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Let $k_2$ be the circumcircle of the triangle $A_0{B}_{0}C$. Since $k_2$ is the image of the circle $k$ under homothety with centre $C$ (with the coefficient $\frac{1}{2}$), circles $k_2$ and $k$ touch at point $C$. Tangents $t_X$ and $t_C$ of circle $k$ through points $X$ and $C$ are respectively radical axes of pairs of circles $(k,k_1)$ and $(k,k_2)$, so the point $R$ which is their intersection is the radical center of circles $k$, $k_1$, $k_2$. Hence the point $R$ lies on the radical axis $A_0{B}_0$ of the pair of circles $(k_1,k_2)$. Let $p$ be the line through points $R$, $A_0$ and $B_0$.



Let $S$ be the intersection of the lines $p$ and $DX$, and the point $P$ be on the line $t_C$ such that the point $C$ lies between $P$ and $R$. Then $\angle CRS=\angle PCD$ (because the lines $CD$ and $p$ are parallel) and $\angle CXS=\angle CXD=\angle PCD$ (because of the chord-tangent theorem using tangent $t_C$). This implies that $\angle CXS=\angle CRS$ and the quadrilateral $CSXR$ is cyclic. Hence $\angle RXC=\angle RSC$. On the other hand, $\angle RSC=\angle SCD$ (again because $CD$ and $p$ are parallel) and $\angle CDS=\angle CDX=\angle RXC$ (because of the chord-tangent theorem using tangent $t_X$). This implies $\angle SCD=\angle RSC=\angle RXC=\angle CDS$, thus $S$ lies on the bisector of the segment $CD$.



It remains to show that the points $D$, $S$, $G$ and $N$ lie on the same line. Note that $CDN$ is a right triangle whose circumcentre is the point $S$, so the point $N$ lies on the line $DS$. Let $C_0$ be the midpoint of $AB$. Let $G^{\prime }$ be the intersection of the lines $DS$ and $C{C}_0$. The triangles $C_{0}N{G}^{\prime }$ and $CD{G}^{\prime }$ are similar and $|C{G}^{\prime }|:|C_0{G}^{\prime }|=|CD|:|N{C}_0|=2:1$. But the point $G$ is the unique point on the segment $C{C}_0$ dividing that segment in the ratio $2:1$, so $G=G^{\prime }$ and $G$ lies on the line $DS$.

This shows that the points $G$, $N$ and $X$ lie on the line $DS$.