Proof
Observe first, that if A and B lie on opposite sides of L, then
all ilines through A and B must meet L. Thus, if there is an
iline through A and B which does not meet L, then A and B must lie
on the same side of L.
We proceed by stages:

If a point P lies outside a circle C , then there is an
extended line through P which does not meet C.
Let Q be the centre of C.
We may choose the line through P perpendicular to PQ.

If a point P lies inside a circle C , then there is a
circle through P which does not meet C.
Let Q be the centre of C.
We may choose the circle on QP as diameter 
all of its points are less than QP from Q.

If C is a circle, and tεI(2), then t maps
all points in the exterior of C to points on one side of t(C), and
all points in the interior of C to points on the other side of t(C).
If P lies in the exterior of C, then, by (1), there is an iline
(an extended line) M through P which does not meet C.
Then t(M) is an iline through t(P) and t(∞) which does
not meet t(C). Thus t(P) lies on the same side of t(C) as t(∞).
Similarly, using (2), any for any Q in the interior of C, t(Q) lies on
the same side of t(C) as t(C), where C is the centre of C.
Since t maps C to t(C), it maps the complement of C to the
complement of t(C). Thus, the images of the interior and
exterior must cover both sides of t(C), so the interior and
exterior must map to opposite sides.

Finally, suppose that L is an extended line. We can choose
an inversion h
which maps L to a circle C, so h^{1}(C) = L.
Applying (3) to C and h^{1}, we see that h^{1} maps
the interior of C to one side of L. Thus, h maps this side
of L to the interior of C, and the other side to the exterior.
Now apply (3) to C and toh^{1}. This maps the interior of C to
one side of (toh^{1})(C) = t(h^{1}(C)) = t(L),
and the exterior to the other side.
Now, applying h then toh^{1}, we see that t = (toh^{1})oh maps
the sides of L to the sides of t(L), as required.

