Here is a computational approach, which is relatively clean. WLOG that $AB < AC.$ Let $\omega_1, \omega_2$ denote the circumcircles of $\triangle BC'D_c, CB'D_b,$ respectively. It suffices to show that $O$ is on the radical axis of these two circles. Let $pow_{\Omega} (P)$ denote the power of point $P$ w.r.t. $\Omega.$ Let $X_b, X_c$ be the points where $\omega_2, \omega_2$ intersect $AC, AB$ again. Let $X = AO \cap BC.$ Let $a, b, c$ denote the lengths of the sides as usual and let $r = \frac{AX}{XA'}.$ Since $O$ is the midpoint of $AA'$, it suffices to show that: $$pow_{\omega_1}(A) - pow_{\omega_2}(A) = -pow_{\omega_1}(A') + pow_{\omega_2}(A').$$Firstly, notice that by Reim we easily obtain that $D_cX_c || D_bX_b||AA'.$ By Menelaus in $\triangle ABX$ with transversal $M_cD_cA'$, we have that $\frac{BD_c}{D_cX} = r.$ Analogously, $\frac{CD_b}{D_bX} = r.$ Therefore, $D_cX_c || D_bX_b || AA'$ gives that: $$pow_{\omega_1}(A) = AX_c \cdot AB = \frac{c^2}{r+1}, pow_{\omega_2}(A) =AX_b \cdot AC = \frac{b^2}{r+1}.$$ Now, by Stewart's Theorem in $\triangle A'BC$, we get that: $$pow_{\omega_1}(A') = A'D_c \cdot A'C' = A'D_c \cdot D_cC' + A'D_c^2 = \frac{A'C^2 \cdot BD_c + A'B^2 \cdot CD_c}{BC}.$$ Analogously, we have that: $$pow_{\omega_2}(A') = \frac{A'C^2 \cdot BD_b + A'B^2 \cdot CD_b}{BC}.$$ Therefore, we can calculate that $pow_{\omega_1}(A') - pow_{\omega_2} (A') = \frac{(AB'^2 - AC'^2)D_bD_c}{BC}.$ Since $AB'^2 - AC'^2 = AC^2 - AB^2$ and $D_bD_c = D_bX + XD_c = \frac{XC}{r+1} + \frac{XB}{r+1} = \frac{a}{r+1}$, we can rewrite this as $\frac{b^2 - c^2}{r+1}.$ However, since we also know that $pow_{\omega_1}(A) - pow_{\omega_2}(A) = \frac{c^2 - b^2}{r+1},$ we're done. $\square$

Draw DDc,DDb//AB,AC then DDcDb and AMcMb perspective hence D on AA', DDb,DDc cuts OC,OB at J,K so (KDcBC') and (JDbCB') cyclic. By OB/OK=OA/OD=OJ/OC so O lies on PQ q.e.d

Let $CO, BO$ meet $A'M_B, A'M_C$ at $P, Q$, respectively. Let $CO$ meets circumcircle of $CD_BB'$ at another point $R$, let $BO$ meets circumcircle of $BD_CC'$ at another point $S$. By Desargue's Theorem on triangle $M_BPC$ and $M_CQB$, we get $PQ\parallel M_BM_C\parallel BC$. So $P,Q$ have same power WRT $\omega$. And $$RP\cdot PC=PB'\cdot PD_B=\frac{PB'}{PA'}\cdot pow_{\omega}(P)=\frac{QC'}{QA'}\cdot pow_{\omega}(Q)=QC\cdot QD_C=QS\cdot QB.$$Therefore $QS=PR$, $OS=OR$, $OR\cdot OC=OS\cdot OB$, so $O$ is on the radical axis of the circumcircles of $CD_BB'$ and $BD_CC'$. $\square$

