TITLE: How do I prove this identity for complex numbers?
QUESTION [6 upvotes]: I stumbled upon following identity I'm having a hard time proving: if the complex numbers $a,b,c,d$ lie on the unit circle in that order and no half-circle contains all of them, the following identity should hold:
$$
|a+d||b+c| + |a+b||c+d| = |a-d||b-c| + |a-b||c-d|
$$
Any idea on how I should approach this / is there some known theorem I'm unaware of?

REPLY [4 votes]: 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. $\square$