Problem involving cross ratio identity Let $z_1, z_2, z_3, z_4$ be distinct complex numbers. Assume that they lie on the same circle, in that order. Prove that $$|z_1 - z_3||z_2 - z_4| = |z_1 - z_2||z_3 - z_4| + |z_2 - z_3||z_4 - z_1|$$
So let F be a fractional linear map which sends $z_2, z_3, z_4$ to $1,0,\infty$ respectively. Note that since any fractional linear map is just a combination of translations, inversions and multiplications, it has to maintain the order of $z_1, z_2, z_3$ and $z_4$ if these points lie on the same circle. So neccesarily, $F(z_1) \in  ]1, \infty[$. Is what I have done useful so far, if so, how do I continue?
Thanks in advance
 A: $\sf Hint$:
$$\large |\color{Green}{z_1-z_3}||\color{Teal}{z_2-z_4}|=|z_1-z_2||z_3-z_4|+|\color{DarkBlue}{z_2-z_3}||\color{Purple}{z_4-z_1}|;$$
$$\large (z_1,z_2;z_3,z_4):=\frac{(\color{Green}{z_1-z_3})(\color{Teal}{z_2-z_4})}{(\color{DarkBlue}{z_2-z_3})(\color{Purple}{z_1-z_4})},~~~~~(z_1,z_3;z_2,z_4)=?$$
A: For any 4 points $z_{1},z_{2},z_{3},z_{4}$ we have $$(z_{2}-z_{1})(z_{4}-z_{3})+(z_{3}-z_{2})(z_{4}-z_{1})=(z_{3}-z_{1})(z_{4}-z_{2})$$
Applying the triangle inequality we get $$|z_{2}-z_{1}||z_{4}-z_{3}|+|z_{3}-z_{2}||z_{4}-z_{1}| \geq |z_{3}-z_{1}||z_{4}-z_{2}|$$
We have equality if $(z_{2}-z_{1})(z_{4}-z_{3})$ and $(z_{3}-z_{2})(z_{4}-z_{1})$ have the same direction.
Let $$\alpha=\textbf{arg}\left(\frac{z_{2}-z_{1}}{z_{4}-z_{1}}\right)\text{  and  }\beta=\textbf{arg}\left(\frac{z_{4}-z_{3}}{z_{3}-z_{2}}\right)$$ 
In a cyclic quadrilateral $$\alpha=-\beta(\text{Property of cyclic quadrilaterals})\\\rightarrow\alpha+\beta=0$$
$$\rightarrow\frac{z_{2}-z_{1}}{z_{4}-z_{1}}\cdot\frac{z_{4}-z_{3}}{z_{3}-z_{2}}=\text{a  positive real number}$$
Therefore,
$$|z_{2}-z_{1}||z_{4}-z_{3}|+|z_{3}-z_{2}||z_{4}-z_{1}| = |z_{3}-z_{1}||z_{4}-z_{2}|$$
A: Hint: There is a theorem in geometry that says that the product of the diagonals of an inscribed quadrilateral is the sum of the products of the opposing sides. 

Here is a proof using trigonometry. I will look for a simpler proof, if no one else does first.
$\hspace{3.6cm}$
Using cross-products to find the area of the quadrilateral yields
$$
\begin{align}
\text{Area of quadrilateral}
&=\frac12(ad+bc)\sin(\phi)\\
&=\frac12(e_1f_1+e_1f_2+e_2f_1+e_2f_2)\sin(\theta)\\
&=\frac12ef\sin(\theta)\tag{1}
\end{align}
$$
Using the Law of Cosines to find $f$ yields
$$
\begin{align}
f^2&=a^2+d^2-2ad\cos(\phi)\\
&=b^2+c^2+2bc\cos(\phi)\\
a^2+d^2-b^2-c^2&=2(ad+bc)\cos(\phi)\tag{2}
\end{align}
$$
Using the Law of Cosines to find $a,b,c,d$ yields
$$
\begin{align}
a^2&=e_1^2+f_2^2+2e_1f_2\cos(\theta)\\
b^2&=e_2^2+f_2^2-2e_2f_2\cos(\theta)\\
c^2&=e_2^2+f_1^2+2e_2f_1\cos(\theta)\\
d^2&=e_1^2+f_1^2-2e_1f_1\cos(\theta)\\
a^2-b^2+c^2-d^2&=2ef\cos(\theta)\tag{3}
\end{align}
$$
Putting together $(1)$, $(2)$, and $(3)$ yields
$$
\begin{align}
e^2f^2&=e^2f^2\cos^2(\theta)+e^2f^2\sin^2(\theta)\\
&=\frac14(a^2-b^2+c^2-d^2)^2+(ad+bc)^2\sin^2(\phi)\\
&=\frac14(a^2-b^2+c^2-d^2)^2+(ad+bc)^2-(ad+bc)^2\cos^2(\phi)\\
&=\frac14(a^2-b^2+c^2-d^2)^2+(ad+bc)^2-\frac14(a^2+d^2-b^2-c^2)\\[4pt]
&=(a^2-b^2)(c^2-d^2)+(ad+bc)^2\\[8pt]
&=(ac+bd)^2\\[9pt]
ef&=ac+bd\tag{4}
\end{align}
$$
