There is nothing to prove if some two of four points $A_1$, $B_1$, $C_1$, $G$ coincide. So, we can assume that all these points are pairwise distinct.

Let $\Gamma$ and $\Gamma(A_1)$ denote the circumcircles of the triangles $ABC$ and $AA_1B$, respectively. Let $M_A$ be the intersection point of the lines $AA_1$ and $BC$. Since $\angle A_1AB = \angle A_1BC$, it follows that $\Gamma(A_1)$ touches $BC$ at $B$. By the Power of a Point Theorem, we have $M_AB^2 = M_AA_1 \cdot M_AA$, Similarly, $M_AC^2 = M_AA_1 \cdot M_AA$, hence $M_AB^2 = M_AC^2$, so $M_AB = M_AC$, i.e., $M_A$ is the midpoint of the side $BC$. Thus, $A_1$ lies on the median of the triangle $ABC$ from the vertex $A$. By the same arguments, $B_1$ and $C_1$ lie on the medians of the triangle $ABC$ from the vertices $B$ and $C$, respectively.

Let $H$ be the orthocenter of the triangle $ABC$. It is well known that the points which are symmetric to $H$ with respect to the sides of the triangle $ABC$ lie on the circumcircle of $ABC$. If $H_A$ is symmetric to $H$ with respect to the side $BC$, then $H_A$ belongs to $\Gamma$ and we have
$$\begin{aligned} \angle BHC &= \angle BH_A C = 180^\circ - \angle BAC = 180^\circ - (\angle A_1 AB + \angle A_1 AC) = \\ &= 180^\circ - (\angle A_1 BC + \angle A_1 CB) = \angle BA_1 C, \end{aligned}$$
hence the quadrilateral $BA_1HC$ is cyclic and $A_1$ belongs to the circle passing through the points $B$, $H$, $C$.

Further,
$$\begin{aligned} \angle HA_1M_A &= \angle HA_1B + \angle BA_1M_A = \angle BCH + (\angle A_1BA + \angle A_1AB) = \\ &= (90^\circ - \angle ABC) + (\angle A_1BA + \angle A_1BC) = 90^\circ - \angle ABC + \angle ABC = 90^\circ. \end{aligned}$$
Therefore, $A_1$ is the projection of $H$ on the median $AM_A$. Similarly, $B_1$ and $C_1$ are the projections of $H$ on the median of the triangle $ABC$ from the vertices $B$ and $C$, respectively.

Since $G$ is the intersection point of the medians of $ABC$, we have $\angle HA_1G = \angle HB_1G = \angle HC_1G = 90^\circ$. Thus $A_1, B_1, C_1$ lie on the circle with the diameter $GH$, so all points $A_1, B_1, C_1, G$ lie on the same circle.