Preselection tests for the full-time training

First solution. Because $A{C}^{\prime }G{B}^{\prime }$ is cyclic, we have $\measuredangle GA{B}^{\prime }=\measuredangle G{C}^{\prime }{B}^{\prime }$. Because $B^{\prime }{C}^{\prime }$ is parallel to $BC$, we have $\measuredangle C{C}^{\prime }{B}^{\prime }=\measuredangle C^{\prime }CB$. We deduce that $$\sin \measuredangle A^{\prime }AC=\sin \measuredangle C^{\prime }CB.$$ But triangles $A{A}^{\prime }C$, $BC{C}^{\prime }$ have the same area which is half of the area of triangle $ABC$. We deduce that $$\frac{1}{2}A{A}^{\prime }\cdot AC\sin \measuredangle A^{\prime }AC=\frac{1}{2}C{C}^{\prime }\cdot BC\sin \measuredangle C^{\prime }CB$$ and therefore $$\frac{AC}{C{C}^{\prime }}=\frac{BC}{A{A}^{\prime }}.$$ We proceed in a similar way, by considering triangles $AB{A}^{\prime }$ and $BC{B}^{\prime }$, obtaining $$\frac{AB}{B{B}^{\prime }}=\frac{BC}{A{A}^{\prime }},$$ and deduce the relation $$AB\cdot C{C}^{\prime }=AC\cdot B{B}^{\prime }.$$

Second solution. Lets us consider the power of the point $B$ with respect to the circumcircle of the cyclic quadrilateral $A{C}^{\prime }G{B}^{\prime }$. We have $$\mathcal{P}(B)=\overline{B{C}^{\prime }}\cdot \overline{BA}=\overline{BG}\cdot \overline{B{B}^{\prime }}.$$ But $\overline{B{C}^{\prime }}=\frac{1}{2}\overline{BA}$ and $\overline{BG}=\frac{2}{3}\overline{B{B}^{\prime }}$. We deduce that $$B{B}^{\prime }=\frac{\sqrt{3}}{2}AB.$$ Similarly, by considering the power of the point $C$ with respect to the circumcircle of the cyclic quadrilateral $A{C}^{\prime }G{B}^{\prime }$, we obtain $$C{C}^{\prime }=\frac{\sqrt{3}}{2}AC$$ From these two relations we deduce easily that $$AB\cdot C{C}^{\prime }=AC\cdot B{B}^{\prime }.$$

Third solution (Contains also a proof for the converse). We can see from the first solution that the quadrilateral $A{C}^{\prime }G{B}^{\prime }$ is cyclic if and only if $$\measuredangle GAC=\measuredangle GCB.$$ This is equivalent to saying that the line $BC$ is tangent to the circumcircle of triangle $AGC$. This property is equivalent to saying that $A^{\prime }{C}^2=\overline{A^{\prime }G}\cdot \overline{A^{\prime }A}$ by using the power of the point $A^{\prime }$ with respect to this circumcircle. But $$A^{\prime }C=\frac{1}{2}BC,\overline{A^{\prime }G}=\frac{1}{3}\overline{A^{\prime }A}\text{and}A{A}^{\prime }=\sqrt{\frac{2A{B}^2+2A{C}^2-B{C}^2}{4}}.$$ We deduce that $A{C}^{\prime }G{B}^{\prime }$ is cyclic if and only if $$2B{C}^2=A{B}^2+A{C}^2$$ Now, using the formulas for the medians $$B{B}^{\prime }=\sqrt{\frac{2B{C}^2+2A{B}^2-C{A}^2}{4}}\text{and}C{C}^{\prime }=\sqrt{\frac{2C{A}^2+2B{C}^2-A{B}^2}{4}},$$ we deduce that the relation $$AB\cdot C{C}^{\prime }=AC\cdot B{B}^{\prime }$$ is equivalent to $$A{B}^2\cdot \left(2C{A}^2+2B{C}^2-A{B}^2\right)=A{C}^2\cdot \left(2B{C}^2+2A{B}^2-C{A}^2\right),$$ which is equivalent to $$\left(A{B}^2-A{C}^2\right)\cdot \left(2B{C}^2-A{B}^2-A{C}^2\right)=0$$ which is equivalent to either $AB=AC$ or $A{C}^{\prime }G{B}^{\prime }$ is cyclic.