Answer Happy

PROJECTS 1. Here is another construction for the common perpendicular between divergently parallel lines / and n. It suffices to locate their symmetry point S, for a perpendicular can then be dropped from S to both lines. Take any segment AB on /. Construct point Con /such that B is the midpoint of AC and lay off any segment A'B' on n congruent to AB. Let M, M', N, and N' be the midpoints of AA', BB', BA', and CB', respectively. Then the lines MM' and NN' are distinct and intersect at S. (The proof follows from the theory of glide reflections, see Exercises 21 and 22 in Chapter 9; also see Coxeter, 1968, p. 269, where it is deduced from Hjelmslev's midline theorem. Beware that Coxeter's description of midlines is partially wrong; e.g., no midline through S cuts / and n.)