Locus of point with constant angle

Question

It's been a few years since I dealt with Euclidean Geometry and I now fell onto this problem: We are given the square $ABCD$, and point $F$ on $AB$. We construct a right angle $FEH$ for which $EH = EF$. We are looking for the locus of point $H$, as $F$ moves between $A$ and $B$. I did it with Geogebra and it seems that the locus is line (or, better, a segment) perpendicular to $AB$ but I can't figure out anything more than this.

Just to clarify, the reason I am trying to figure out this locus, is because I want to construct a square, with one side $EF$ and the opposite side passing from $C$. So the perpendicular at $EH$ at point $H$ (which moves on this blue line), must pass from $C$.

Any help is most welcome.

EDIT:

To clarify: Point $E$ on $AD$ is fixed. Point $F$ is not fixed; it moves on $AB$ and angle $FEH$ is always right; therefore point $H$ is moving, as $F$ moves between $A$ and $B$.

Answer 1

If $EA=a$ and $\angle EFA=\theta$, then $EF=EH=a\csc\theta$.

Therefore the horizontal displacement of $H$ from the line $AD$ is $a\csc \theta*\sin\theta=a$.

Therefore $H$ moves on a vertical straight line at the same distance from $AD$ as $E$ is from $A$.

Answer 2

Since $E$ is fixed, we see that $H$ is a picture of $F$ under rotation around $E$ for $90^{\circ}$. So, as $F$ describes line $AB$ we see that $H$ describes also line which is perpendicular to $AB$ (i.e. this is line $AB$ rotated around $E$ for $90^{\circ}$).

The condition that $ ABCD $ is a square is redundant.