How do I prove this identity for complex numbers?

we use geometry and trigonometry of course.

$|A+D|=2r\sin{\theta_{1}}$, $|A+B|=2r\sin{\theta_{2}}$, $|B+C|=2r\sin{\theta_{3}}$, $|A+D|=2r\sin{\theta_{4}}$

$|A-D|=2r\cos{\theta_{1}}$, $|A-B|=2r\cos{\theta_{2}}$, $|B-C|=2r\cos{\theta_{3}}$, $|C-D|=2r\cos{\theta_{4}}$

We also know that $\theta_{1}+\theta_{2}+\theta_{3}+\theta_{4}=\pi$

$$ \begin{aligned} \cos{(\theta_{1}+\theta_{3})}+\cos{(\theta_{2}+\theta_{4})}&=0\\ \cos{\theta_{1}}\cos{\theta_{3}}-\sin{\theta_{1}}\sin{\theta_{3}}+\cos{\theta_{2}}\cos{\theta_{4}}-\sin{\theta_{2}}\sin{\theta_{4}}&=0 \end{aligned} $$

multiply by $4r^{2}$ to get

$$ |A-D||B-C|-|A+D||B+C|+|A-B||C-D|-|A+B||C+D|=0 $$

note

"no half circle contains all 4 points" requirement is required to have the origin inside the cyclic quadrilateral

The sides and diagonals of cyclic quadrilateral $\square ABCD$ (whose vertices we can think of as complex numbers or position vectors) are related by Ptolemy's Theorem:

$$|A-B||C-D|+|B-C||D-A|=|AC||BD| \tag{1}$$

Dissecting the four isosceles triangles $\triangle AOB$, $\triangle BOC$, $\triangle COD$, $\triangle DOA$ into pairs of congruent right triangles and rearranging, we have cyclic $\square A'B'C'D'$ whose sides are the "other" diagonals of the parallelograms determined by $A$, $B$, $C$, $D$. By Ptolemy again, $$|A+B||C+D|+|B+C||D+A|=|A'C'||B'D'| \tag{2}$$

Note that $\overline{OA}\cong\overline{O'A'}$, so that the circles are congruent. Moreover, a little angle chasing shows that $\angle AOC\cong\angle A'O'C'$ and $\angle BOD\cong\angle B'O'D'$, so that $\overline{AC}\cong\overline{A'C'}$ and $\overline{BD}\cong\overline{B'D'}$. Consequently, the right-hand sides of $(1)$ and $(2)$ are equal, so that the left-hand sides are also equal, giving the result. $\square$