Cross ratio of 4 points on circle I'm in a bit of trouble I want to calculate the cross ratio of 4 points $ A B C D $ that are on a circle 
Sadly "officially" it has to be calculated with A B C D as complex numbers and geometers sketchpad ( the geomerty program I am used to) don't know about complex numbers
Now I am wondering
The cross ratio of 4 points on a circle is a real number
And is there  so much difference between
$$ \frac { (A-C)(B-D) }{ (A-D)(B-C) } \text{  (complex method)  }$$ 
And 
$$ \frac { |AC||BD| }{ |AD||BC| } \text{  (distance method). }$$
Where X-Y is the complex difference between X and Y
and |XY| is the distance between X andY
If allready is given that the four points are on a circle?
I think that the absolute values of the calculations are the same but am I right (and can we prove it)
ADDED LATER :
on Cave 's  suggestion I used a circle inversion to move the four points on the circle to four points on a line and then the formulas give the  same value ( if I did it correctly)
What I did:
I took a diameter of the circle 
This diameter intersects the circle in O and U ,
u is the line through U perpendicular to OU
Project the 4 points onto  line u with centre O (draw a ray through O and the point, the new point is where this ray intersects line u)
And calculate the cross ratio of the four new points.
This "projection" method  and the earlier "distance" method  give the same value but does this prove anything?

For $A,B,C,D$ on the unit circle,
$$
\begin{align}
\frac{(A-C)(B-D)}{(A-D)(B-C)}
&=\frac{\left(e^{ia}-e^{ic}\right)\left(e^{ib}-e^{id}\right)}{\left(e^{ia}-e^{id}\right)\left(e^{ib}-e^{ic}\right)}\\
&=\frac{\left(e^{i(a-c)}-1\right)\left(e^{i(b-d)}-1\right)}{\left(e^{i(a-d)}-1\right)\left(e^{i(b-c)}-1\right)}\\
&=\frac{e^{i\frac{a-c}2}2i\sin\left(\frac{a-c}2\right)e^{i\frac{b-d}2}2i\sin\left(\frac{b-d}2\right)}{e^{i\frac{a-d}2}2i\sin\left(\frac{a-d}2\right)e^{i\frac{b-c}2}2i\sin\left(\frac{b-c}2\right)}\\
&=\frac{\sin\left(\frac{a-c}2\right)\sin\left(\frac{b-d}2\right)}{\sin\left(\frac{a-d}2\right)\sin\left(\frac{b-c}2\right)}\tag{1}
\end{align}
$$
which is indeed real.

For $A,C$ on the unit circle,
$$
\begin{align}
|A-C|
&=\sqrt{\left(e^{ia}-e^{ic}\right)\left(e^{-ia}-e^{-ic}\right)}\\[3pt]
&=\sqrt{2-2\cos(a-c)}\\
&=2\sin\left(\frac{a-c}2\right)\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{|A-C||B-D|}{|A-D||B-C|}
&=\frac{2\sin\left(\frac{a-c}2\right)2\sin\left(\frac{b-d}2\right)}{2\sin\left(\frac{a-d}2\right)2\sin\left(\frac{b-c}2\right)}\\
&=\frac{\sin\left(\frac{a-c}2\right)\sin\left(\frac{b-d}2\right)}{\sin\left(\frac{a-d}2\right)\sin\left(\frac{b-c}2\right)}\tag{3}
\end{align}
$$
which is obviously real, but not so obviously, the same as $(1)$.

When is the Complex Cross-Ratio Real?

$$
\begin{align}
\frac{A-C}{A-D}
&=\frac{C-A}{D-A}\\
&=\frac{|C-A|}{|D-A|}\,e^{-i\alpha}
\end{align}
$$
$$
\begin{align}
\frac{B-D}{B-C}
&=\frac{D-B}{C-B}\\
&=\frac{|D-B|}{|C-B|}\,e^{-i\beta}
\end{align}
$$
Therefore,
$$
\frac{(A-C)(B-D)}{(A-D)(B-C)}
=\frac{|C-A||D-B|}{|D-A||C-B|}\,e^{-i(\alpha+\beta)}
$$
$\alpha+\beta$ is an integer multiple of $\pi$ ($\pi$) if and only if $A$ and $B$ are on opposing arcs of a circle with $C$ and $D$.


$$
\begin{align}
\frac{A-C}{A-D}
&=\frac{C-A}{D-A}\\
&=\frac{|C-A|}{|D-A|}\,e^{-i\alpha}
\end{align}
$$
$$
\begin{align}
\frac{B-D}{B-C}
&=\frac{D-B}{C-B}\\
&=\frac{|D-B|}{|C-B|}\,e^{i\beta}
\end{align}
$$
Therefore,
$$
\frac{(A-C)(B-D)}{(A-D)(B-C)}
=\frac{|C-A||D-B|}{|D-A||C-B|}\,e^{-i(\alpha-\beta)}
$$
$\alpha-\beta$ is an integer multiple of $\pi$ ($0$) if and only if $A$ and $B$ are on the same arc of a circle with $C$ and $D$.