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.92, xmax = 8.92, ymin = -5.24, ymax = 5.24; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */ draw(circle((-1.7836607462152814,-0.5008046251043033), 3.7894302412830574), linewidth(0.8) + zzttff); draw(circle((-3.2493003577126625,-0.5527868905540388), 2.54741471661165), linewidth(0.8) + zzttff); draw(circle((0.6775559606875308,1.0165128090193547), 3.563475829289608), linewidth(0.8) + zzttff); draw((-3.8751747034397392,-3.6607652135113575)--(-3.68,2.78), linewidth(0.8) + qqwuqq); draw((-5.14,-2.26)--(1.46,-2.46), linewidth(0.8) + blue); draw((1.46,-2.46)--(-3.68,2.78), linewidth(0.8) + blue); draw((-3.68,2.78)--(-5.14,-2.26), linewidth(0.8) + blue); draw((0.11267850756943742,-3.7816092502086063)--(-5.256197546800665,1.0161890209373659), linewidth(0.8) + red); draw((-3.8751747034397392,-3.6607652135113575)--(0.3078532110091765,2.6591559633027506), linewidth(0.8)); draw((0.11267850756943742,-3.7816092502086063)--(-2.0769664071675535,3.277257489631758), linewidth(0.8) + red); draw((-1.4654424259831296,-2.3713502295156625)--(-1.358600715425324,1.1544262188919205), linewidth(0.8)); draw((-0.31395684680535435,-2.406243731914989)--(-0.10488807862493901,4.49302561803871), linewidth(0.8)); draw((0.3078532110091765,2.6591559633027506)--(0.11267850756943742,-3.7816092502086063), linewidth(0.8) + qqwuqq); draw((0.3078532110091765,2.6591559633027506)--(-5.14,-2.26), linewidth(0.8) + linetype("4 4") + ffvvqq); draw((-0.10488807862493901,4.49302561803871)--(1.46,-2.46), linewidth(0.8) + linetype("4 4") + ffvvqq); /* dots and labels */ dot((-3.68,2.78),dotstyle); label("$A$", (-3.6,2.98), NE * labelscalefactor); dot((-5.14,-2.26),dotstyle); label("$B$", (-5.06,-2.06), NE * labelscalefactor); dot((1.46,-2.46),dotstyle); label("$C$", (1.54,-2.26), NE * labelscalefactor); dot((-1.7836607462152814,-0.5008046251043033),linewidth(4pt) + dotstyle); label("$O$", (-1.7,-0.34), NE * labelscalefactor); dot((0.11267850756943742,-3.7816092502086063),dotstyle); label("$A'$", (0.2,-3.58), NE * labelscalefactor); dot((-4.41,0.26),linewidth(4pt) + dotstyle); label("$M_{C}$", (-4.32,0.42), NE * labelscalefactor); dot((-1.11,0.16),linewidth(4pt) + dotstyle); label("$M_{B}$", (-1.02,0.32), NE * labelscalefactor); dot((-1.4654424259831296,-2.3713502295156625),linewidth(4pt) + dotstyle); label("$D_{C}$", (-1.8,-2.92), NE * labelscalefactor); dot((-0.31395684680535435,-2.406243731914989),linewidth(4pt) + dotstyle); label("$D_{B}$", (-0.76,-2.66), NE * labelscalefactor); dot((-5.256197546800665,1.0161890209373659),linewidth(4pt) + dotstyle); label("$C'$", (-5.18,1.18), NE * labelscalefactor); dot((-2.0769664071675535,3.277257489631758),linewidth(4pt) + dotstyle); label("$B'$", (-2,3.44), NE * labelscalefactor); dot((-3.8751747034397392,-3.6607652135113575),dotstyle); label("$H_{1}$", (-3.8,-3.46), NE * labelscalefactor); dot((-3.833926605504588,-2.299577981651376),linewidth(4pt) + dotstyle); label("$H_{3}$", (-3.76,-2.14), NE * labelscalefactor); dot((0.3078532110091765,2.6591559633027506),dotstyle); label("$A_{1}$", (0.38,2.86), NE * labelscalefactor); dot((-0.10488807862493901,4.49302561803871),linewidth(4pt) + dotstyle); label("$E$", (-0.02,4.66), NE * labelscalefactor); dot((-1.358600715425324,1.1544262188919205),linewidth(4pt) + dotstyle); label("$F$", (-1.28,1.32), NE * labelscalefactor); dot((0.15392660550458862,-2.420422018348624),linewidth(4pt) + dotstyle); label("$G$", (0.24,-2.26), NE * labelscalefactor); dot((-2.7621269194987086,1.9476098338695886),linewidth(4pt) + dotstyle); label("$P$", (-2.68,2.1), NE * labelscalefactor); dot((-1.1730427732325066,-2.028753076243345),linewidth(4pt) + dotstyle); label("$Q$", (-1.1,-1.86), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let the $A$-altitude meet $BC$ and $(ABC)$ at $H_3$, $H_1$ respectively. Let $A_1$ be the reflection of $H_1$ over $O$. Claim 1. $H_1B$ and $H_1C$ are tangent to $(BC'D_C)$ and $(CB'D_B)$ respectively. Proof. $\angle H_1BC=\angle A'CB=\angle A'C'B$ so $H_1B$ is tangent to $(BC'D_C)$, the other assertion is similar. $\blacksquare$ Claim 2. Let the perpendicular at $D_C$ to $BC$ meet $(BC'D_C)$ at $E$ and the perpendicular at $D_B$ to $BC$ meet $(CD_BB')$ at $F$. Then $B,F,A_1$ and $B,C,B_1$ are collinear Proof. $\angle EBD_C=90^{\circ}-\angle BC'A'=90^{\circ}-\angle BA_1A'=\angle D_CBA_1$ so $B,F,A_1$ are collinear, the other assertion is similar. $\blacksquare$ Claim 3. $A_1C\times XC=A_1B\times AY$ Proof. From Claim 2, $$A_1C\times XC=\frac{CD_B\cdot CG}{\cos^2{B}}$$Similarly $$A_1Y\times A_1B=\frac{GB\times BD_C}{\cos^2C}$$Meanwhile, $$\frac{CG}{GB}=\frac{BH_3}{CH_3}=\frac{\tan C}{\tan B}$$Therefore it suffices to show $$\frac{CD_B}{BD_C}=\frac{\sin 2B}{\sin 2C}\hspace{30pt}(1)$$Let $AA'\cap BC=K$, by Menelaus we have $$\frac{BD_C}{D_CK}=\frac{AA'}{A'K}=\frac{CD_B}{D_BK}$$Hence$$\frac{CD_B}{BD_C}=\frac{CK}{BK}=\frac{CK}{AK}\cdot\frac{AK}{BK}=\frac{\sin\angle CAK}{\sin\angle ACK}\cdot\frac{\sin\angle ABK}{\sin\angle BAK}=\frac{\sin 2B}{\sin 2C}$$$\blacksquare$ Therefore define $f(X)=Pow(X,(C'BD_C))-Pow(X,(B'CD_B))$. We have $$f(O)=\frac{f(H_1)+f(A_1)}{2}=\frac{H_1B^2-H_1C^2+A_1B(A_1B-FB)-A_1C(A_1C-EC)}{2}=\frac{H_1B^2-H_1C^2+A_1C^2-A_1B^2}{2}=0$$

