Proof
We observe that the definition of inversion depends only on length,
so does not depend on the choice of axes.
Thus, if C is an extended line, then we may assume that it is the real axis, so i_{C}(z)=z*.
If C is a circle, we may assume it is z=r, so i_{C}(z)=r^{2}/z*.
Suppose that A and B are inverse with respect to L,
and have complex coordinates a and b respectively.
Then, from Apollonius Theorem, L = A_{k}(A,B) for some k.
Hence L has equation za = kzb.
If C is the real axis, then i_{C}(L) has equation
z*a = kz*b
For any complex w, w = w*. If we apply this to both sides, we see that
i_{C}(L) has equation za* = kzb*,
i.e. zi_{C}(a) = kzi_{C}(b).
Thus i_{C}(L) belongs to A_{k}(i_{C}(A),i_{C}(B)).
It follows from Apollonius theorem that i_{C}(L) is an iline
and that i_{C}(A) and t(B) are inverse with respect to it.
If C is the circle z=r, then i_{C}(L) has equation
r^{2}/z*a = kr^{2}/z*b.
Now if we multiply each side by z*/a, and apply conjugation to each side,
we get
r^{2}/a*z=kb*/a*r^{2}/b*
z.
Now, since w = w and i_{C}(z)=r^{2}/z*, the equation becomes
zi_{C}(A) = kb*/a*zi_{C}(B).
Once again, i_{C}(L) is an iline, and i_{C}(A) and i_{C}(B) are inverse with respect to it.
Suppose that B is the inverse of A with respect to L, i.e. that B = i_{L}(A).
The second part of the theorem says that i_{C}(B) = i_{L'}(i_{C}(A))
Thus, as B = i_{L}(A),
i_{C}(i_{L}(A)) = i_{L'}(i_{C}(A)).
Since this holds for any A, the third part follows.

