Euclidean Geometry: Diagonals of Cyclic Hexagon Convex Hexagon $ABCDEF$ is cyclic. Prove that its three main diagonals $AD$, $BE$ and $CF$ are concurrent iff:
$$|AB| \cdot |CD| \cdot |EF|=|BC| \cdot |DE| \cdot |FA|$$
This seems like it might have something to do with Ptolemy's Theorem about Cyclic quadrilaterals. But I've got no idea how to apply it.
Hints would be  greatly appreciated.
 A: Label the generic convex cyclic hexagon as shown, with $P$, $Q$, $R$ the (not-necessarily-distinct) points where pairs of diagonals meet. Note that we take $\triangle PQR$ to "point toward" segments $AB$, $CD$, and $EF$.

Consider $\triangle PAB$ and $\triangle PED$. Here, $\angle A$ and $\angle E$ subtend the same arc, $\stackrel{\frown}{BD}$, and are therefore congruent; also, $\angle B \cong \angle D$. Thus, $\triangle PAB \sim \triangle PED$, and (because of how $\triangle PQR$ "points"), we can write:
$$\frac{a}{e+q} = \frac{b}{d+r} = \frac{|AB|}{|DE|}$$
with $q$ and $r$ (and, below, $p$) non-negative. Likewise, 
$$
\frac{c}{a+r} = \frac{d}{f+p} = \frac{|CD|}{|FA|} \qquad\qquad
\frac{e}{c+p} = \frac{f}{b+q} = \frac{|EF|}{|BC|}
$$
so that
$$\frac{|AB|^2}{|DE|^2}\cdot\frac{|CD|^2}{|FA|^2}\cdot\frac{|EF|^2}{|BC|^2} =
\frac{a}{a+r} \cdot \frac{b}{b+q} \cdot \frac{c}{c+p} \cdot \frac{d}{d+r} \cdot \frac{e}{e+q} \cdot \frac{f}{f+p}$$
which equals $1$ if and only if $p=q=r=0$.
Therefore, $|AB|\cdot|CD|\cdot|EF| = |DE|\cdot|FA|\cdot|BC|$ if and only if the diagonals concur. $\square$
A: 
        
A: The proof is given in Jens Cartensen's "About Hexagons". For completeness I will reproduce it here.

Suppose that the diagonals intersect at $O$. Then $\triangle ABO$ and $\triangle EDO$ are similar with
$$\frac{AB}{DE} = \frac{AO}{EO}$$
Similarly, the other two pairs of similar triangles gives
$$\frac{CD}{FA}=\frac{CO}{AO},\ \ \ \ \frac{EF}{BC}=\frac{EO}{CO}$$
Multiplied together, these give the desired product.
Conversely, suppose the product holds, i.e.
$$ace = bdf$$ 
Let $BE$ and $AD$ intersect at $O$. Extend $CO$ to intersect the circle at $F_1$. From the previous discussion, we get
$$ace_1 = bdf_1$$
The two equalities so far provide
$$acef_1 = bdff_1 = acfe_1$$
which cancel to provide
$$ef_1 = fe_1$$
From Ptolemy's theorem on Cyclic quadrilaterals on $AEFF_1$ we get
$$ef_1 = e_1f + FF_1\cdot AE \implies FF_1 \cdot AE = 0 \implies FF_1 = 0$$
so that $CF$ also passes through $O$ as required. $\square$
A: Let us prove one direction. 
Assume that $AD, BE, CF$ are concurrent at a point $P$. Consider $\triangle BFD$. From Trig Ceva, we have $$\frac{\sin \angle FDA}{\sin \angle ADB} \cdot \frac{\sin \angle DBE}{\sin \angle EBF} \cdot \frac{\sin \angle BFC}{\sin \angle CFD} = 1$$
Let the circumradius be $R$. From Law of Sines, 
\begin{align*}
AB &=  2R\sin \angle ADB \\
AF & = 2R \sin \angle ADF \\
FE & = 2R \sin \angle FBE \\
ED & = 2R \sin \angle DBE \\
CD & = 2R \sin \angle CFD \\
BC & = 2R \sin \angle BFC \\
\end{align*}
Indeed, 
$AB \cdot CD \cdot EF = BC \cdot DE \cdot AF$, holds true because it is just the equation derived from Trig Ceva. 
Since the steps are reversible, we have other proven the other direction, as desired. 
