SELECTION EXAMINATION

Let $D$, $E$, $F$ be the middles of the sides $BC$, $AC$, $AB$, respectively. Let, also $B_1{C}_1$, $A_1{C}_1$, $A_1{B}_1$ 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 $A_{1}D$, $B_{1}E$ and $C_{1}F$ 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 $A_{1}D$, $B_{1}E$ and $C_{1}F$ are the medians of the triangle $A_1{B}_1{C}_1$. Let the extension of $A_{1}D$, meets $B_1{C}_1$ at $N$. We will prove that $N$ is the middle of the line segment $B_1{C}_1$.

From the inscribe quadrilateral $AME{B}_1$ ($\hat{M}=\hat{E}=90^{\circ }$), we have $M\hat{A}E=M{\hat{B}}_{1}E=\hat{\omega }$. Also, from the inscribe quadrilateral $DOEC$ ($\hat{D}=\hat{E}=90^{\circ }$), we get $E\hat{C}D=E\hat{O}N=\hat{\varphi }$. Therefore the triangles $ADC$ and $B_{1}NO$ are similar, and so $$\frac{N{B}_1}{NO}=\frac{AD}{CD}.(1)$$ From the inscribe quadrilateral $AMF{C}_1$ ($\hat{M}=\hat{F}=90^{\circ }$), we have $M\hat{A}F=M{\hat{C}}_{1}F=\hat{x}$ 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 $C_{1}NO$ are similar and therefore: $$\frac{N{C}_1}{NO}=\frac{AD}{BD}.(2)$$ From (1) and (2) we get $N{B}_1=N{C}_1$. In a similar way we prove that $B_{1}E$, $C_{1}F$ are the other two medians of the triangle $A_1{B}_1{C}_1$.