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

47]

CORRESPONDING POINTS

79

Then SM=SN.
Through S draw Sf2 parallel to Pf2.
It will also be parallel to Rf2, and it will bisect L MSN,
since there is only one angle of parallelism for a given distance.
Let S' be any point upon the parallel through S to (i) and (ii).
From S' draw S'M' and S'N' perpendicular to these lines.
By congruence theorems, it is easy to show that S'M' =S'N',
and that S'2 bisects LzM'S'N'.
From P draw PL perpendicular to S2, and from L draw
Lm and Ln perpendicular to (i) and (ii). (Cf. Fig. 53.)
Cut off nQ,=mP on the opposite side of n from Q2, and
join LQ.
Then it follows that PLQ is a straight line, and that Q
corresponds to P.
It is easy to show that there can only be one point on the
second line corresponding to P on the first.
2. If P and Q are corresponding points on the lines (i) and (ii),
and Q and R corresponding points on the lines (ii) and (iii), the
three lines being parallel to each other, then P, Q, and R cannot
be in the same straight line.
P
Q.
R
FIG. 54.
If possible, let PQR be a straight line.
By the definition of corresponding points, we have
L. 2PQ =- 2QP,

L 2QR = LRQ.