Let $D$, $E$, $F$ be the middles of the sides $BC$, $AC$, $AB$, respectively.

Let, also $B_1C_1$, $A_1C_1$, $A_1B_1$ be the perpendicular bisectors of the line segments $GA$, $GB$ and $GC$, respectively. Then the points $A_1$, $B_1$ and $C_1$ are the circumcenters of the triangles $GBC$, $GAC$ and $GAB$, respectively. Hence $A_1D$, $B_1E$ and $C_1F$ are the perpendicular bisectors of the sides $BC$, $AC$ and $AB$, respectively, and therefore they will pass through the circumcenter $O$ of the triangle $ABC$.

Next, we will show that $A_1D$, $B_1E$ and $C_1F$ are the medians of the triangle $A_1B_1C_1$. Let the extension of $A_1D$, meets $B_1C_1$ at $N$. We will prove that $N$ is the middle of the line segment $B_1C_1$.

From the inscribe quadrilateral $AMEB_1$ ($\hat{M} = \hat{E} = 90^\circ$), we have $M\hat{A}E = M\hat{B}_1E = \hat{\omega}$. Also, from the inscribe quadrilateral $DOEC$ ($\hat{D} = \hat{E} = 90^\circ$), we get $E\hat{C}D = E\hat{O}N = \hat{\phi}$. Therefore the triangles $ADC$ and $B_1NO$ are similar, and so
$$\frac{NB_1}{NO} = \frac{AD}{CD}. \qquad (1)$$
From the inscribe quadrilateral $AMFC_1$ ($\hat{M} = \hat{F} = 90^\circ$), we have $M\hat{A}F = M\hat{C}_1F = \hat{x}$ and similarly from $DOFB$ ($\hat{D} = \hat{F} = 90^\circ$), we obtain that $F\hat{B}D = F\hat{O}N = \hat{y}$. From the above equalities the triangles $ADB$ and $C_1NO$ are similar and therefore:
$$\frac{NC_1}{NO} = \frac{AD}{BD}. \qquad (2)$$
From (1) and (2) we get $NB_1 = NC_1$. In a similar way we prove that $B_1E$, $C_1F$ are the other two medians of the triangle $A_1B_1C_1$.