The elements of non-Euclidean plane geometry and trigonometry, by H. S. Carslaw.

70

NON-EUCLIDEAN GEOMETRY

[OH. III.

It follows from ~ 26 (6) that AD = DB.
Therefore we have z DAB =/ DBA.
But L BAC =L ABC.
Therefore we have L DAB=L CAB, which is absurd.
Thus, the rays AE and AF cannot be parallel.
Similarly the rays EA, FB produced through A and B cannot
be parallel.
We have now shown that the lines a' and b' neither intersect
nor are parallel.
They must, therefore, have a common perpendicular (~ 32).
We shall now show that this common perpendicular is
parallel to both 02 and 02'.
Let it cut the lines AE and BF at U and V.
Then AU = BV, by ~ 29.
If VU is not parallel to A2, draw through U the ray U2
parallel to A2, and through V the ray V2 parallel to A2.
Then, by ~ 26 (4), L AU2 =zL BVg2.
Also the angles AUV and BVU are right angles, so the exterior
angle at U would be equal to the interior and opposite angle
2VU, which is impossible (~ 26 (3)).
Thus we have shown that the ray VU is parallel to O02.
The same argument applies to the ray UV and 02'.
Therefore we have proved that there is a common parallel
to the two given intersecting rays, and we have shown how to
construct it.
COROLLARY. A common parallel can be drawn to any two
given coplanar lines.
If the given lines intersect when produced, the previous
proof applies.
If they do not intersect, take any point A on the line (i) and
draw a parallel from A to the line (ii).
We can now draw a common parallel to the two rays through
A, and by ~ 25 this line will also be parallel to the two given
lines.
~ 45. Construction of the Straight Line which is perpendicular to one of two Straight Lines containing an
Acute Angle, and parallel to the other.
Let a(OA) and b(OB) be the two rays containing an acute
angle.