Sum of distances for vertices lying on a circle 
If the consecutive vertices $z_1,z_2,z_3,z_4$ of a quadrilateral lie on a circle, prove that $|z_1-z_3|\cdot|z_2-z_4|=|z_1-z_2|\cdot|z_3-z_4|+|z_2-z_3|\cdot|z_1-z_4|$.

I know that when the four points $z_1,z_2,z_3,z_4$ lie on a circle, the cross ratio $(z_1,z_2,z_3,z_4)$ is real. (In fact, the relation is 'if and only if'.) Therefore, the ratio $\dfrac{\left(\dfrac{z_1-z_3}{z_1-z_4}\right)}{\left(\dfrac{z_2-z_3}{z_2-z_4}\right)}$ is real, so there exists a real number $t$ such that $\left(\dfrac{z_1-z_3}{z_1-z_4}\right) = t\cdot \left(\dfrac{z_2-z_3}{z_2-z_4}\right)$. (And similarly for the other permutations of $z_1,z_2,z_3,z_4$ as well.) Is it possible to obtain the desired relation from here?

You can show that if the cross ratio of $(ABCD)$ is $m$, then the cross ratio of $(ACBD)$ is $1-m$. This is independent of where the points are, and follows directly from writing it out.
You mentioned that $ABCD$ is concyclic if and only if the cross ratio is real. There is an additional fact that if $AB$ does not separate $CD$, then the cross ratio is positive.
Hence, we have $0 < m < 1$. Applying absolute value, Ptolemy's Theorem follows.
