First solution. Because $AC'GB'$ is cyclic, we have $\measuredangle GAB' = \measuredangle GC'B'$. Because $B'C'$ is parallel to $BC$, we have $\measuredangle CC'B' = \measuredangle C'CB$. We deduce that
$$\sin \measuredangle A'A C = \sin \measuredangle C'C B.$$

But triangles $AA'C$, $BCC'$ have the same area which is half of the area of triangle $ABC$. We deduce that
$$\frac{1}{2} AA' \cdot AC \sin \measuredangle A'A C = \frac{1}{2} CC' \cdot BC \sin \measuredangle C'C B$$
and therefore
$$\frac{AC}{CC'} = \frac{BC}{AA'}.$$
We proceed in a similar way, by considering triangles $ABA'$ and $BCB'$, obtaining
$$\frac{AB}{BB'} = \frac{BC}{AA'},$$
and deduce the relation
$$AB \cdot CC' = AC \cdot BB'.$$

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

Third solution (Contains also a proof for the converse). We can see from the first solution that the quadrilateral $AC'GB'$ 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 $A'C^2 = \overline{A'G} \cdot \overline{A'A}$ by using the power of the point $A'$ with respect to this circumcircle. But
$$A'C = \frac{1}{2} BC, \quad \overline{A'G} = \frac{1}{3} \overline{A'A} \quad \text{and} \quad AA' = \sqrt{\frac{2AB^2 + 2AC^2 - BC^2}{4}}.$$
We deduce that $AC'GB'$ is cyclic if and only if
$$2BC^2 = AB^2 + AC^2$$
Now, using the formulas for the medians
$$BB' = \sqrt{\frac{2BC^2 + 2AB^2 - CA^2}{4}} \quad \text{and} \quad CC' = \sqrt{\frac{2CA^2 + 2BC^2 - AB^2}{4}},$$
we deduce that the relation
$$AB \cdot CC' = AC \cdot BB'$$
is equivalent to
$$AB^2 \cdot \left(2CA^2 + 2BC^2 - AB^2\right) = AC^2 \cdot \left(2BC^2 + 2AB^2 - CA^2\right),$$
which is equivalent to
$$\left(AB^2 - AC^2\right) \cdot \left(2BC^2 - AB^2 - AC^2\right) = 0$$
which is equivalent to either $AB = AC$ or $AC'GB'$ is cyclic.