Given Circle $E$ inscribed in square $ABCD,$ prove that its radius is equal to half the length of the square's side length.

Question

Given Circle $E$ inscribed in square $ABCD$, prove that its radius is equal to half the length of the square's side length. I'm stuck on this proof but I've got a few ideas. If anyone could help that would be awesome.

1. Draw radii $EF$ and $EG$ so that $EF$ is perpendicular to $CD$ and $EG$ is perpendicular to $AD$
2. With this, we now have a quadrilateral $EFDG$.
3. The presence of congruent and opposite angles implies that the new quadrilateral formed is a rectangle.
4. $EF$ is congruent to $GD$ and $EG$ is congruent to $FD$.