What's the ratio between the segments $\frac{AF.BG}{FG}$ in the figure below? For reference: In the figure below the trapezoid has height $13$ and is inscribed in a circle of radius $15$. Point $E$ is on the minor arc determined by $A$ and $B$, the points $F$ and $G$ intersect with $ED$, $E$C and $AB$. Then the ratio between the segments $\frac{AF \times BG}{FG}$ is?

My progress:
I'm lost in this question...Perhaps using some of the segment $AB$ ratio with the diameter $DC$ ??
$\triangle EFG\sim HJC: \frac{JC}{FG} = \frac{JH}{FE}=\frac{HC}{EG}\\
\triangle BJG \sim \triangle HJC: \frac{JC}{JG}=\frac{JH}{BJ}=\frac{HC}{BG}\\
\triangle KID \sim \triangle KAF: \frac{JC}{FK}=\frac{JH}{AF}=\frac{HC}{AK}$
...???


Can you see that  $~\triangle AFE \sim  \triangle CBE~$ and $~\triangle BGE \sim \triangle DAE~$?
That leads to
$~ \displaystyle \frac{BC}{AF} = \frac{CE}{AE}~,  \frac{AD}{BG} = \frac{AE}{GE}$
Multiplying, $\displaystyle \frac{BC \times AD}{AF \times BG} = \frac{CE}{GE} \tag1$
As $ \displaystyle \triangle FEG \sim \triangle DEC, ~\frac{CE}{GE} = \frac{CD}{FG}$
Plugging into $(1)$,
$\displaystyle \frac{AF \times BG}{FG} = \frac{BC \times AD}{CD}$
As it is a cyclic trapezium, it must be isosceles and $BC = AD$. Applying Pythagoras,
$OH^2 = OB^2 - BH^2 = 15^2 - 13^2 = 56$
So, $CH = 15 - \sqrt{56}$
$BC \times AD = BC^2 = BH^2 + CH^2 = 13^2 + (15 - \sqrt{56})^2 = 30 (15 - 2 \sqrt{14})$
So the answer is $~(15 - 2 \sqrt{14})$
