Ultraparallel theorem

In hyperbolic geometry, two lines are said to be ultraparallel if they do not intersect and are not limiting parallel.

The ultraparallel theorem states that every pair of (distinct) ultraparallel lines has a unique common perpendicular (a hyperbolic line which is perpendicular to both lines).

From any two distinct points A and C on s draw AB and CB' perpendicular to r with B and B' on r. If it happens that AB = CB', then the desired common perpendicular joins the midpoints of AC and BB' (by the symmetry of the Saccheri quadrilateral ACB'B).

If not, we may suppose AB < CB' without loss of generality.

Let E be a point on the line s on the opposite side of A from C. Take A' on CB' so that A'B' = AB.

Construct a point D on ray AE so that AD = A'D'.

Then D' ≠ D. They are the same distance from r and both lie on s. So the perpendicular bisector of D'D (a segment of s) is also perpendicular to r.[1] (If r and s were asymptotically parallel rather than ultraparallel, this construction would fail because s' would not meet s. Rather s' would be limiting parallel to both s and r.) Let be four distinct points on the abscissa of the Cartesian plane.

be semicircles above the abscissa with diameters

Then in the Poincaré half-plane model HP,

Compose the following two hyperbolic motions: Then

The unique semicircle, with center at the origin, perpendicular to the one on

The right triangle formed by the abscissa and the perpendicular radii has hypotenuse of length

, the common perpendicular sought has radius-square The four hyperbolic motions that produced

above can each be inverted and applied in reverse order to the semicircle centered at the origin and of radius

to yield the unique hyperbolic line perpendicular to both ultraparallels

In the Beltrami-Klein model of the hyperbolic geometry: If one of the chords happens to be a diameter, we do not have a pole, but in this case any chord perpendicular to the diameter it is also perpendicular in the Beltrami-Klein model, and so we draw a line through the pole of the other line intersecting the diameter at right angles to get the common perpendicular.

The proof is completed by showing this construction is always possible:

Alternatively, we can construct the common perpendicular of the ultraparallel lines as follows: the ultraparallel lines in Beltrami-Klein model are two non-intersecting chords.

The polar of the intersecting point is the desired common perpendicular.