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

128

NON-EUCLIDEAN GEOMETRY

[IH. VI.

take the place of the Parallel Hypothesis of Euclid and his
implicit assumption that the line is infinite.
Let A and B be any two points on a given line L.
The perpendiculars at A and B to the line must intersect,
by assumption (i).
Let them meet at the point 0.
Since LOAB=LOBA, we have OA=OB.
At O make z BOQ=/ AOB (Fig. 87), and produce OQ to
cut the line L at P.
0
A B L
FIG. 87.
Then AB=BP and L OPA is a right angle.
By repeating this construction, we show that if P is a point
on AB produced through B, such that AP =m. AB, the line OP
is perpendicular to L and equal to OA and OB. The same
holds for points on AB produced through A, such that
BP= m. AB. In each case m is supposed to be a positive
integer.
Now, let C be a point on AB, such that AB =m. AC, m being
a positive integer. The perpendicular at C to L must pass
through the point 0, since if it met OA at 0' the above argument shows that O'B must be perpendicular to L and coincide
with OB.
It follows that if P is any point on the line L, such that
AP=. AB, m and n being any two positive integers, OP is
n
perpendicular to the line L and equal to OA and OB.
The case when the ratio AP: AB is incommensurable would
be deduced from the above by proceeding to the limit.