My solution : Let $ A_1 $ be the reflection of $ A $ in $ XO $ and $ T \equiv AA_1 \cap XO $ . Let $ E, F, K $ be the midpoint of $ CA, AB, XA $, respectively . Since $ A, T, X, B_1, C_1 $ lie on the circle $ \odot (K) $ (with diameter $ AX $) , so $ \measuredangle XC_1T=\measuredangle XAA_1=\measuredangle AA_1X=\measuredangle C_1BX \Longrightarrow XB \perp TC_1 $ , hence $ KF \perp TC_1 $ ($\because KF \parallel XB $) $ \Longrightarrow KF $ is the perpendicular bisector of $ TC_1 $ , so $ FC_1=FT \Longrightarrow $ $ T $ lie on the circle $ \odot (F, FC_1) \equiv \omega_C $ with center $ F $ and radius $ FC_1 $ . Similarly, we can prove $ T \in \odot (E, EB_1) \equiv \omega_B \Longrightarrow \omega_B $ and $ \omega_C $ has a common point $ T $ on $ XO $ . Q.E.D

My solution: Let $N$ be the midpoint of $AB$. $OX$ cuts $\Gamma$ again at $X'$. Draw $X'C_{2}\perp AB$ at $C_{2}$ and $AA'\perp OX$ at $A'$. From $ON\parallel XC_{1}\parallel X'C_{2}$ and $O$ is midpoint of $XX'$, we have $N$ is midpoint of $C_{1}C_{2}$ $\Rightarrow C_{1}C_{2}$ is diameter of $\omega _{C}$. From $AA'XC_{1}$ and $AA'XC_{2}$ are concyclic, we have $\angle AXX'=\angle AXA'=\angle AC_{1}A'=\angle C_{2}C_{1}A'$ and $\angle AX'A'=\angle AC_{2}A'=\angle C_{1}C_{2}A'$ $\Rightarrow \angle C_{1}C_{2}A'+\angle C_{2}C_{1}A'=angle AX'A'+\angle AXX'=90^{\circ}$ ($\because \angle XAX'=90^{\circ}$) $\Rightarrow C_{1}A'C_{2}=90^{\circ}$ $\Rightarrow A'\in \omega _{C}$. Similarly $A'\in \omega _{B}$ $\Rightarrow OX\cap \omega _{B}\cap \omega _{C}=A'$. DONE

Attachments:

Let $Y$ be the projection of $A$ onto $XO$. We claim that $\omega_B$, and by symmetry $\omega_C$, passes through $Y$. Note that $AB_1YC_1X$ and $ANYO$ are cyclic, where $N$ is the midpoint of $AB$. Let the projection of $O$ onto $AX$ be $Z$, so $Z$ lies on the second circle. Then by spiral similarity about $Y$, $YNC_1\sim YZX\sim AOX$. This implies $NY=NC_1$, as desired.