Hexagon inscribed in circle Given a cyclic hexagon $ABCDEF$ such that main diagonals intersects at S and $|AB|=|CD|=|EF|$. Let $P=AD \cdot CE$. Prove that $\frac{|CP|}{|PE|}=(\frac{|AC|}{|CE|})^{2}$. 
I tried with Ptolemy's theorem and power of point but that was not enough to prove this equality. My second attempt was inscribing $ABCDEF$ in triangle such that $ABCDEF$ was Lemoin's hexagon but that also didn't make it. 
Any help would be greatly appreciated.
 A: As it turns out, this was just a matter of considering similar triangles, which arise everywhere naturally in this configuration based on the principle that equal arcs subtend equal angles.
Let $Q$ denote the point where the diagonals intersect. Define $\alpha:=\angle BAQ, \beta:=\angle QBA$, and $\gamma:=\angle AQB$ and prove that the intuitive (I hope) angle correspondences in the image below are right, i.e. the blue angles are all the same, and so on.

 Hint. $\angle BAD=\angle ADC$ since $AB=CD$, which also implies that $\angle BED=\angle AFC=\alpha$. Repeat the process for the remaining outer angles. For the central angles around $Q$ consider the fact that the angles of a triangle add up to $180^\circ$, just as opposite angles in a cyclic quadrilateral.


Now, observe the triangles $\triangle PQC, \triangle PDE$ where $\angle CQP=\angle EDP$ implies that they are similar. Hence $CP:PE = QC:DE$. Finally
$$\frac{CP}{PE}=\frac{QC}{DE}=\frac{QC}{DQ}\cdot \frac{DQ}{DE}=\frac{\sin\alpha}{\sin\gamma}\cdot \frac{\sin\alpha}{\sin\gamma} =\left(\frac{\sin\alpha}{\sin\gamma}\right)^2 =\left(\frac{AC}{CE}\right)^2$$
Where we have employed many times the Sine Law.
