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

75, 76]

THE POLE OF A LINE

129

Now, all points on the line are included in this argument,
so that the perpendiculars at all points of the line L pass
through the same point.
Now, let L' be another line and A', B' two points upon it,
such that the segment AB=A'B'.
The perpendiculars at A', B' meet in a point, which we shall
call O'.
0'
A' B'
FIG. 88.
The triangles AOB and A'O'B' have a side of the one equal
to a side of the other, and the two angles adjacent to the sides
are equal, each to each.
It follows that O'A' = OA.
Thus we have shown that the perpendiculars at all points on
any line meet at a point which is at a constant distance from
the line.
The point will be called the Pole of the Line.
~76. Now, in Fig. 89, produce OA to O0, where O1A=OA.
Join 01B.
Then, from the triangles OAB and O1AB, it follows that
L O1BA=L OBA=a right angle.
Thus OB and O1B are in a straight line.
Also, AO1 produced must intersect AB at a point C, such that
OC is perpendicular to AB, and OC will be also perpendicular
to AB.
Thus OAO1 produced returns to 0, and the line is endless or
unbounded.
Its length is four times the distance of the pole of the line
from the given line.
N.-EG. I