Let $k_2$ be the circumcircle of the triangle $A_0B_0C$. 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 $A_0B_0$ 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 $G'$ be the intersection of the lines $DS$ and $CC_0$. The triangles $C_0NG'$ and $CDG'$ are similar and $|CG'| : |C_0G'| = |CD| : |NC_0| = 2 : 1$. But the point $G$ is the unique point on the segment $CC_0$ dividing that segment in the ratio $2 : 1$, so $G = G'$ and $G$ lies on the line $DS$.

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