Let $G$ be the centroid of $\triangle ABC$. Let $D=\overline{A'G}\cap\overline{BC}$. Let $H_A$ be point on $\odot(ABC)$ such that $\overline{AH_A}\perp\overline{BC}$. Let $A_1$ be point on $\odot(ABC)$ such that $\overline{AA_1}\parallel\overline{BC}$. Let $X=\overline{A'G}\cap \odot(ABC)$, let $Y=\overline{A_1D}\cap \odot(ABC)$, let $Z=\overline{BB'}\cap\overline{CC'}$. Let $R=\overline{BC}\cap\overline{B'C'}$, let $S=\overline{BC'}\cap\overline{B'C}$. [asy][asy]import olympiad;import geometry; size(12cm);defaultpen(fontsize(10pt)); pair O,A,B,C,G,Mb,Mc,A0,B0,C0,Ha,X,Y,A1,D,Z,P,Q,Db,Dc; O=(0,0);A=dir(115);B=dir(205);C=dir(335);Mb=midpoint(A--C);Mc=midpoint(A--B);G=(A+B+C)/3;A0=-A; B0=intersectionpoints(line(A0,Mb),circumcircle(A,B,C))[0];C0=intersectionpoints(line(A0,Mc),circumcircle(A,B,C))[0]; Ha=intersectionpoints(perpendicular(A,line(B,C)),circumcircle(A,B,C))[1]; X=intersectionpoints(line(A0,G),circumcircle(A,B,C))[0];A1=intersectionpoints(perpendicular(A0,line(B,C)),circumcircle(A,B,C))[0];D=extension(A0,G,B,C);Y=intersectionpoints(line(A1,D),circumcircle(A,B,C))[1];Z=extension(X,Y,A,Ha); P=extension(B,C,B0,C0);Q=extension(B,C0,B0,C);Db=extension(B,C,A0,Mb);Dc=extension(B,C,A0,Mc); draw(A--B--C--cycle);draw(circumcircle(A,B,C)); draw(A--Ha--A0--cycle);draw(Y--A1);draw(C0--A0--B0);draw(B--P--B0);draw(X--A0);draw(X--Y--P--A);draw(C--C0);draw(B--B0); draw(B--Q--C);draw(Y--Q--Ha);draw(D--Q,dashed);draw(circumcircle(B,C0,Dc)^^circumcircle(C,B0,Db),grey);draw(Ha--A1); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$O$",O,dir(90)); dot("$G$",G,dir(G)); dot("$A'$",A0,dir(A0)); dot("$B'$",B0,dir(B0)); dot("$C'$",C0,dir(C0)); dot("$H_A$",Ha,dir(Ha)); dot("$M_B$",Mb,dir(Mb)); dot("$M_C$",Mc,dir(Mc)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$A_1$",A1,dir(A1)); dot("$R$",P,dir(P)); dot("$S$",Q,dir(Q)); dot("$D_B$",Db,dir(Db)); dot("$D_C$",Dc,dir(Dc)); [/asy][/asy] Claim. $D$ lies on the radical axis of $\odot(BD_CC')$ and $\odot(CD_BB')$. Proof. By DDIT on complete quadrilateral $BCM_BM_CGP_\infty$, $(\overline{A'B},\overline{A'C}),(\overline{A'M_B},\overline{A'M_C})$ and $(\overline{A'G},\overline{A'P_\infty})$ are reciprocal pairs of involution. Projecting onto $\overline{BC}$, we get that there exists point $K$ such that $KB\cdot KC=KD_B\cdot KD_C=KD\cdot KP_\infty$, hence $D=K$ and we get desired. $\square$ Claim. $Z$ lies on $\overline{AH_A}$. Proof. By DDIT on complete quadrilateral $BCM_BM_CAP_\infty$, $(\overline{A'B},\overline{A'M_B}),(\overline{A'C},\overline{A'M_C})$ and $(\overline{A'A},\overline{A'P_\infty})$ are reciprocal pairs of involution. Projecting onto $\odot(ABC)$, we get that $\overline{BB'},\overline{CC'}$ and $\overline{AH_A}$ are concurrent, i.e. $Z$ lies on $\overline{AH_A}$. $\square$ Claim. $R$ lies on $\overline{AX}$. Proof. By DDIT on complete quadrilateral $BCM_BM_CAG$, $(\overline{A'B},\overline{A'C}),(\overline{A'M_B},\overline{A'M_C})$ and $(\overline{A'A},\overline{A'G})$ are reciprocal pairs of involution. Projecting onto $\odot(ABC)$, we get that $\overline{BC},\overline{B'C'}$ and $\overline{AX}$ are concurrent, i.e. $R$ lies on $\overline{AX}$. $\square$ Claim. $R$ lies on $\overline{H_AY}$. Proof. Pascal's theorem on $H_AYA_1AXA'$ yields collinearity of $R,H_A,Y$. $\square$ Claim. $S$ lies on $\overline{H_AX}$. Proof. By DDIT on complete quadrilateral $BCM_BM_CGP_\infty$, $(\overline{A'B},\overline{A'M_C}),(\overline{A'C},\overline{A'M_B})$ and $(\overline{A'G},\overline{A'H_A})$ are reciprocal pairs of involution. Projecting onto $\odot(ABC)$, we get that $\overline{BC'},\overline{B'C}$ and $\overline{H_AX}$ are concurrent, i.e. $S$ lies on $\overline{H_AX}$. $\square$ By radical axis thereom, $S$ lies on the radical axis of $\odot(BD_CC')$ and $\odot(CD_BB')$. Also, by Brokard, $Z$ lies on $\overline{XY}$ and $S$ lies on $\overline{AY}$. Finally, by Pascal's theorem on $A_1H_AXA'AY$, we get that $O=\overline{A_1H_A}\cap\overline{A'X}$, $S=\overline{H_AX}\cap\overline{AY}$ and $D=\overline{A_1Y}\cap\overline{XA'}$ are collinear, hence $O$ lies on the radical axis of $\odot(BD_CC')$ and $\odot(CD_BB')$. $\blacksquare$

I like this problem. Here's my generalization for it. Let $\omega$ and $O$ be respectively the circumcircle and the circumcenter of a triangle $ABC$. The line $AO$ intersects $\omega$ second time at $A'$. $M_B$ and $M_C$ are points on $AC$ and $AB$, respectively such that $M_BM_C$ is parallel to $BC$. The lines $A'M_B$ and $A'M_C$ intersect $\omega$ secondly at points $B'$ and $C$, and also intersect $BC$ at points $D_B$ and $D_C$, respectively. The circumcircles of $CD_BB'$ and $BD_CC'$ intersect at points $P$ and $Q$. Prove that $O$, $P$, $Q$ are collinear.

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My solution to the generalization (it seems that my solution is shorter then the above ones, using only single trig, without Desargue or involution )

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