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 iC(z)=z*.
If C is a circle, we may assume it is |z|=r, so iC(z)=r2/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 = Ak(A,B) for some k.
Hence L has equation |z-a| = k|z-b|.
If C is the real axis, then iC(L) has equation
|z*-a| = k|z*-b|
For any complex w, |w| = |w*|. If we apply this to both sides,
we see that
iC(L) has equation |z-a*| = k|z-b*|,
i.e. |z-iC(a)| = k|z-iC(b)|.
Thus iC(L) belongs to Ak(iC(A),iC(B)).
It follows from Apollonius theorem that iC(L) is an i-line
and that iC(A) and t(B) are inverse with respect to it.
If C is the circle |z|=r, then iC(L) has equation
|r2/z*-a| = k|r2/z*-b|.
Now if we multiply each side by |z*|/|a|, and apply conjugation to each side,
we get
|r2/a*-z|=k|b*/a*||r2/b*
-z|.
Now, since |w| = |-w| and iC(z)=r2/z*, the equation becomes
|z-iC(A)| = k|b*/a*||z-iC(B)|.
Once again, iC(L) is an i-line, and iC(A) and iC(B) are inverse with respect to it.
Suppose that B is the inverse of A with respect to L, i.e. that B = iL(A).
The second part of the theorem says that iC(B) = iL'(iC(A))
Thus, as B = iL(A),
iC(iL(A)) = iL'(iC(A)).
Since this holds for any A, the third part follows.