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?

 A: First, show that they're similar. Each triangle has one angle of 90°, one of 60°, and one of 30°. That makes them similar.
(More specifically, for GAS and FBE, the 60° angle is the one they share with the equilateral triangle ABC. For the other two, the 30° angle is half of the angle at C. And all of them have a right angle too.)
To show congruence, check that they have one side the same length. For example, the sides you've marked with one line are each the hypothenuse of one of these triangles. So, if you can show that they're the same size, you're finished!
A: 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 $AC$.

