Problem from Ahlfors concerning points on a circle. Let $z_1,z_2,z_3,z_4$ be consecutive vertices of a quadrilateral which lie on a circle. Show that
$$|z_1-z_3|•|z_2-z_4| = |z_1-z_2|•|z_3-z_4|+|z_2-z_3|•|z_1-z_4|$$. 
Sometime these geometric problems from Ahlfor's mystify me. I know that the cross ratio of $z_1,z_2,z_3,z_4$ must be real, but I'm not finding the right idea.  I divided boths sides by $|z_1-z_4|•|z_2-z_3|$ to get $|(z_1,z_2,z_3,z_4)|=|(z_1,z_3,z_2,z_4)|+1$, but this doesn't help much. Need a hint.
 A: In context, Ahlfors is probably expecting you to use the question that immediately precedes the one that is puzzling you. This asks you to investigate how $(z_1, z_2, z_3, z_4)$ transforms under permutations of the $z_i$. See Cross ratios of permutations of four points if you have problems with that question.
You can now argue as follows. Put:
$$
a = (z_1 - z_3)(z_2 - z_4)\\
b = (z_1 - z_2)(z_3 - z_4)\\
c = (z_1 - z_4)(z_2 - z_3)
$$
so that what we want to prove is $|a| = |b| + |c|$. Consider the following two cross ratios that we can express in terms of $a$, $b$ and $c$:
$$
\lambda = (z_1, z_2, z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)} = \frac{a}{c}\\
\mu = (z_2, z_3, z_4, z_1) = \frac{(z_2 - z_4)(z_3 - z_1)}{(z_2 - z_1)(z_3 - z_4)} = \frac{a}{b}.
$$
So the cross ratio $\mu$ is obtained from $\lambda$ by the cyclic permutation $(1\,2\,3\,4)$ of the indices, from which it follows, using the results on permutations, that $\mu = \frac{\lambda}{\lambda - 1}$ (which you can verify by evaluating the linear fractional transformation $z \mapsto \frac{\lambda}{\lambda - (z, z_2, z_3, z_4)}$ on the $z_i$).
Hence we have:
$$
c = \frac{a}{\lambda}\\
b = \frac{a}{\mu} = \frac{(\lambda -1)}{\lambda}a = \left(1 - \frac{1}{\lambda}\right)a
$$
and we would have what we need, namely $|a| = |b| + |c|$, if we knew that $\lambda$ and $1 - \frac{1}{\lambda}$ were positive reals, or equivalently that $\lambda$ belongs to the real interval $(1, \infty)$. But, by the definition of the cross ratio, $\lambda$ is the image of $z_1$ under the linear fractional transformation that maps $z_2$, $z_3$ and $z_4$ to $1$, $0$ and $\infty$ respectively and we are given that $z_1$, $z_2$, $z_3$ and $z_4$ are listed consecutively, i.e., that $z_1$ does not lie on the arc from $z_4$ to $z_2$ via $z_3$ whose image is $[-\infty, 1]$, so $z_1$ is mapped to a point of $(1, \infty)$ and we are done.
