If we have an equilateral triangle with a square inscribed in it, could we prove that the triangles we get are congruent? 
(I forgot to add a last point. Let $X$ be the midpoint of $\overline{GF}$.)

Could we prove that triangles $CGX$, $CFX$, $GAS$, and $FBE$ are all congruent?


Let $M$ be the midpoint of $\overline{AB}$. As $C,G$ and $A$ are collinear, $C,X$ and $M$ are collinear and $\overline{GX}$ is parallel to $\overline{AM}$, $\triangle CXG$ must be similar to $\triangle CMA$. Hence, $\frac{GX}{XC}=\frac{AM}{MC}$. Then, observe that $XC=MC-2GX$ and, by Pythagoras', $MC=\sqrt{(2AM)^2-AM^2}=\sqrt{3}AM$.
Hence, $\frac{GX}{\sqrt{3}AM-2GX}=\frac{AM}{\sqrt{3}AM}$ and $GX=\frac{\sqrt{3}}{\sqrt{3}+2}AM$. Given that $\triangle CXG$ and $\triangle GDA$ are oriented the same way, if they were congruent, we would have $GX=AD=\frac{AM}{2}$ but this is not the case as $\frac{\sqrt{3}}{\sqrt{3}+2}\neq \frac12$. Hence the triangles are not congruent and furthermore, $G$ is not the midpoint of $AC$.

