Why does the circumcentre of a triangle inscribed in a square lie on the diagonal of the square?

Question

There are a few more conditions that I did not mention in the title which can be seen in the sketch below where n is the perpendicular bisector of l, as p is of j. The angle alpha is always 45°. Why does G lie on the diagonal of the square?

I do not know where to start this proof, the only thing I have discovered that might be useful is that the triangle AGH is similar to ABE and AKG is similar AFD but I do not know how to prove that either.

Answer 1

Central angle $\angle EGF=90°$, because it is the double of inscribed angle $\angle EAF$. Hence point $G$ lies on the circle $c$ having $EF$ as diameter, and at the same time it is equidistant from $E$ and $F$.

But the intersection of $c$ with diagonal $AC$ is equidistant from $E$ and $F$: just consider congruent triangles $EGI$ and $FGH$ in diagram below.

Answer 2

Hint: The lines to draw in are $EG$ and $EF$. What can we say about the angles in triangle $EFG$ and quadrilateral $ECFG$?