Angle bisector on a piece of paper?

Question

Let's draw $\overline{AB}$ and $\overline{CD}$ (not parallel) on a piece of paper (rectangular). The intersection of the lines AB and CD is off the paper. Is it possible to construct the section of the angle bisector falling on the piece of paper without going off the paper? If I mirror $\overline{AB}$ (at least a short enough section of which should exist since the rectangle is convex and it's not possible that AB and CD both lie on a side each since they are not parallel and the intersection is on the outside of the triangle) to the midpoint of $\overline{AC}$ and then construct the angle bisector of the resulting two lines then I get a line parallel to the bisector but I do not know how to finish from here. (Mirroring back and drawing the line halving the distance is not it.)

Answer 1

If you keep taking parallels, you are just folding the whole triangle over your sheet, and soon or later you will have a small triangle in your sheet in which you will be able to find the bisector. By the bisector theorem, that will be the wanted bisector, too. Another rather simple approach is the following: consider the internal angle bisector of the quadrilateral $ABCD$.

Given that $X$ is the intersection of $AB$ and $CD$, the two small red points are the incenter of $BDX$ and the $X$-excenter of $XAC$ (the centres of two circles tangent to both $AB$ and $CD$), hence they both lie on the bisector from $X$.

Answer 2

Draw two lines offset parallelly by the same amount (using ruler/compass) into the enclosed angular space but cutting and draw its bisector ... which is same as bisector of lines cutting outside of paper limits.