ lies on $(ACA’)$ Our goal is to prove that $\angle A’X’X= \frac{\angle B}{2}$. Notice that $\angle A’AX= \angle A’AB’=\angle A’BB’= \frac{\angle B}{2}$ This is equivalent to proving that $\angle A’AX=\angle A’X’X \iff A,X’,A’,X $ are concyclic $\iff X’I \times IX = A’I \times IA$ Observe that $AI \times IA’ = C’I \times IC$. So, we’re left to prove that $X’I \times IX= CI \times IC’$ This is the same as proving $C,X,C’,X’$ are concylic $(\spadesuit)$ Claim $I,A’,X,C$ are concyclic Proof Let $I_A$ be the $A$-excenter of $\triangle ABC$ Notice that $\angle ICX=\angle ICB + \angle BCX =\frac{\angle C}{2} + \angle BCX = \frac{\angle C}{2} + \angle BAI +\angle IAX =\frac{\angle C + \angle A}{2}+\angle A’AB’= \frac{\angle C + \angle A}{2}+\angle A’BB’= 90^\circ $ $\implies C,X,I_A$ are collinear which means $X=I_AC \cap (ABC)$ Apply angle bisector theorem twice yields $AI=2IA’=2x$ for some x Let $M$ be the midpoint of arc $BC$ not containing $A$ Consider power of point $A’$ wrt. $(ABC)$ and $(IBI_AC)$, we obtain $AA’ \times A’M =BA’ \times A’C = IA’ \times A’I_A$ This means $MI_A=2\times MA’=2y$ for some y Since $IM=MI_A \implies x+y=2y \implies x=y$ which means $AI:IA’:A’M:MI_A=2:1:1:2$ Note that $3 \times 4 = 2 \times 6$, we conclude that $I_AA’ \times I_AI = I_AM \times I_AA= I_AX \times I_AC $ By converse of PoP we’re done $\square$ By the claim, we obtain $\angle X’XC = \angle IXC = \angle IA’C = \angle AA’C = \angle AC’C = \angle X’C’C$ So, $(\spadesuit)$ is true which means that the problem is solved $\blacksquare$