Proof 1: as $\displaystyle \arg\left(\frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}\right)=\arg\left(\frac{z_1-z_3}{z_1-z_4}\right)-\arg\left(\frac{z_2-z_3}{z_2-z_2}\right)$ ( because $\displaystyle \frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}=\frac{z_1-z_3}{z_1-z_4}:\frac{z_2-z_3}{z_2-z_4}$ ) , from basic geometry
we get that $\displaystyle z_i$ are on a circle iff the above is either the difference of two inscribed angles subtended on the same arc, and thus this difference is zero, or else
this is the difference of two inscribed angles subtended over complementary arcs and thus the difference is $\displaystyle \pm \pi$ (opposite angles of an quadrilateral inscribed
in a circle sum 180 degrees).
Proof 2: We use that the cross ratio is invariant under fractional transformations and that the cross ration is well defined if three points out of the four are distinct.
Now, the frac. trans. $\displaystyle T(z):=\frac{(z-z_2)(z_3-z_1)}{(z-z_1)(z_3-z_2)}$ maps $\displaystyle z_1,\,z_2,\,z_3$ to $\displaystyle \infty,\,0,\,1$ , resp., so $\displaystyle T(z_4)=
z_1,z_2,z_3,z_4)$ is real iff $\displaystyle z_1,z_2,z_3$ lie on the circle (or on the
straight line) on which $\displaystyle z_1,z_2,z_3$ lie
Tonio