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

56

NON-EUCLIDEAN GEOMETRY

[CH. III.

Therefore h' must cut ac.
Let it cut this line at T.
Since a' is parallel to B'T, it follows that the ray a' must
cut PT (Pasch's Axiom).
Let these rays a,, a' intersect at Q.
From Q draw QR perpendicular to the line b, and from the
line b cut off BR' equal to B'R and on the opposite side of B
from B'.
In the same way, from the line a cut off AQ' equal to A'Q,
and on the opposite side of A from P.
In this way we obtain a quadrilateral ABR'Q' congruent with
A'B'RQ.
Thus QRR'Q' is a Saccheri's Quadrilateral, and the line
joining the middle points of QQ', RR' is perpendicular to a
and b.
~ 33. Two parallel lines approach each other continually, and
their distance apart eventually becomes less than any assigned
quantity.
Let a and b be two parallel lines.
Upon a take any two points P and Q, PQ being the direction
of parallelism for the lines.
-fiN~ b
M P N
FIG. 34.
From P and Q draw the perpendiculars PM and QN to b.
Bisect MN at H, and draw the perpendicular at H to the
line b.
This must intersect the segment PQ; let it do so at K.
At K draw the ray a' parallel to b in the other direction.
This ray must intersect PM, since it enters the triangle
PKM at the vertex K.
Let it cut PM at P'.