Let $P=BG\cap AC$. Then, by Power of Point, $G,B,C,A'$ are concyclic if and only if $PC*PA'=PG*PB$. Now, it's a well known fact that $PG=\frac{1}{3}PB$, so $PG*PB=3PG^2$. One the other hand, since $CA'=AC$, we know that $PA'=PC+CA'=PC+AC=PC+2PC=3PC$, so $PC*PA'=3PC^2$. Thus $3PG^2=3PC^2\Rightarrow PG=PC=PA$. Then, $P$ is the center of the circle $(AGC)$, so $\angle{AGC}=\frac{1}{2}\angle{APC}=90^o$, QED.