Proof

 The Mirror Property Suppose that L and M are distinct i-lines, then iL(M) = M if and only if L and M are orthogonal. Proof Since L, M are distinct, M must contain points not on L. Since inversion in L maps points on one side of L to points on the other side, M must contain points on either side of L. It follows that, if iL(M) = M, then M must cut L twice in E+. Suppose that they meet in A and B. As in the proof of the Uniqueness Theorem, we invert in a circle C with centre A. Let us write P' for the inverse of any object P. Then L' and M' are extended lines through B'. As inversion in an extended line is just reflection, we see that the result holds in this case. Now consider the general case: iL(M) = M if and only if iC(iL(M)) = iC(M). By the third part of the Algebraic Inversion Theorem, iC° iL = iL'° iC, so iL(M) = M if and only if iL'(iC(M) = iC(M) , i.e. iL'(M') = M'. As we have already observed, this is equivalent to L'and M' being orthogonal. But inversion preserves angle, and hence orthogonality. Thus iL(M) = M if and only if L and M are orthogonal.