EDIT: I was wrong, the angle equality is indeed true, I confused $A$ with $A_1$. SORRY! To prove the common point $X$, note that $A'I\cap (O)=A_1$, the midpoint of arc$BAC$, similarly define $B_1,C_1$. Apply inversion at $I$ with radius $\sqrt {IA'*IA_1}$, as a result $\omega_A$ is mapped to a line through $A_1$ parallel to $AI$, and symmetric result for $B_1,C_1$. Clearly the three lines concur at the orthocenter of $A_1B_1C_1$ or the circumcenter of the excenter triangle of $ABC$ denoted by $H$, which lies on $IO$, hence the existance of a common point is proven. Now Let $A$ be sent to $A_2$ by the same inversion, clearly it's the midpoint of arc $BC$ without $A$, Since $(A_1,H,A',X),(A,X,A_2,H)$ are concyclic, therefore $\angle AXO=\angle AA_2H=\angle HA_1A'=\angle OXA'$ since $A_1IA_2H$ is clearly a parallelogram because $A_1A_2,IH$ bisect each other at $O$ so we are done.

I will fill in some intermediate steps in XmL's post. Lemma 1: If $ X $, $ Y $ are the tangency points of the $ A $-mixtilinear incircle with $ AB, AC $ respectively then $ I $ is the midpoint of $ XY $. Proof: This is a well-known result, and moreover can be proved immediately with a limiting case of Sawayama's Theorem, but I shall provide a more elementary proof. Let $ B_2, C_2 $ be the midpoints of arcs $ CA, AB $ respectively. Then by Archimedes' Lemma we have that $ A', X, C_2 $ and $ A', Y, B_2 $ are collinear. Then by Pascal's Theorem on hexagon $ A'C_{2}CABB_{2} $ we have that $ X, I, Y $ are collinear. But since $ AX = AY $ and $ AI \perp XY $ this immediately implies the desired result. Lemma 2: If $ A_1 $ denotes the midpoint of arc $ BAC $ then $ A', I, A_1 $ are collinear Proof: Consider the homothety centered at $ A' $ that takes the $ A $-mixtilinear incircle to the circumcircle of triangle $ ABC $. This takes $ X $ to $ C_2 $ and $ Y $ to $ B _2 $ so by Lemma 1 it takes $ I $ to $ M $, the midpoint of $ B_{2}C_{2} $. So it suffices to show that $ A_1, I, M $ are collinear. Some quick angle-chasing yields the fact that $ A_{1}C_{2}IB_{2} $ is a parallelogram which implies the desired result. Now, continuing as in XmL's post, taking the inversion centered at $ I $ with radius $ \sqrt {A'I * A_{1}I} = \sqrt {B'I * B_{1}I} = \sqrt {C'I * C_{1}I} $, we have that $ \omega_A $ is mapped to the line parallel to $ AI $ passing through the reflection of $ A_1 $ over $ I $. We obtain two similar lines, and then reflecting them about $ I $ we obtain the configuration in XmL's post. So, letting $ l_A $ be the line through $ A_1 $ parallel to $ AI $ and defining $ l_B, l_C $ similarly it suffices to show that these three lines concur at the circumcenter of the excentral triangle of triangle $ ABC $. Let $ I_A, I_B, I_C $ be the $ A, B, C $-excenters of triangle $ ABC $ respectively. It is well-known that $ A_1 $ is the midpoint of $ I_{B}I_{C} $ and that $ AI \perp I_{B}I_{C} $ so $ l_A $ is the perpendicular bisector of $ I_{B}I_{C} $. This immediately implies the desired result, and so we are done.

I wish I made MOP...

Step 1: I'll prove that circles $AOA'$, $BOB'$, $COC'$ have a common point $X$, other than $O$ which lies on the line $OI$. Perform an inversion at $O$ which leaves the circumcircle of $\triangle{ABC}$ invariant. Then, it remains to prove that the lines $AA', BB', CC' $ and $OI$ are concurrent, which is a known fact. (a proof can be seen on "cut-the-knot"). Step 2: I'll prove that $X$ is the common point of the circles $ \omega_A,\omega_B,\omega_C $ too. Let $X'$ be the point of intersection between the line $IO$ and the circle $ \omega_A$. We'll prove that $X=X'$ and to do that it suffices to prove that $X'$ lies on $(AOA')$. Let $M$ be the midpoint of the big arc $BAC$, $N$ be the midpoint of the little arc $BAC$ and $T$ the tangency point between $ \omega_A$ and $AB$. It's well-known that $M$, $I$, $A'$ are collinear and $TI \perp AI$. To prove this step, it's enough to show that $\angle{A'X'I} = \angle{A'IN}= \angle{A'AO}$. The first equality is obvious as $ \omega_A$ is tangent to line $AI$. All that remains to show is $\angle{A'IN}= \angle{A'AO}$. Let $\angle{BAA'}= \alpha$. Easy angle-chase shows that $BTIA'$ is cyclic so $\angle{TIA'}= 180 - (\angle{B}+\angle{A}-\alpha)$$\implies$ $\angle{A'IN}= (\angle{B}+\angle{A}-\alpha) - 90$ as $ \omega_A$ is tangent to $AI$ at $I$. On the other side, $\angle{A'AO} = \angle{A}-\alpha - (90-\angle{B}) = (\angle{B}+\angle{A}-\alpha) - 90$ so $\angle{A'X'I} = \angle{A'AO}$ $\implies$ $X'AOA'$ is cyclic $\implies$ $X=X'$ so $ \omega_A$ passes through $X$. Similarly $\omega_B,\omega_C $ pass through $X$. The angle equality follows immediately. $\Box$

