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

164

NON-EUCLIDEAN GEOMETRY

[CH. VIII.

Since the circle AA'C touches the circle of inversion at C,
its inverse also touches that circle at C. But the points
A, A', B, B' are concyclic, and the radical axes of the three
circles AA'C, BB'C and AA'B'B are concurrent.
FIG. 110.
Therefore BB' passes through O and OB. OB'=OC2. Thus
the circle AA'B'B is orthogonal to the fundamental circle and
also to the circle of inversion.
It follows that if any nominal point A is changed by inversion
with regard to a circle of the system into the point B, the nominal
line AB is perpendicular to the nominal line with which the
circle of inversion coincides.
We shall now prove that it is " bisected " by that nominal
line. Let the circle through A, A', B and B' meet the circle
of inversion at M and the fundamental circle at U and V
(Fig. 111). It is clear that U and V are inverse points with
regard to the circle of inversion.
BV CV AV CV
Then we have =
AU CA BU CB
AV BV CV2 CV2 MV\2
AU BU CA. CB CM2 = 'U
AV / MV MV / BV
AU/ MU MU/ BU
Thus the nominal length of AM is equal to the nominal length
of BM,