Is it possible to construct the angle bisector of two lines with only a right-angled ruler? I have a problem with a special construction.

Is it possible to construct the angle bisector of two lines with only a right-angled ruler?

Here with a right-angled ruler one can performe the following constructions:

*

*draw line through two points

*draw perpendicular from any point to any line

I already know that with such a ruler one can divide a segment into half, double a segment, but I couldn’t construct an angle bisector. I am starting to believe that it is not possible. I am familiar for example with the impossibility of halving a segment with a ruler only, but I couldn’t generalize it. Any idea?
 A: Identify the plane with $\Bbb R^2$ and let $Q$ be a subfield of $\Bbb R$.
We call a point $(x,y)$ nice if $x\in Q$ and $y\in Q$. We call a line nice if it passes through two nice points.
Claim. If all given points and lines are nice then all points and lines that are constructible with the right-angled ruler from these givens are also nice.
Proof: (By induction) We need to show that every construction step produces only nice objects from nice givens.

*

*drawing the line through two nice points: Okay by the definition of nice line

*drawing a line perpendicular to a nice line (through nice points $(x_A,y_A)$ and $(x_B,y_B)$) and through a nice point $(x_C,y_C)$: This line passes through the tow nice points $(x_C,y_C)$ and $(x_C+y_B-y_A,y_C+x_A-x_B)$.

*intersecting two constructed lines: The coordinates of the intersection point are found by solving a system of linear equations based on the coordinates of some nice points. The solutions are in the same firld $Q$, hence the intersection point is nice

*"Pick any point" (that does not alter the final outcome): An often overlooked construction step. But as $Q$ is dense in $\Bbb R$, we may assume that a picked point is nice

*"Pick any point on a nice line": Again, points from $F\times F$ are dense on nice lines so that we may assume that a nice point is picked

I think those are all allowed construction steps, hence the claim follows. $\square$
The points $A=(1,0)$, $B=(0,0)$, $C(1,1)$ are nice if we let $Q=\Bbb Q$.
Let $\ell$ be the bisector of $\angle ABC$ (which is an angle of $45^\circ$).
As $\tan 22\frac12^\circ = \frac{1}{1+\sqrt 2}$ is irrational, we see that $(0,0)$ is the only nice point on $\ell$, i.e., $\ell$ is not nice. According to the claim, the bisector cannot be constructed.