My solution: Let $ D. E. F $ be the intersection of $ (I) $ and $ BC, CA, AB $ , respectively. Let $ A_1, B_1, C_1 $ be the midpoint of $ EF, FD, DE $ , respectively. It's well-known that $ \angle IA'B =\angle IA'C $ so $ A'I $ passes through the midpoint of arc $ BAC $ . i.e. $ (AIA') $ and $ (ABC) $ are orthogonal hence the image of $ A' $ under the inversion WRT $ (I) $ is the reflection of $ A_1 $ in $ I $ . Since $ \omega_A $ is tangent to $ AI $ , so the image $ A'' $ of $ \omega_A $ under the inversion WRT $ (I) $ is the D-altitude of $ DEF $ . Similarly, the image of $ \omega_B $ , $ \omega_C $ under this inversion is E-altitude, F-altitude of triangle $ DEF $ , so $ \omega_A,\omega_B,\omega_C $ pass through the image of the orthocenter $ T $ of $ DEF $ under the inversion WRT $ (I) $ . Since $ OI $ is the Euler line of triangle $ DEF $ , so $ T, X, I, O $ are collinear. Since $ A''IA_1T $ is a parallelogram, so $ \angle AXO =\angle AXI =\angle IA_1T =\angle TA''I =\angle IXA' =\angle OXA' $ Q.E.D

Let $Be$ be Bevan point of $\triangle ABC$ let $A_{1},A_{2}$ be the midpoints of smaller and larger arc $\widehat{BC}$ respectively. Consider the $\mathcal{I}_{\sqrt {IA\cdot IA_{1}}}$ along with with the central symmetry around $I$.$\mathcal{I}$ obviously fixes $\odot ABC$ and hence $A$ is taken to $A_{1}$.$A',I,A_{2}$ are collinear (well-known) and hence $\omega_{A}$ is taken to a line passing thru $A_{2}$ and parallel to $AI$.Consider the triangle $\triangle I_{A}I_{B}I_{C}$,in which $A_{2}$ is the midpoint of $I_{B}I_{C}$ and $I_{A}A_{1}$ is an altitude which implies the intersection of the three lines is actually the circumcenter of $\triangle I_{A}I_{B}I_{C}$ i.e Be of the $\triangle ABC$.The previous ,besides proving the concurrency ,proves that $X$ is on $IO$.By rewriting the angle condition with this knowledge with respect to $\mathcal{I}$ ,it is equivivalent to proving $\angle IA_{1}Be=\angle IA_{2}Be$.Having in mind that $A_{2}$ is the midpoint of $I_{B}I_{C}$ and $A_{1}$ is the midpoint of $I_{A}I$ $\implies$ $A_{1}IA_{2}Be$ is a parallelogram from which the angle condition follows.$\clubsuit$

Solution Let $DEF$ be the contact triangle. Let $X,Y,Z $ be the midpoints of the sides $EF,FD,DE$ respectively. Let $H$ be the orthocenter of $DEF$. Suppose that $M,N,P$ are the midpoints of $DH,EH,FH$ respectively. Now, observe that points $I,H,O$ are collinear. We see that since $A'I$ bisects arc $BAC$, an inversion about $I$ which fixes the incircle maps $A'$ to $M$. This inversion sends $A$ to $X$ and so, $\omega_A$ is sent to the line through $M$ parallel to $IX$. Thus, the circles $\omega_A,\omega_B,\omega_C$ intersect at $H'$ and $I$ where $H'$ is the inverse of $H$ in the incircle. Since $H',I,O$ are collinear, we need to show that $\angle IH'A'=\angle IH'A$. This is equivalent to $\angle IXH=\angle IMH$ which follows since $HXIM$ is a parallelogram as $DH=2IX$. The result follows.

