= Solution
Write $z_j=c+Re^{i\theta_j}$ with $\theta_1<\theta_2<\theta_3<\theta_4<\theta_1+2\pi$. In the identity
$$
z_j-z_\ell=2iR e^{i(\theta_j+\theta_\ell)/2}
\sin\frac{\theta_j-\theta_\ell}{2},
$$
the phase factors cancel in the <cross-ratio>. In our convention the two numerator sine factors are negative, while the denominator factors have opposite signs. Thus \b[$r<0$], and the exchanged cross-ratio $1-r$ is \b[greater than one]. Consequently $|r|+1=1-r=|1-r|$.
Taking absolute values of the two formulas in part (a) gives
$$
|r|=\frac{|z_1-z_4||z_2-z_3|}{|z_1-z_2||z_3-z_4|},\qquad
|1-r|=\frac{|z_1-z_3||z_2-z_4|}{|z_1-z_2||z_3-z_4|}.
$$
Multiplying $|r|+1=|1-r|$ by the common positive denominator proves <Ptolemy theorem>:
$$
\boxed{|z_1-z_3||z_2-z_4|
=|z_1-z_2||z_3-z_4|+|z_1-z_4||z_2-z_3|.}
$$
The product of the diagonals equals the sum of the products of opposite sides.
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