If $z_1,z_2,z_3,z_4$ are consecutive vertices of a quadrilateral that lie on a circle prove the following: $|z_1-z_3||z_2-z_4|=|z_1-z_2||z_3-z_4|+|z_1-z_4||z_2-z_3|$. I know that you can prove this without using the cross-ratio, but I would like to complete this proof using it. Here are some things I know:
- The cross-ratio is real if and only if $(z_1,z_2,z_3,z_4)$ lie on the same circle.
- The cross-ratio is invariant under linear transformations, i.e. $(z_1,z_2,z_3,z_4)=(Tz_1,Tz_2,Tz_3,Tz_4)$.
I am not sure if (2) will be of any use, but I definitely think (1) would be helpful. Using (1) I can only get as far as: $|z_1-z_3||z_2-z_4|=C|z_1-z_4||z_2-z_3|$, where $C \geq 0$.
Any help is appreciated.
Edit: After playing with the equation for a while this is what I concluded: Of course by simple verification, one indeed finds that $(z_1-z_3)(z_2-z_4)=(z_1-z_2)(z_3-z_4)+(z_1-z_4)(z_2-z_3)$. Then we have $$\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}=\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)}+1.$$ Now our result is proven if we can show that both of these quotients are real and positive. Well, the real part follows from the fact that a cross-ratio is real if and only if all four points lie on the same circle. Now for the positive part: The quotient on the left is the cross ratio $(z_1,z_2,z_3,z_4)$ where $z_2 \rightarrow 1$, $z_3 \rightarrow 0$, $z_4 \rightarrow \infty$. Now this transformation, which is bijective, takes the circle that passes through these points to the real line. Therefore the arc from $z_4$ to $z_2$ is taken to $[-\infty,1]$, hence $(z_1,z_2,z_3,z_4) > 1$. A similar argument works to show that the other cross-ratio is also positive.
Is this reasoning correct? I guess we use the fact that continuous functions take connected sets to connected sets, so the image of these arcs must be connected.
Best Answer
Your reasoning is correct. You have demonstrated that $$ \frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}=\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)}+1 $$ and the quotient on the left is $> 1$. That implies that the quotient on the right is $> 0$. Therefore all expressions are positive, and it follows that $$ \left|\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}\right|=\left|\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)}\right|+1 \, . $$
Instead of a connectedness argument you can also use that Möbius transformation preserve orientation: The quotient on the left is equal to $T(z_1)$ where $T$ is the Möbius transformation which maps $z_2, z_3, z_4$ to $1, 0, \infty$, respectively. $z_1,z_2,z_3,z_4$ are consecutive points on a circle. It follows that $T(z_1), 1, 0, \infty$ are consecutive points on the extended real axis (a generalized circle) and that is only possible if $T(z) > 1$.