We claim that the desired point is $T= AOA' \cap BOB' \cap COC'$. If $S=AA'\cap BB' \cap CC'$, let $T_A= AOA'\cap OS$. We have $T_AS\cdot SO=AS\cdot SA'$ which implies $T_A=T_B=T_C$ which shows $T= AOA' \cap BOB' \cap COC'$. Observe that $T$ satisfies the angle condition. Let $AA' \cap (I)=U$. We have $UI\parallel AO$ from the homothety taking $I$ to $O$ centered at $S$ which shows $UIA'T$ is a cyclic quadrilateral. Performing inversion centered at $A$ with radius $AI$ then reflecting wrt $AI$ takes $(I)$ to $A$-mixtilinear which implies $AU\cdot AA'=AI^2$. Thus $\omega_A=UIA'T \implies T \in \omega_A$ by symmetry $T= \omega_A \cap \omega_B \cap \omega_C$.

Let $M_a$ and $M_1$ denote the midpoints of arcs $BC$ and $BAC$ respectively, the incircle touch $BC, CA, AB$ at $D, E, F$ respectively, $K_a = M_aA' \cap BC$, $S_a$ be the second intersection between $(AEIF)$ and $(ABC)$, i.e. the $A$-Sharky Devil point, and $\gamma$ denote the incircle of $(ABC)$. Trivially, we have $M_a \in AI$. Now, the Shooting Lemma and Fact $5$ yields $$M_aA' \cdot M_aK_a = M_aB^2 = M_aI^2 = Pow_{\omega_a}(M_a)$$so $K_a$ lies on $\omega_a$. It's well-known that $A', I, M_1$ are collinear. Hence, we have $$\angle IDK_a = 90^{\circ} = \angle M_1A'K_a = \angle IA'K_a$$so $D \in \omega_a$. But it's well-known that $M_a, D, S_a$ are collinear, so the Shooting Lemma and Fact $5$ once again yield $$M_aD \cdot M_aS_a = M_aB^2 = M_aI^2 = Pow_{\omega_a}(M_a)$$implying $S_a \in \omega_a$. Now, we know $$\angle IS_aK_a = \angle IDK_a = 90^{\circ} = \angle IFA = \angle IS_aA$$so $A, S_a, K_a$ are collinear. Now, let $P$ denote the projection of $D$ onto $EF$, and consider the inversion about $\gamma$. It's well-known that $P$ and $S_a$ are inverses wrt $\gamma$. Trivially, we have $I^* = P_\infty$ and $D^* = D$, so $\omega_a$ goes to $DP$, i.e. the $D$-altitude of $DEF$. Analogous reasoning implies $\omega_b, \omega_c$ map to the $E$-altitude and $F$-altitude respectively, so $\omega_a^*, \omega_b^*, \omega_c^*$ concur at the orthocenter of $DEF$, which we denote by $X^*$. Inverting back gives the first desired result. Now, let $D'$ be the $D$-antipode and $\Omega$ be the Nine-Point Circle of $DEF$. It's well-known that $$A'^* = (ABC)^* \cap \omega_a^* = \Omega \cap DP.$$Since $P$ is the inverse of $S_a$, we know $A'^*$ must be the other intersection, i.e. the midpoint of $DX^*$. Now, it's easy to see $IA'^* \parallel D'X^*$. Claim: $O, I, X$ are collinear. Proof. Because $I, X, X^*$ are collinear from inversion, it suffices to show $O \in IX^*$, i.e. $O$ lies on the Euler Line of $DEF$. Now, since $(ABC)$ goes to $\Omega$ under incircle inversion, we know the center of $\Omega$, i.e. the Nine-Point Center of $DEF$ lies on $OI$. But the Nine-Point Center clearly lies on the Euler Line, so we're done. $\square$ Trivially, we know $A^*$ is the midpoint of $EF$. Now, the Orthocenter Reflection Lemma yields $A^* \in D'S^*$, so $$\angle AXO = \angle AXI = \angle IXA = \angle IA^*X^* = \angle PX^*A^*$$$$= \angle PA'^*I = \angle IA'^*X^* = \angle IXA' = \angle OXA'$$which finishes. $\blacksquare$ Remark: Let $DP$ meet $\Omega$ again at $H_d$. Then, $H_d \in \omega_a$ from inversion. Claim: $X$ lies on $(AOA'), (BOB'), (COC')$. Furthermore, $X$ is the inverse of $X_{56}$ wrt $(ABC)$. Proof. By our original proof, we know $XO$ bisects $\angle AXA'$. But $OA = R = OA'$, i.e. $O$ lies on the perpendicular bisector of $AA'$, so $O$ must the the midpoint of arc $AA'$ wrt $(AXA')$. Analogous processes imply $X \in (BOB'), (COC')$. Now, inverting about $(ABC)$ implies $$X^* = AA' \cap BB' \cap CC' = X_{56}$$as desired. $\square$ Second Approach: Let $T$ be the inverse of $X_{56}$ wrt $(ABC)$, i.e. $T = (AOA') \cap (BOB') \cap (COC')$. Since $OA = OA'$, we know $\angle ATO = \angle OTA'$ is true. (Similar results hold for $B, C$.) Now, we show $T \in \omega_a$, which finishes by symmetry. Let $AA'$ meet first the incircle at $K, Z$ such that $AK < AZ$, the $A$-mixtilinear incircle touch $AB$ at $Q$, and $AA'$ meet the $A$-mixtilinear incircle again at $T$. Then, $$AKF \overset{+}{\sim} ATQ \overset{-}{\sim} AQA'$$so $A'QFK$ is cyclic, yielding $$AK \cdot AA' = Pow_{(A'QFK)}(A) = AF \cdot AQ = AI^2 = Pow_{\omega_a}(A)$$which implies $K \in \omega_a$. Now, observe a homothety at $X_{56}$ takes $I$ to $O$ and $K$ to $A$, so $AO \parallel KI$. Hence, Reim's implies $KIA'T$ is cyclic, so $T \in (A'IK) = \omega_a$, as desired. $\blacksquare$

