Please enable Java for an interactive construction (with Cinderella).

Above is an interactive version of figure 6.32. You can drag any of the red points around using the mouse.
Recall, you are to show that if lines l and m (green) are parallel but not limiting parallel, we construct AA' and BB' (purple) to be perpendicular to m. Then, assuming that AA' is longer than BB', we can create ray EF (dark green) so that EA' is congruent to BB', and so that angles A'EF and B'BG are congruent.
You are to show that EF intersects line l at a point H.

In order to do this exercise, you may use Major exercises 2, 3, 4, and 5. Indeed, you almost certainly must use them.

---

This exercise is needed to show that Hilbert's construction of the mutual perpendicular to lines l and m exists. Below is that construction, which works as follows. (see page 263 of the text).

First, we create ray EF as above, and from the exercise, we know it intersects AB at a point H. Now we find the unique point K on ray AB so that EH is congruent to BK. Drop perpendiculars HH' and KK', and observe that they are congruent (since Lambert quadrilaterals A'H'HE and B'K'KB are congruent).
Since we have two congruent sides perpendicular to the base, quadrilateral H'K'KH is a Saccheri quadrilateral. We showed earlier that the line joining the midpoint M of the summit HK to the midpoint M' of the base H'K' will be perpendicular to both lines.

Please enable Java for an interactive construction (with Cinderella).