the inversive invariant and i-lines

We show how the inversive cross-ratio can be used to characterize i-lines
and i-segments. Some of our results can be interpreted as familiar theorems
of euclidean geometry.

Theorem II2
(1) The points A,B,C,D lie on an i-line if and only if (A,B,C,D) is real,
(2) The points C,D lie on the same i-segment AB if and only if (A,B,C,D) > 0.

 To understand some of the interpretations, we consider first ratios of the form (u-z)/v-z) for complex numbers u,v and z. Let U,V,Z be the points with complex coordinates u,v,z. Then u-z = r.exp(iα), where r = |UZ|, and α is as shown. and  v-z = s.exp(iβ), where s = |VZ|, and β is as shown. Thus, (u-z)/(v-z) = (r/s).exp(i(α-β), so that the modulus of (u-z)/(v-z) is the ratio |UZ|/|VZ|, and the argument of (u-z)/(v-z) is the signed angle VZU. Now (U,V,Z,W) = ((u-z)/(v-z))/((u-w)/(v-w)), so a determination of the argument of the cross-ratio is the difference of the signed angles VZU and VWU. We are now in a position to prove two of euclid's theorems: Theorem A Suppose that C,D are points not on the line AB. Then the points C and D lie on the same arc AB of a circle if and only if the signed angles BCA and BDA are equal. This is just a thinly disguised version of Theorem II2(2). Since C is not on AB, the i-line through A,B and C must be a circle. The cross-ratio is positive if and only if its argument is zero. By the remarks just before the theorem, this is equivalent to the signed angles being equal. Theorem B The vertices of a convex quadrilateral ACBD lie on a circle if and only if the unsigned angles BCA and BDA add to π. Again, this comes quickly from Theorem II2. Since the quadrilateral is convex, the vertices would have to lie on opposite arcs of a circle. From (1) and (2), this happens if and only if (A,B,C,D) is negative. Thus, the argument of the cross-ratio must be ±π. Let the unsigned angles be α =