Attachments:

Let $X_A\in BC$ satisfy that $X_AI\perp AI.$ We will now show the following Claim. $X_A\in \omega_A.$ Proof This is equivalent to showing that $\angle X_AA'I=90^\circ.$ This is in fact well known, and for a proof see my post (#83) in here: https://artofproblemsolving.com/community/u930637h1176478p27448377. $\hfill \blacksquare$ Now deffine $X_B,X_C$ similarly. By homothety of $\frac{1}{2}$ around $I$ then the centers of $\omega_A,\omega_B,\omega_C$ are collinear iff $X_A,X_B,X_C$ are collinear. In order to show this, which as mentioned will finish the problem, notice that $X_AI^2=XB\cdot XC,$ as $X_AI$ is tangent to $(BIC)$ at $I.$ This implies that $X_1$ lies on the radical axis of $(ABC)$ and $I.$ This is also true for $X_B$ and $X_C,$ showing the desired collinearity. We've now shown that $X$ exists and want are left to show the angle condition. First notice that $X$ is collinear with $X_AX_BX_C$ trivially as $\angle X_AXI=\angle X_BXI=\angle X_CXI=90^\circ.$ Now as $OI\perp X_AX_BX_C $ because of the radical axis condition, and $IX\perp X_AX_BX_C$ we also have that $O-I-X.$ We now deffine $A_1=XA\cap BC$ and $A_2=XA'\cap (ABC).$ If we show that $XA_2=XA$ we will be done. In order to do this consider the inversion centered at $X$ with radius $XI,$ which we know sends $(ABC)\to (ABC)$ as $X$ lies on the radical axis of $I$ and $(ABC).$ Then we have that $A\to A_1$ and $A'\to A_2.$ Notice how $\omega_A\to A_2I$ and $AI\to (XA_1I),$ were $A_2I$ is tangent to $(XA_1I)$ at $I.$ Now we are already done as $\angle AIX=\angle XIA_2 $ follows by moving angles with the tangency conditions.

The condition $\angle AXO = \angle OXA'$ is equivalent to $A,O,A',X$ are cyclic because of $|AO|=|OA'|$ So we will prove that $X= \omega_A \cap (AOA')$. Let $IO \cap \omega_A = X'$, $M$ be the midpoint of arc $BC$ and $N$ be the midpoint of arc $BAC$. $\angle A'AO = \angle A'IM = \angle A'XI = \angle A'XO$. Therefore, $A,O,A',X$ are cyclic. Now we just need to prove that the lenght $|IX|$ is symmetric for $A,B,C$. Let $AI \cap (AOA') = D \Rightarrow |XI|=\frac{|AI||ID|}{|IO|}$. $\angle A'IM= \angle A'AO$ and $\angle IDA'= \angle ADA'= \angle AOA'$. So $IDA'$ and $AOA'$ are similar $\Rightarrow |ID|= |AO| \frac{|A'I|}{|AA'|}=|AO| \frac{|IM|}{|NM|} $ $\Rightarrow |XI|=\frac{|AI||ID|}{|IO|}=\frac{|AO|}{|NM||IO|}|IM||AI|=\frac{1}{2|IO|}.Pow_{\omega}(I) $ which is symmetric for $A,B,C$. $\Rightarrow X=X' \blacksquare$