perpendiculars and common perpendiculars

 Our first result is to generalize the euclidean theorem that guarantees a unique line through a given point and perpendicular to a given line. the existence of perpendiculars Suppose that K is a complete hyperbolic line, and that P is a point of E, not asymptotic for K, then there is a unique hyperbolic line through P perpendicular to K. proof Let K have asymptotic points A and B, so P ≠ A,B, and K,C are in B(A,B). We require an i-line through P orthogonal to K and to C. By theorem O2, The family of i-lines orthogonal to K,C is A(A,B). There is exacly one member of this family through P. Notice that this result applies even when P is on C. Also notice that, if K has asymptotic points A,B and H is the perpendicular through P, then as H is orthogonal to K, inversion in H maps K to itself. It follows that this inversion swaps A and B. It also fixes P which lies on H. Thus the inversion maps (A,B,P) to (B,A,P). For P on C, the existence of a hyperbolic transformation with this effect is implicit in the fundamental theorem of weird geometry. What is new is that this hyperbolic transformation is an inversion. In euclidean geometry, two lines have a common perpendicular if and only if they are parallel. Two parallel lines have an infinite family of perpendiculars. In hyperbolic geometry, the situation is quite different, as we shall show. In fact, both results are really consequences of results in inversive geometry. We shall leave the euclidean result as an exercise. the common perpendicular theorem Two complete hyperbolic lines have a common perpendicular if and only if they are ultraparallel. When the lines are ultraparallel, the perpendicular is unique. proof Suppose that K,L are complete hyperbolic lines. In other words, they are arcs of i-lines K*,L* orthogonal to C. An i-line H* will give a hyperbolic line perpendicular to both if and only if H* is orthogonal to C (to give a hyperbolic line) as well as to K* and L* By theorem O3, such an H* will exist if and only if K* and L* are disjoint. This is equivalent to saying that K,L are ultraparallel. The theorem also shows that the common perpendicular is unique when K,L are ultraparallel.