By I. Dolgachev

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Example text

In the same way, on the other side of the plane a perpendicular can be drawn parallel to BA produced. If AB does not meet M N , then at least in one direction it diverges from M N . Through H, any point of the projection of AB on the plane, we can draw a line, HK, parallel to AB towards that part of AB which diverges from M N , and then draw CD parallel to this line and perpendicular to the plane. Unless AB is parallel to M N it will meet the plane at some point, or the plane and line will have a common perpendicular, and the line will diverge from the plane in both directions.

To H on CD draw, AH and EH. As H moves off indefinitely, AH approaches the position of AB, and the plane EAH the position of the plane EAB. Therefore, the limiting position of EH is the intersection of the planes ECD and EAB. The intersection of these planes is, then, parallel to CD, and in the same way we prove that it is parallel to AB. Now, if EF is given as parallel to one of these two lines towards the part towards which they are parallel, it must be the intersection of the two planes determined by them and the point E, and therefore parallel to the other line also.

Tan iρ1 Now, tan iρ1 = ∴ tan ip . cos(θ − α) sin ip sin iρ = sin iρ cos ip cos(θ − α), tan iρ1 and sin iδ = cos iρ cos ip [tan iρ cos(θ − α) − tan ip] . δ being fixed, this may be regarded as the polar equation of an equidistantcurve. CHAPTER 5. ANALYTIC NON-EUCLIDEAN GEOMETRY 60 5. The angle between two lines: φ being the angle which a line makes with the radius vector at any point, we have cos φ = cos ip sin(θ − α), sin ip . sin φ = sin iρ For two lines intersecting at this point, sin φ1 sin φ2 = sin ip1 sin ip2 sin2 iρ = sin ip1 sin ip2 + sin ip1 sin ip2 .